\(\newcommand{\bb}[1]{\mathbb{#1}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\F}{\mathbb{F}} \newcommand{\E}{\mathcal{E}} \newcommand{\H}{\mathcal{H}} \newcommand{\l}{\mathcal{l}} \newcommand{\iso}{\cong} \newcommand{\orb}[1]{\text{Orb}({#1})} \newcommand{\stab}[1]{\text{Stab}({#1})} \newcommand{\tleq}{\trianglelefteq} \newcommand{\tgeq}{\trianglerighteq} \newcommand{\tf}{\therefore} \newcommand{\qed}{ // Q.E.D } \newcommand{\sub}{\subseteq} \newcommand{\sup}{\supseteq}\) The following is a cleaned up version of an exported Obsidian file. The original Markdown is available here. It will look better in Obsidian.

Chapter 1 - Sylow (“See-Low”) Theory

Prop [Cauchy’s Theorem for Finite Abelian Groups] If \(G\) is a finite abelian group and \(p \in \bb{N}\) is prime with \(p \mid |G|\) then \(G\) has an element of order \(p\).

Definition [Sylow Groups] \(G\) a group.

  1. We say \(G\) is a \(p\)-group, \(p\) prime, if \(|G| = p^n, n \in \bb{N}\)
  2. We say \(H \leq G\) is \(p\)-subgroup if \(H\) is a \(p\)-group.
  3. Say \(|G| = p^n m, n \in \bb{N}, p \nmid m, p\) prime. Any subgroup \(H \leq G\) of order \(p^n\) is called a Sylow \(p\)-subgroup

Recall[Group Actions]

  1. Suppose a finite group \(G\) acts on a finite set \(X\), i.e. \(\bullet: G \times X \to X\) where
    1. \(\forall x, e \cdot x = x\)
    2. \(\forall g, h \in X, \ g(hx) = (gh)x\)
  2. For \(x \in X\), \(\orb{X}\)
  3. Orbit Stabilizer Theorem: \(\forall x \in X\)
    $$|G| = |\stab{X}| \cdot |\orb{X}|$$
  4. If \(x, y \in X\), then either \(\orb{X} \cap \orb{Y} = \varnothing\) or \(\orb{X} = \orb{Y}\).
    Therefore \(X = \sqcup \orb{X_i}\) where \(X_i\) are distinct orbit reps.
  5. Assume \(X = G\), and \(G\) acts on \(X\) by conjugation, i.e. \(gx = gxg^{-1}\)
    For \(x \in X\), \(\orb{x}\) = Conjugacy Class, and \(\stab{X} = C(x)\), the centralizer (aka the things that commute with \(x\))
  6. \(|\orb{X}|=1 \iff \orb{X} = \{x\} \iff \forall g, gxg^{-1} = x \iff Z(G)\)
  7. The Class Equation \(|G| = |Z(G)| + \sum[G:C(a_i)]\)

Theorem [Sylow’s 1st Theorem]
Let \(G\) be a finite group of order \(p^n\)m where \(p\) prime, \(n \in \bb{N}\), \(p \nmid m\).
There exists a subgroup \(H \leq G\) s.t. \(|H| = p^n\)

Proof (by Induction):
if \(|G| = 2\), take \(H=G\). Proceeding inductively, assume \(G = p^n m, p \nmid m\)

Case 1: \(p \mid |Z(G)|\). By Cauchy, \(\exists a \in Z(G)\) s.t. \(|a| = p\). Take \(N = <a>\)

If \(n=1, H=N\). Assume \(n > 1\). Since \(N \subseteq Z(G)\), \(N \tleq G\). Note \(|G/N| = p^{n-1}m\).

By induction, \(\exists \ \bar{p} \leq G/N\) such that \(|\bar{p}| = p^{n-1}\). By PMATH347, \(\bar{P} = P/N\) where \(N \leq P \leq G\).
\(\tf p^{n-1} = |\bar{P}| = \frac{|P|}{|N|} = \frac{|P|}{p}\)
\(\implies |P| = p^n\)

Case 2: \(p \nmid |Z(G)|\)
\(p^n m = |G| = |Z(G)| + \sum [G : C(a_i)]\). \(\tf \exists s.t. p \nmid [G:C(a_i)]\) i.e. \(p \nmid \frac{|G|}{|C(a_i)|}. Hence \)p^n \mid |C(a_i)|.
By induction, \(\exists H \leq C(a_i) \leq G\). such that \(|H| = p^n\) \(\qed\)

Corollary [Cauchy’s General Theorem]
\(G\) a finite group, \(p\) prime. If \(p \mid |G|\) then \(\exists a \in G\) s.t. \(|a|=p\)

\(|G| = p^n m, p \nmid m\). \(\exists |H| = p^n, e \neq a \in H, |a| = p^k\). if \(k=1\), we done. Otherwise \(b=a^{p^{k-1}}, b\neq e\)
\(b^p = a^{p^k} = e \implies |b| = p\)

Definition [Normalizer]
\(G\) a group, \(H \leq G\), \(N_G(H) = \{g \in G: gHg^{-1} = H\}\)
called the normalizer in \(G\).
–> the largest subgroup of \(G\) in which \(H\) is normal, \(H \tleq N_G(H)\)

Theorem [Sylow’s 2nd Theorem]
If \(P, Q\) are Sylow \(p\)-subgroups of \(G\), then \(\exists \ g Pg^{-1} = Q\). I.e. once you find one of these, conjugate and you’ll find them all

Theorem [Sylow’s 3rd Theorem]
\(|G| = p^n m, p \nmid m\). Let \(n_p\) be the number of Sylow \(p\)-subgroups of \(G\).

  1. \(n_p \equiv 1 \mod{p}\)
  2. \(n_p \mid m\)

\(n_p = [G : N_G(P)]\) where \(P\) is any Sylow \(p\)-subgroup of \(G\)

Definition [Simple Groups] G a group.
\(G\) is simple \(\iff G\) has no proper non-trivial normal subgroups

\(n_p = 1 \iff P \tleq G\), where \(P\) is a Sylow \(p\)-subgroup of \(G\).

Exercise Prove there is no simple group of order 56
Proof 56 = \(2^3 \times 7\)
\(n_2 \equiv 1 \mod{2}, n_2 \mid 7 \implies n_2 \in \{1, 7\}\)
\(n_7 \equiv 1 \mod{7}, n_7 \mid 8 \implies n_7 \in \{1, 8\}\)
Suppose \(n_2 = 7, n_7 = 8\). This acounts for \(8 \cdot 6 = 48\) elements of order 7. This leaves 56-48 = 8 other elements. Hence \(n_2 =1\). Contradiction!
Therefore \(n_2=1\) or \(n_7 = 1\) and such a group is not simple. \(\qed\)

Counting Arguments in Sylow Theory
\(p, q \mid |G|, p, q\) are distinct primes

  1. \(H_p\) Sylow \(p\)-subgroup, \(H_q\) Sylow \(q\)-subgroup, \(H_p \cap H_q = \{e\}\)
  2. \(|G| = pm, p \nmid m\), if \(H_1 \neq H_2\) are Sylow \(p\)-subgroups, then \(H_1 \cap H_2\) = {e}
  3. Suppose \(H_p \tleq G\) or \(H_q \tleq G\), \(\tf H_pH_q \leq G\) and
    $$|H_pH_q| = \frac{|H_p| \cdot |H_q|}{|H_p \cap H_q|} = |H_p| \cdot |H_q|$$
  4. \(|H_p \cup H_q| = |H_p| + |H_q| -1\)

Exercise \(|G| = pq\), \(p < q\) are primes where \(p \nmid (q-1)\). Prove \(G\) is cyclic.
Proof \(n_p \equiv 1 \mod{p}\), \(n_p \mid q \implies n_p = 1\)
Then \(n_q \equiv 1 \mod{q}, n_q \mid p \implies n_q = 1\)
So \(H_p, H_q \tleq G\). We know that \(H_p\) and \(H_q\) are abelian (prime order). Let \(a \in H_p, b \in H_q\), then \(aba^{-1}b^{-1} \in H_p \cap H_q = \{e\}, \ \tf ab = ba\). So \(H_pH_q\) is abelian! By the Fundamental Theorem of finite abelian groups,
$$H_p H_q \iso \Z_p \cdot \Z_q \equiv \Z_{pq}$$ and therefore cyclic.
Note: \(H_p H_q \leq G\) where \(|H_pH_q| = pq = |G|\), therefore \(H_pH_q = G \ \qed\)

Proposition \(|G|=30\). Then there exists \(H \tleq G\) such that \(H \iso \Z_{15}\)

$$30 = 15 \times 2 = 2 \cdot 3 \cdot 5$$ \(n_2 \equiv 1 \mod{2}, n_2 \mid 15\)
\(n_3 \equiv 1 \mod{3}, n_3 \mid 10 \implies n_3 \in \{1, 10\}\)
\(n_5 \equiv 1 \mod{5}, n_5 \mid 6 \implies n_5 \in \{1, 6\}\)

Suppose \(n_3 = 10, n_5 = 6\). This accounts for 10(3-1)+6(5-1)+1=20+24=45 elements in \(G\). Contradiction!
$$\tf \ n_3 = 1 \text{ or } n_5 = 1$$ Let \(H_3\) be a Sylow 3-subgroup and let \(H_5\) be Sylow 5-sub. \(\tf H_3 \tleq G \text{ or } H_5 \tleq G\)
\(\implies H_3 H_5 \leq G\)
$$|H_3H_5| = \frac{|H_3| \cdot |H_5|}{|H_3 \cap H_5|} = \frac{3 \cdot 5}{1} = 15$$ Since \(3\mid(5-1)\) from last time \(H_3H_5\) is cyclic, \(\tf H_3 H_5 \iso \Z_{15}\)
Since \([G:H_3H_5]=2, H_3H_5 \tleq G \ \qed\).

\(|G|= 60\), if \(n_5 > 1\), then \(G\) is simple. Note \(60 = 2^2 \times 3 \times 5\)
\(n_5 \equiv 1 \mod 5, n\mid 12 \implies n_5 = 6\)
This gives us 6(5-1) = 24 elements of order 5.
Assume \(G\) has a proper non-trivial \(H \tleq G\).

Case 1: \(5 \mid |H|\)
Since \(H \tleq G\) and \(H\) contains a Sylow 5-subgroup of \(G\), then \(H\) contains ALL Sylow 5
$$\tf \ |H| \mid 60 \text{ and } |H| \geq 24+1$$ \(\implies |H| = 30\)
We know \(\exists \ H_0 \tleq H\) such that \(H_0 \iso \bb{Z}_{15}\)
Again \(H_0\) contains all Sylow 5-subgroups of \(G\). Since \(H_0\) is abelian, \(n_5 = 1\). Contradiction!

Case 2: \(5 \nmid |H|\)
\(|H| \in \{2, 3,4,6, 12\}\)

  1. \(|H| = 12 = 2^2 \times 3\). HOMEWORK to show \(n_2 = 1\) or \(n_3 = 1\)
    \(\tf \ H\) contains a Sylow 2-or-3 subgroup which is normal. Call it \(K\). \(|K| \in \{3, 4\}\)
  2. HOMEWORK, If \(|H|=6\), then \(n_3=1\). Let \(K \tleq\) Sylow 3-subgroup of \(H\)
    Note: by Sylow 2nd theorem, in either case \(K\) is normal in \(G\) (see Case 1 Argument)
    By replacing \(H\) with \(K\) (if necessary), we may assume \(|H|\in\{2,3,4\}\)
    \(\bar{G} := G/H\), then \(|\bar{G}| \in \{15, 20, 30\}\). HOMEWORK in any case above, \(\exists \bar{P} \tleq \bar{G}, |\bar{P}| = 5\)

So then \(\bar{P} = P/H, \text{ where } H \leq P \tleq G\) (by the correspondence theorem)
\(\tf P \tleq G \text{ s.t. } 5 = \frac{|P|}{|H|}\)
\(\implies 5 \mid |P|\). This contradicts that Case 1 is impossible. Therefore \(G\) is simple \(\qed\)

Corollary: \(A_5\) is simple.

Chapter 2 - Irreducibility Criteria

\(\F\) a field, \(p[x] \in \F[x]\). Let \(I\) be a non-zero proper ideal of \(\F[x]\). Say \(I = <p(x)>\).
Then \(\F[x]/<p(x)>\) is a field \(\iff\) \(<p(x)>\) maximal \(\iff p(x)\) irreducible.

\(R\) integral domain (“ID”). Then \(p(x) \in R[x]\) is irreducible \(\iff\)

  1. \(p(x) \neq 0\)
  2. \(p(x) \notin R[x]^\times = R^\times\)
  3. Whenever \(p(x) = a(x)b(x), a(x), b(x) \in R[x]\), then \(a(x)\) or \(b(x)\) is a unit.

We say \(p(x) \in R[x]\) is reducible \(\iff\)

  1. \(p(x) \neq 0\)
  2. \(p(x) \notin R^\times\)
  3. \(p(x)\) NOT irreducible

example: \(p(x) = 2x+2\). Then \(p\) is reducible in \(\Z[x]\) but irreducible in \(\Q[x]\)

Motivating Question Given \(p(x) \in R[x]\) how can we decide if \(p(x)\) is irreducible?

Proposition \(\F\) a field. \(f(x) \in \F[x], a \in \F\)
The remainder when \(f(x)\) is divided by \(x-a\) is \(f(a)\)

\(f(x) = (x-a)q(x)+r(x)\) where \(r(x)=0\) or \(\deg r(x) < \deg (x-a)\) (from the Division Algorithm, also another way to say \(r(x)\) is constant)
Therefore \(r(x) = r \in \F\).
$$\tf f(a) = 0+r \implies r =f(a)$$ \(\qed\)

Proposition \(\F\) a field, \(f(x) \in \F[x]\), \(\deg (f(x)) \geq 2\). If \(f(x)\) has a root in \(\F\), then \(f(x)\) is reducible.

$$f(a) = 0$$ $$f(x) =(x-a) q(x) + 0 = (x-a)q(x)$$ example: \(f(x) = x^4+2x^2+1 = (x^2+1)^2 \in \R[x]\)
no roots, reducible.

Proposition [Irreducible Means No Roots] \(\F\) field, \(f(x) \in \F[x]\), \(\deg f(x) \in [2, 3]\)
Then \(f(x)\) is irreducible \(\iff\) \(f(x)\) has no roots.

\(f(x) \text{ reducible} \iff \text{linear factor } \iff \text{root}\)

Theorem [Gauss’s Lemma]
\(R\) UFD, \(\F = \operatorname{Frac}(R)\), let \(f(x) \in R[x]\)
If \(f(x) = A(x)B(x)\) where \(A(x), B(x) \in \F[x]\) are non-constant then \(\exists\) \(a(x)b(x) \in R[x]\) such that
$$a(x) = rA(x),b(x) = sB(x)$$ with \(r, s \in \F^\times\) and \(f(x) = a(x)b(x)\)
–> i.e. if \(f\) is reducible over its field of fractions, it reduces over its integral domain
In particular \(\deg a(x) = \deg A(x)\) and \(\deg b(x) = \deg B(x)\)

Proposition [Mod-\(p\) Irreducibility Test]
Let \(f(x) \in \Z[x], p \in \N\) is prime. Let \(\bar{f}(x) \in \Z_p[x]\) be obtained by reducing each coefficient of \(f(x)\) modulo \(p\)

  1. \(\deg f(x) = \deg \bar{f}(x)\) and
  2. \(\bar{f}(x) \in \Z_p[x]\) is irreducible

Then \(f(x) \in \Z[x]\) is irreducible (over \(\Q\) too by Gauss)

example: \(f(x) = 2x^2+x\) reducible, \(\bar{f}(x) = x\) irreducible in \(Z_2[x]\)

Suppose \(f(x)\) is reducible over \(\Q\). Say \(f(x) = g(x) h(x)\) where \(g(x)h(x) \in \Q[x]\)
\(\deg g(x), \deg h(x) < \deg f(x)\)
–> just a cleaner way of saying that neither \(g\) nor \(h\) are constants.

By Gauss’s Lemma, we may assume \(g(x), h(x) \in \Z[x]\)
Then \(\bar{f}(x) = \bar{g}(x) \bar{h}(x) \in \Z_{p}[x]\). Since \(\bar{f}(x)\) is irreducible, we may assume \(\bar{g}(x)\) is constant. Therefore \(\deg \bar{h}(x) = \deg \bar{f}(x)\)

$$\tf \deg h(x) < \deg f(x)$$ $$= \deg \bar{f}(x)$$ $$= \deg \bar{h}(x)$$ $$\leq \deg h(x)$$ Contradiction! \(\deg h(x) < \deg h(x)\) makes no sense. \(\qed\)

example \(f(x) = 23x^3 + 15x^2 - 1 \in \Z[x]\)
So \(\bar{f}(x) = x^3+x^2+1 \in \Z_2[x]\). Note \(\deg f(x) = \deg \bar{f}(x)\)
\(\bar{f}(0)=1, \bar{f}(1)=1\)
Since \(\deg \bar{f}(x) = 3\) and \(\bar{f}(x)\) has no roots, \(\bar{f}(x)\) is irreducible. By the Mod-2 Irred. Test, \(f(x)\) is irreducible.

Proposition [Generalized Mod-P] \(R\) an integral domain, \(I \neq R\) ideal, \(p(x) \in R[x]\) non-constant, monic.
If \(\bar{p}(x)\) cannot be factored as two smaller degree polynomials in \((R/I)[x]\), then \(p(x) \in R[x]\) is irreducible.

Proof: Exercise

Proposition [Eisenstein’s Criteria] \(R\) integral domain, \(P \sub R\) is a prime ideal. Let
$$f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_{1}x + a_0 \in R[x]$$ If:

  1. \(a_{n-1}, \cdots, a_1, a_0 \in P\)
  2. \(a_0 \notin P^2\)

Then \(f(x)\) is irreducible over \(R\).

Recall [Pairwise Ideals] \(R\) a ring, \(I, J \sub R\) ideals
$$IJ = \{\sum a_i b_i : a_i \in I, b_i \in J\}$$ and \(IJ\) itself is an ideal.

Proof. Suppose \(f(x) = g(x) h(x)\) where \(\deg g(x), \deg h(x) < \deg f(x)\).
Then \(\bar{f}(x) = \bar{g}(x) \bar{h}(x) = x^n \in (R/P)\)
Since \(R/P\) is an integral domain (\(\star\))
\(\bar{g}(0) = \bar{h}(0) = 0, \tf g(0), h(0) \in P\)
\(\implies a_0 = f(0) = g(0)h(0) \in P^2\). This is a contradiction \(\qed\)

example \(f(x, y) = y^2 + x^2 - 1 \in \Q[x, y]\). Claim: \(f(x, y)\) is irreducible.
Consider \(g(y) = y^2 + (x^2-1) \in \Q[x][y]\).
Note: \(x^2-1 = (x-1)(x+1) \in <x-1>\). Since \(x-1\) is irreducible in \(\Q[x]\), \(P\) is maximal (prime). Also \(P^2 = <x-1>^2 = <(x-1)^2>\)
Clearly \((x-1)^2 \nmid x^2-1\), so \(x^2-1 \notin P^2\) and so \(g(f)\) is irreducible by Eisenstein.

$$f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in \Z[x]$$ \(p \in \N\) prime. If \(p\) divides \(a_{n-1},\cdots, a_1, a_0\) but \(p^2 \nmid a_0\) then \(f(x)\) is irreducible over \(\Q\).
Proof Let \(P = <p>\)

Exercises: Are these irreducible over \(Q\)?

  1. \(f(x) = x^7 + 21x^5 + 15x^2 + 9x+6\)
    Yes irreducible by 3-Eisenstein.
  2. \(g(x) = x^3+2x+16\)
    Notice that \(\bar{g}(x) = x^3+2x+1 \in \Z_3[x]\), so \(\deg g(x) = \deg \bar{g}(x)\)
    \(\bar{g}(0)=1, \bar{g}(1)=1, \bar{g}(2)=1\)
    Since \(\deg \bar{g}(x) = 3\), \(\bar{g}(x)\) is irreducible. By the Mod-3 irreducibility test, \(g(x)\) is irred.
  3. \(p(x) = x^4+5x^3+6x^2-1\)
    \(\bar{p}(x) = x^4+x^3+1 \in \Z_{2}[x]\), with \(\bar{p}(0)=1, \bar{p}(1)=1\)
    check: the only irreducible quadratic is \(x^2+x+1\)
    Moreover \((x^2+x+1)^2 = x^4+x^2+1 \neq \bar{p}(x)\)
    \(\implies\) \(\bar{p}(x)\) is irreducible, and by the Mod-2 irred. test, \(p(x)\) is irred.
  4. \(q(x) = x^{p-1} + x^{p-2} + \cdots + x^2 + x + 1\)

Chapter 3 - Field Extensions

Definition [Field Extension] \(F\), \(K\) are fields. We say \(K\) is a (field) extension of \(F\) if \(F\) is a subfield of \(K\)
–> subfield == subring that’s also a field
Notation: \(K/F\)
–> “K” over F. Remember if \(F \neq K\), K/F isn’t quotient ring. B/c fields only have two ideals: themselves and 0

Example \(\C/\R\) and \(\C / \Q\), \(\R / \Q\), \(\Q / \Q\)

Example \(F\) a field, \(F(x) = \{ {f(x)}/{g(x)}, f, g \in F[x], g \neq 0 \}\)
–> field of rational functions over \(F\)
Note then \(F(x)/F\) extension

Example \(\Z_p(x) / \Z_p\). (one way to extend \(\Z_p\))

Example Note \(\Q(\sqrt{2}) = \{a+b\sqrt{2} : a, b \in \Q\}\)
–> field extension of \(\Q\)
Example \(\Q\) is NOT an extension of \(\Z_p = \{0, 1, \cdots, p-1\}\)
–> different operations!

Example \(F\) field, \(f(x) \in F[x]\) is irreducible.
\(K = F[x]/<f(x)>\) field. \(K/F\) is an extension
Note: \(F \iso \{a+<f(x)> : a \in F\} \sub K\)

Definition [Characteristic of a Field] \(F\) a field
The characteristic of \(F\), denoted \(Char(F)\), is the least positive \(n \in \N\) such that \(n \cdot 1= 1+1+\cdots+1 (n times) = 0\)
If no such \(n\) exists, we say \(Char(F)=0\)
–> basically the additive order of 1

Example \(Char(\Z_p)\) = \(Char(\Z_p(x)) = p\)
Example \(Char(\R) = 0\)

**HOMEWORK: ** \(F\) a field, \(Char(F) = 0\) or prime.
–> think zero-divisors, as soon as you have a composite, you’ve got elements that will multiply to 0

Example \(Char(F) = p\)
Then \(F/\Z_p\) extension. \(\Z_p = \{0, 1, 2, \cdots, p-1\} \sub F\)
–> isomorphic copy generated by 1

Example \(Char(F)=0\), \(F/\Q\) extension
\(\Q \iso \{nm^{-1}: \ n, m \in \Z, m \neq 0\}\)

Definition \(K/F\), \(\alpha_1, \cdots, \alpha_n \in K\)
$$F(\alpha_1, \cdots, \alpha_n) = \{f(\alpha_1, \cdots, \alpha_n)/g(\alpha_1, \cdots, \alpha_n): f, g \in F[x_1, \cdots, x_n], g \neq 0\}$$ This is called the extension field of F generated by \(\alpha_1, \cdots, \alpha_n\) in \(K\)
\(F\) adjoin \(\alpha_1, \cdots, \alpha_n\) “.

HOMEWORK \(F(\alpha_1, \cdots, \alpha_n)\) field. via operations of \(K\).

Suppose \(L\) is a subfield of \(K\) s.t. \(F \sub L\) and \(\alpha_1, \cdots, \alpha_n \in L\).
Then \(F(\alpha_1, \cdots, \alpha_n) \in L\)
i.e. \(F(\alpha_1, \cdots, \alpha_n)\) is the smallest subfield of \(K\) which contains \(F\) and \((\alpha_1, \cdots, \alpha_n)\)

Example: Prove that \(\Q(\sqrt{2}, \sqrt{3}) = \Q(\sqrt{2}+\sqrt{3})\)

  1. \(\sqrt{2}+\sqrt{3} \in \Q(\sqrt{2}, \sqrt{3})\)
    By minimality, \(Q(\sqrt{2}+\sqrt{3}) \sub Q(\sqrt{2}, \sqrt{3})\)
  2. \(1/(\sqrt{2}+\sqrt{3}) (\sqrt{2}-\sqrt{3})/(\sqrt{2}-\sqrt{3}) = \sqrt{3}-\sqrt{2} \in \Q(\sqrt{2}+\sqrt{3})\)

Therefore \(\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2} = 2\sqrt{3} \in \Q(\sqrt{2}, \sqrt{3})\)
\(\implies \sqrt{3} \in \Q(\sqrt{2}+\sqrt{3})\)
\(\implies \sqrt{2} \in \Q(\sqrt{2}+\sqrt{3})\)

By minimality, \(\Q(\sqrt{2}, \sqrt{3}) \sub \Q(\sqrt{2}+\sqrt{3})\) \(\qed\)

Exercise \(K/F\), \(\alpha, \beta \in K\). Prove \(F(\alpha, \beta) = F(\alpha)(\beta)\)
Since \(\alpha, \beta \in F(\alpha)(\beta)\),
\(F(\alpha, \beta) \sub F(\alpha)(\beta)\) by minimality
Note: \(F(\alpha) \sub F(\alpha, \beta)\) and \(\beta \in F(\alpha, \beta)\).
\(\tf F(\alpha)(\beta) \sub F(\alpha, \beta)\)

Proposition \(K/F\), \(\alpha \in K\)
Assume \(\alpha\) is a root of an irreducible \(f(x) \in F[x]\)
Then \(F(\alpha) \iso F[x]/<f(x)>\)
If \(\deg f(x) = n\), \(F(\alpha) = \Span_F\{1, \alpha, \alpha^2, \cdots, \alpha^{n-1}\} = \{c_0 + c_1\alpha + \cdots + c_{n-1}\alpha^{n-1} : c_i \in F \}\)

Exercise: \(\Q(\sqrt{2}) = \{f(\sqrt{2})/g(\sqrt{2}) : f, g \in \Q[x], g(\sqrt{2}) \neq 0\}\)
\(= \Span_{\Q} \{1, \sqrt{2}\}\) \((f(x) = x^2-2)\)
\(=\{a+b\sqrt{2} : a, b \in \Q\}\)

Notation: In \(R/I\), \(\bar{x} = x+I\).
Recall: \(R=F[x] / <f(x)>, \deg f(x)=n\)
Take \(\bar{g}(x) \in R\).
We know \(g(x) = f(x)q(x)+ r(x)\), \(r(x)=0\) or \(\deg r(x) < n\)
\(\tf \bar{g}(x) = \bar{f}(x) \bar{q}(x) + \bar{r}(x)\)
\(= \bar{r}(x)\)
\(\tf R = \{\overline{c_0 + c_1x+\cdots+c_{n-1}x^{n-1}} : c_i \in F\}\)

Proof: Consider the ring homomorphism \(\varphi: F[x] \to F(\alpha)\), \(\varphi(g(x)) = g(\alpha)\)
Then, \(I = \ker \varphi = \{g(x): g(\alpha) = 0\}\)
Since, \(f(x) \in I\), \(<f(x)> \sub I\)
Since \(F[x]\) is a PID, \(I = <g(x)>\) for some \(g(x) \in F[x]\)
\(\tf f(x)=g(x)h(x)\) for some \(h(x) \in F[x]\)
Since \(I \neq F[x]\) and \(f(x)\) irreducible, \(h(x)\) is a unit.
Hence, \(I = <g(x)> = <f(x)>\)
–> in an integral domain, two elements generate the same ideal iff they are associates (347 result)
By the first iso theorem, \(F[x]/<f(x)> \iso \varphi(F[x])\)

By definition, \(\varphi(F[x]) \sub F(\alpha)\) (image contained in co-domain)
Also \(\varphi(F[x])\) is a field and \(\varphi(x) = \alpha\)
By minimality, \(F(\alpha) \sub \varphi(F[x])\)

Finally, \(F[x]/<f(x)> = \{\overline{c_0+c_1x+\cdots+c_{n-1}x^{n-1}} : c_i \in F\}\)
and so \(F(\alpha) = \psi (F[x]/<f(x)>)\)
\(=\overline{c_0+c_1\alpha+\cdots+c_{n-1}\alpha^{n-1}: c_i \in F}\)
\(\psi(\bar{g}(x)) = \varphi(g(x)) = g(\alpha)\)
is the isomorphism afforded by the FIT.

\(K/F\), \(\alpha \in K, g(x) \in F[x], g(\alpha)=0, g(x)\neq0\)
–> note, no mention of irreducibility yet
Since \(F[x]\) is a UFD, \(\alpha\) is a root of an irreducible factor \(f(x)\) of \(g(x)\)

By our proof: \(<f(x)> = \{h(x) : h(\alpha) = 0\}\)

  1. If \(h(x) \in F[x]\) such that \(h(\alpha)=0\), then \(f(x) | h(x)\)
  2. Let \(h(x) \in F[x]\) be irreducible such that \(h(\alpha) = 0\)
    \(\tf <f(x)> = < h(x)>\)$\implies\( \)f(x) = c h(x)\(, \)c \in F^\times$
  3. There exists a unique irreducible, monic \(m(x) \in F[x]\) such that \(m(\alpha)=0\)

Definition [Minimal Polynomial]
\(K/F\) , \(\alpha \in K\).
Suppose \(\alpha\) is a root of a non-zero \(g(x) \in F[x]\)
The unique irreducible, monic, \(m(x) \in F[x]\) with \(m(\alpha)=0\) is called the minimal polynomial of \(\alpha\) over \(F\)

If \(\deg m(x) = n\), we write \(\deg_F (\alpha) = n\)

Example \(p\) prime, \(\sqrt{p} = \alpha\)
\(m(x) = x^2 - p \in \Q[x]\) (irred by \(p\)-Eisenstein)
\(\deg_{\Q} (\sqrt{p})=2\)

Example \(\alpha = \sqrt{1+\sqrt{3}} \in \R\)
\((\alpha^2-1)^2 = 3\)
\(\implies \alpha^4 - 2\alpha^2=0\)
\(m(x)=x^4-2x^2-2\) (irred by 2-Eisenstein)

\(\tilde{m}(x) = x - \alpha\), \(\deg_{\R} (\alpha)=1\)
Example \(a \in F\), \(\deg_F (a)=1\) \((x-a)\)

\(K/F\). Then \(K\) is an \(F\)-vector space.

Proposition \(K/F\), \(\alpha \in K\)
\(\alpha\) is the root of a non-zero \(g(x) \in F[x]\)
Let \(m(x)\) be the minimal polynomial (“min poly”) of \(\alpha\) over \(F\)

Then, \(\{1, \alpha, \alpha^2, \cdots, \alpha^{n-1}\}\), where \(n=\deg m(x) = \deg_F (\alpha)\) is a basis for \(F(\alpha)/F\).
In particular, \(|F(\alpha)| = |F|^n\)

\(F(\alpha) = \Span_F \{1, \alpha, \cdots, \alpha^{n-1}\}\)
Suppose, \(c_0 + c_1 \alpha + \cdots + c_{n-1}\alpha^{n-1}=0\) (\(c_i \in F\))
Say \(f(x) = c_0 + c_1 x + \cdots + c_{n-1}x^{n-1}\)
\(\tf m(x) | f(x)\)
By degrees, \(f(x) = 0\)
\(\implies c_0 = c_1 = \cdots = c_{n-1} = 0\)
Hence, \(\{1, \alpha, \cdots, \alpha^{n-1}\}\) is a basis for \(F(\alpha)\) over \(F\).

Then every \(\beta \in F(\alpha)\) can be uniquely written as \(\beta = c_0 + c_1\alpha+ \cdots + c_{n-1} \alpha^{n-1}\) (\(c_i \in F\))
Hence \(|F(\alpha)| = |F|^n\) \(\qed\)
\(\dim_F F(\alpha) = \deg_F (\alpha)\)

\(F\) as \(\Z_p\) btw for assignment

Proposition \(K/F\)
If \(\alpha, \beta \in K\) have the same min poly over \(F\) then \(F(\alpha) \iso F(\beta)\)

\(F(\alpha) \iso F(\beta) \iso F[x]/<m(x)>\)

Example \(\alpha = \sqrt[3]{2}, \beta = \sqrt[3]{2} \zeta_3\)
–> primitive third root of 3 is \(\zeta_3\)
\(m(x)=x^3-2\) (2-eisenstein)

  1. \(\Q(\alpha) \iso\Q(\beta)\)
  2. \(\Q(\alpha) \neq \Q(\beta)\) since \(\Q(\alpha) \sub \R\) and \(\Q(\beta) \not\subseteq \R\)
  3. \(e^{2\pi i}/3\) for example is \(\zeta_3\)

Goal for Today \(K/F\), Explore \(K\) as an \(F\)-vector space.

Definition [Finite Field Extension and Degrees] \(K/F\). We say \(K/F\) is finite iff \(K\) is a finite dimensional \(F\)-vector space

We call \([K:F] = \dim_F(K)\) the degree of \(K/F\).

Example \([\C:\R] = 2\), \([\R : \Q] = \infty\)
Example \(K/F\), \(\alpha \in K\) is a root of \(0 \neq f(x) \in F[x]\). Let \(m(x)\) be the min poly of \(\alpha\).
\([F(\alpha):F] = \deg_F(\alpha) = \deg m(x)\)
Example \([F:F] = 1\), \([K:F] = 1 \iff F=K\)
Example \([\Q(\sqrt{1+\sqrt{3}}): \Q] = 4\)
–> when working with \(F(\alpha)\) things, find the min poly, and we’ll get the degree of the extension

Definition [Tower] \(K/E\), \(E/F\) are field extensions.
We call \(K/E/F\) a tower of fields.

Proposition [Tower Theorem]
\(K/E/F\) tower of fields. If \(K/E, E/F\) are finite, then \(K/F\) is finite.
Moreover, \([K:F] = [K:E][E:F]\)

Let \(\{v_1, \cdots, v_n\}\) be a basis for \(K/E\) and let \(\{w_1, \cdots, w_n\}\) be a basis for \(E/F\)

Claim: \(\{v_i w_j : 1 \leq i \leq n, 1 \leq j \leq m\}\)
is a basis for \(K/F\)
–> note if this is true, there are \(nm\) elements so theorem is done.

Linear Independence: \(\sum\limits_{i, j} c_{ij} v_i w_j = 0, c_{ij} \in F\)
\(\implies \sum\limits_{i, j} (c_{ij} w_j) v_i = 0\)
\(\implies \sum\limits_i (\sum\limits_j c_{ij} w_j) v_i =0\)
Note \((\sum\limits_j c_{ij} w_j) \in E\). Since \(\{v_i : 1 \leq i \leq n\}\) is LI

\(\forall i, (\sum\limits_j c_{ij} w_j)=0\)
Since \(\{w_j : 1 \leq j \leq m\}\) is LI,
\(\forall i, \forall j, c_{ij}=0\)

Spanning: Let \(\alpha \in K\). \(\implies \alpha = \sum\limits_{i} c_i v_i\) where \(c_i \in E\)
For all \(i, c_i = \sum\limits_j d_{i,j} w_j\), \(d_{i,j} \in F\)
$$\tf \alpha = \sum\limits_{i, j} d_{i,j} v_i w_j$$ \(\qed\)

Example: \([\Q(\sqrt[3]{5}, i):\Q] = [\Q(\sqrt[3]{5})(i):\Q(\sqrt[3]{5}] \cdot [\Q(\sqrt[3]{5}):\Q]\)
\(p(x) = \text{ min poly } i \text{ over } \Q(\sqrt[3]{5})\)
\(q(x) = \text{ min poly } \sqrt[3]{5} \text{ over } \Q\)

\(q(x)=x^3-5\) (5-Eis)
\(p(x)=x^2+1\) (irred. b/c no roots, \(\Q(\sqrt[3]{5}) \sub \R\))

\(\tf [\Q(\sqrt[3]{5}, i):\Q] = 3 \cdot 2 = 6\)

Note: Basis for \(\Q(\sqrt[3]{5})(i)\) over \(Q(\sqrt[3]{5})\)
is \(\{1, i\}\)
Basis for \(\Q(\sqrt{3}[5])\) over \(\Q\) is \(\{1, \sqrt[3]{5}, (\sqrt{3}[5])^2\}\)

Therefore a basis for \(Q(\sqrt[3]{5}, i)\) over \(\Q\) is:
$$\{1, \sqrt[3]{5}, (\sqrt[3]{5})^2, i, i \sqrt[3]{5}, i(\sqrt[3]{5})^2\}$$

Goal Investigate why \(\alpha \in K\) being a root of \(0 \neq f(x) \in F[x]\) is important.

[Definitions of \(\alpha\)] \(K/F\)

  1. We say \(\alpha \in K\) is algebraic over \(F\) iff there exists \(0 \neq f(x) \in F[x]\) s.t. \(f(\alpha)=0\)
  2. We say \(K/F\) is algebraic iff \(\alpha\) is algebraic over \(F\) for all \(\alpha \in K\)

If \(K/F\) is finite, then \(K/F\) is algebraic.

\([K:F] = n < \infty\). Take \(\alpha \in K\)
Consider \(1, \alpha, \alpha^2, \cdots, \alpha^{n}\).
\(\exists \ c_0, c_1, \cdots, c_n \in F\) (not all 0)
s.t. \(c_0 + c_1 \alpha + \cdots + c_n \alpha^n = 0\)
\(f(x)=c_0+c_1 x + \cdots + c_n x^n\)
\(f(\alpha)=0\) \(\qed\)

Example (converse is not true)
\(p_1 < p_2 < \cdots\) primes
\(\Q(\sqrt{p_1}) \sub \Q(\sqrt{p_1}, \sqrt{p_2}) \sub \cdots\)
\(K = \cup_{n=1}^{\infty} \Q(\sqrt{p_1}, \cdots, \sqrt{p_n})\) is a field extension of \(\Q\)
\(K/\Q\) algebraic, but NOT finite.

$$K/F \text{ finite } \iff [K:F] < \infty$$ Finite \(\implies\) Algebraic, Algebraic \(\nrightarrow\) Finite

Proposition \(K/F\)
If \(\alpha_1, \cdots, \alpha_n \in K\) are algebraic over \(F\), then \([F(\alpha_1, \cdots, \alpha_n):F] < \infty\)
–> if the alphas are algebraic, and you adjoin to a field, extension is finite and algebraic
–> adjoin algebraic elements, everything in the field you just created is algebraic

Proof Induction on \(n\)
If \(n=1\) and \(\alpha_1 \in K\) is algebraic over \(F\)
\([F(\alpha_1):F] = \deg_F(\alpha_1) < \infty\)

Assume the result for \(n-1\). Let \(\alpha_1, \alpha_2, \cdots, \alpha_{n-1}\) be alg over \(F\)

Then, \([F(\alpha_1, \cdots, \alpha_n):F]\)
\(=[F(\alpha_1, \cdots, \alpha_{n-1})(\alpha_n):F(\alpha_1, \cdots, \alpha_{n-1})] \cdot [F(\alpha_1, \cdots, \alpha_{n-1}):F]\)
the first term is finite b/c base case, the second term is finite bc inductive hypothesis. So \(\qed\)

\(K/E, E/F\) are algebraic, then \(K/F\) algebraic

Let \(\alpha \in K\), and suppose we have \(f(\alpha)=0\) where \(0 \neq f(x)=x^n+c_{n-1}x^{n-1} + \cdots + c_1 x + c_0 \in E[x]\)

Then, \(\alpha\) is alg over \(F(c_{n-1}, c_{n-2}, \cdots, c_1, c_0)\)

Thus, \([F(c_{n-1}, \cdots, c_1, c_0)(\alpha):F(c_{n-1}, \cdots, c_0)] \cdot [F(c_{n-1}, \cdots, c_0) : F]\)
\(=[F(c_{n-1, \cdots, c_1, c_0, \alpha}):F] < \infty\)

Proposition \(K/F\)
\(L = \{\alpha \in K : \alpha \text{ is alg over } F \}\)
Then \(K/L/F\) is a tower of fields.

\(\alpha, \beta \in L\), \(\alpha \neq 0\)
\(\alpha-\beta, \alpha \beta, \alpha^{-1} \in F(\alpha, \beta)\)
\([F(\alpha, \beta):F] < \infty\) and therefore algebraic


\(p_1 < p_2 < \cdots\) primes
\(K = \cup_{n=1}^\infty \Q(\sqrt{p_1}, \cdots, \sqrt{p_n})\)

Claim: \(K/\Q\) algebraic
\(\alpha \in K \implies \alpha \in \underbrace{\Q(\sqrt{p_1}, \cdots, \sqrt{p_n})}_{\text{alg ext of} \Q}\)

Claim: \(K/\Q\) NOT finite.
\([K : \Q] = [K : \Q(\sqrt{p_1}), \cdots, \sqrt{p_n}] \cdot \underbrace{[\Q(\sqrt{p_1}), \cdots, \sqrt{p_n} : \Q]}_{2^n}\)

Chapter 4 - Splitting Fields

Goal Given \(f(x) \in F[x]\), find an extension \(K/F\) such that \(f(x)\) completely factors over \(K\)

Example \(f(x)=x^2-2 \in \Q[x]\), \(K=\Q(\sqrt{2})\)

Definition [Splits] \(K/F\)
Let \(f(x) \in F[x]\) be non-constant. We say \(f(x)\) splits over \(K\) if there exists \(u \in F, \alpha_1, \cdots, \alpha_n \in K\) such that \(f(x)=u(x-\alpha_1) \cdots (x-\alpha_n)\)

Theorem [Kronecker’s Theorem]

Let \(F\) be a field and let \(f(x) \in F[x]\) be non-constant.
Then there exists \(K/F\) such that \(f(x)\) has a root in \(K\).

Proof We may assume \(f(x)\) is irreducible.
Let \(K=F[t]/<f(t)>\), we know \(K\) is a field.

Then, \(f(\bar{t}) = 0\) \(\qed\)

By applying Kronecker repeatedly , \(\exists K/F\) such that \(f(x)\) splits over \(K\)

Definition \(F\) field, \(f(x) \in F[x]\) non-const.
We say \(K\) is a splitting field for \(f(x)\) over \(F\) iff

  1. \(K/F\)
  2. \(f(x)\) splits over \(K\)
  3. Whenever \(f(x)\) splits over \(F \sub L\), then \(K \sub L\)

\(F, f(x) \in F[x]\) as before. Let \(K/F\) be an ext. such that \(f(x)\) splits over \(K\)
\(f(x) = u(x-\alpha_1) \cdots u(x-\alpha_n) \in K[x]\)
Then, \(F(\alpha_1, \cdots, \alpha_n)\) is a splitting field for \(f(x)\)

Problem (picture drawn of different extensions \(K\) and \(E\) we could split over)

Goal \(F(\alpha_1, \cdots, \alpha_n) \iso F(\beta_1, \cdots, \beta_n)\)
i.e. splitting fields are unique up to isomorphism

Example \(f(x)=x^4+x^2-6 \in \Q[x]\)
“The” splitting field of \(f(x)\) is \(\Q(i \sqrt{3}, \sqrt{2})\)

\(F, F'\) are fields. \(\varphi: F \to F'\) is an isomorphism
The natural map \(\tilde{\varphi} : F[x] \to F'[x]\) is an isomorphism

We write \(\tilde{\varphi} = \varphi\)

Test 1: Chapter 1-3(Irreducibles, Irreducibility, Field Extensions)
  1. a)[5 marks] (Sylow 3), b)[5 marks] Sylow 2
  2. a)[5 Marks] (Irreducibility of a polynomial), b)\(\deg_F(\alpha)\)
  3. a)[5 Marks] Proof \(F(\alpha_1, \cdots, \alpha_n)\), b) [5 Marks] Proof \([K:F]\)
    Test is out of 30 + 2/30 for showing up
Lemma [Isomorphism Extension Lemma]

\(F, F'\) fields, \(\varphi : F \to F'\) isomorphism. \(f(x) \in F[x]\) irreducible. Let \(\alpha\) be a root of \(f(x)\) in some extension of \(F\).

Let \(\beta\) be a root of \(\varphi(f(x))\) in some extension of \(F'\)

Then \(\exists \ \psi: F(\alpha) \to F'(\beta)\) such that:

  1. \(\psi |_F = \varphi\)
  2. \(\psi(\alpha) = \beta\)

\(\rho_1 (g(\alpha)) = \overline{g(x)} = g(x) + <f(x)>\)
\(\rho_2 (\overline{h(x)}) = h(\beta)\)
–> afforded by the 1st iso theorem

\(\sigma(\overline{g(x)}) = \overline{\varphi(g(x))}\)
are all isomorphisms.

$$\psi = \rho_2 \circ \sigma \circ \rho_1$$ isomorphism.

  1. \(a \in F\).
    \(\begin{aligned} \psi(a) &= \rho_2(\sigma(\rho_1(a))) \\ &=\rho_2(\sigma(\bar{a})) \\ &= \rho_2(\overline{\varphi(a)}) \\ &= \varphi(a) \end{aligned}\)
  2. \[\begin{aligned} \psi(\alpha) &= \rho_2(\sigma(\rho_1(\alpha))) \\ &= \rho_2 (\sigma(\bar{x})) \\ &= \rho_2 (\overline{\varphi({x})}) \\ &= \rho_2(\bar{x}) \\ &= \beta \end{aligned}\]

\(F\) field. \(f(x) \in F[x]\) is irreducible. \(\alpha, \beta\) are roots of \(f(x)\) in some extension of \(F\).
Then \(\exists\) isomorphism \(\psi : F(\alpha) \to F(\beta)\)
such that \(\psi |_F = \text{id}, \psi (\alpha)=\beta\)
–> fixing the constants, and send the roots to each other

Why? \(\varphi: F \to F \ \text{ id }\)


\(F\) field, \(f(x) \in F[x]\) non-constant. Let \(K\) splitting field for \(f(x)\) over \(F\). Let \(\varphi: F \to F'\) be an isomorphism
Let \(K'\) be a splitting field for \(\varphi(f(x))\) over \(F'\)

There exists an isomorphism, \(\psi : K \to K'\) such that \(\psi |_F = \varphi\)

Why? Induction

Theorem[Splitting Fields are Unique]

\(F\) field, \(f(x) \in F[x]\) is non-constant. Let \(K, K'\) be two splitting fields for \(f(x)\) over \(F\)
\(\exists\) isomorphism \(\psi : K \to K'\) such that \(\psi |_F = \text{ id }\)

Why? \(\varphi = \text{ id }\)

Question: \(F\) a field. Does there exists \(K/F\) such that every \(f(x)\) non-constant splits over \(K\).

Definition [Algebraically Closed] \(F\) field. We say \(F\) is algebraically closed iff every non-constant \(f(x) \in F[x]\) has a root (splits) in \(F\).

Exercise \(\bb{C}\)

Definition [Algebraic Closure] \(F\) field.
A field \(\bar{F}\) is called an algebraic closure of \(F\) if

  1. \(\bar{F}/F\) algebraic extension
  2. Every non-constant \(f(x) \in F[x]\) splits over \(\bar{F}\)

Example \(\C\) alg closure of \(\R\)
Example \(\C\) not an alg closure of \(\Q\)
(\(\pi \in \C\) not algebraic over \(\Q\))

Proposition Suppose \(\bar{F}\) is an algebraic closure of \(F\). Then \(\bar{F}\) is algebraically closed.

\(f(x) \in \bar{F}[x]\) non constant.
\(f(\alpha) = 0\) where \(\bar{F} \sub K, K/\bar{F}\)
\([\bar{F}(\alpha) : \bar{F}] < \infty\)

\(\bar{F}(\alpha)/\bar{F}\), \(\bar{F}/F\) algebraic
\(\implies \bar{F}(\alpha) / F\) algebraic
\(\implies \alpha\) alg. over \(F\)
\(\alpha \in \bar{F}\), \(\qed\)

Proposition \(F\) field.
There exists an alg. closed field \(K \sup F\).

Proof A4

Proposition \(F\) field
An algebraic closure of \(F\) exists.

Let \(K \sup F\) be algebraically closed.
and let \(L = \{\alpha \in K: \alpha \text{ algebraic over } F\}\)
We know \(K/L/F\) is a tower of fields.

Claim: Let \(f(x) \in F[x]\) be non-constant. Then \(f(x)\) has a root in \(L\).

Let \(\alpha \in K\) such that \(f(\alpha)=0\). (we know this exists since \(K\) is algebraically closed)

By definition, \(\alpha \in L\).

Fact: Algebraic closures are unique up to isomorphism

Notation \(F\) field, \(\bar{F}\) will denote the algebraic closure of \(F\)

Chapter 5 - Cyclotomic Extensions

Question: What is the splitting field of \(f(x)=x^n-1\) over \(\Q\)

The complex roots of \(f(x)=x^n-1\) are called the nth roots of unity.

$$1, \zeta_n, \zeta_n^2, \cdots, \zeta_n^{n-1}$$ \(\zeta_n = e^{\frac{2\pi i}{n}}\)
\(=\cos(\frac{2\pi}{n}) + i \sin(\frac{2\pi}{n})\)

Therefore the SF of \(x^n-1\) over \(\Q\) is \(\Q(\zeta_n)\)

We call \(\Q(\zeta_n)/\Q\) a cyclotomic extension.

Question: What is the minimal polynomial of \(\zeta_n\) over \(\Q\)?

Example \(n=p\) prime.
\(x^p-1 = (x-1)(\underbrace{x^{p-1}+x^{p-2} + \cdots + x + 1}_{\phi_p (x)}\)

\(\phi_p (x)\) minimal polynimial for \(\zeta_p\)


We know that \(G=\{1, \zeta_n, \zeta_n^2, \cdots, \zeta_n^{n-1}\}\)
is a cyclic subgroup of \(\C^\times\)

We have \(G=<\zeta_n>\). and \(G=<\zeta_n^{k}>\) iff
\(\gcd(k, n)=1\).

We call such a generator a primitive nth root of unity

i.e. \(\zeta \in \C\) is a primitive \(n\)th root of unity iff

  1. \(\zeta^n=1\)
  2. \(\zeta^k \neq 1\), \(1 \leq k \leq n\)
    i.e. order of \(\zeta\) = n

\(\tf\) The number of primitive nth roots of unity is \(\phi(n)=|\{1 \leq k \leq n: \gcd(k, n)=1\}|\)
–> Euler totient function

Definition [Cyclotomic Polynomial]
\(n \in N\), \(\alpha_1, \cdots, \alpha_{\phi(n)}\) are primitive nth ROUs (roots of unity).

\(\phi_n(x)=(x-\alpha_1) \cdots (x-\alpha_{\phi(n)})\)
–> nth cyclotomic polynomial


  1. \(\{z \in \C: z^n =1\}\)
    \(= \cup_{d|n} \{z \in \C: z \text{prim dth rou} \}\)

  2. \(x^n-1\)
    \(=\Pi_{\text{ nth roots of unity }} (x-\alpha_i)\)
    \(= \Pi_{d|n}\Pi_{\text{ prim dth }} (x-\alpha_i)\)
    \(= \Pi_{d|n} \phi_d(x)\)

Example Compute \(\phi_6(x)\)
\(x^6-1 = \phi_1(x)\phi_2(x)\phi_3(x)\phi_6(x)\)
\(\implies \phi_6(x) = \frac{x^6-1}{(x-1)(x+1)(x^2+x+1)}\)
\(= x^2-x+1\)

Prove \(\phi_n(x)\) is the minimal polynomial of \(\zeta_n\) over \(\Q\)

Proposition \(\phi_n(x) \in \Z[x]\)

Proof Induction on \(n\).
Clearly \(\phi_1(x) = x-1 \in \Z[x]\)

Assume the result holds for \(k < n\)

By the investigation, \(x^n-1 = \phi_n(x) f(x)\), where
\(f(x)=\Pi_{d | n, d<n}\phi_d (x)\)

By induction, \(f(x) \in \Z[x]\)

Let \(F=\Q(\zeta_n)\). Note:\(\phi_n(x) \in F[x]\)

By the division algorithm, \(\exists\) unique \(x^n-1=f(x)q(x)+r(x)\) where \(q(x), r(x) \in F[x]\),
\(r(x)=0\) or \(\deg r(x) < \deg f(x)\)

Similarly, \(\exists\) unique \(x^n-1=f(x)\tilde{q}(x) + \tilde{r}(x)\), \(\tilde{q}(x), \tilde{r}(x) \in \Q[x]\)
\(0=\tilde{r}(x)\) or \(\deg \tilde{r}(x) < \deg f(x)\)

By uniqueness,
\(q(x)=\tilde{q}(x)=\phi_n(x) \in \Q[x]\)

By Gauss, \(\phi_n(x) \in \Z[x]\), \(\qed\)

\(\phi_n(x)\) is irreducible over \(\Q\).

Let \(g(x)\) be the minimal polynomial for \(\zeta_n\) over \(\Q\).

We show \(g(x) = \phi_n(x)\)

Since \(g(\zeta_n) = \phi_n(\zeta_n) = 0\),
\(g(x) | \phi_n(x)\).

Say \(\phi_n(x) = g(x)h(x)\), \(h(x) \in \Q[x]\)

To show \(\phi_n(x) | g(x)\), we prove that \(g(\zeta_n^k)=0\) whenever \(\gcd(k, n)=1\).
–> every root of \(\phi_n(x)\) is a root of \(g(x)\).

Say \(k = p_1 p_2 \cdots p_r\) where \(p_i\) prime, \(p_i \nmid n\)

We will prove that \(g(\zeta_n)=0 \implies g(\zeta_n^{p_1})=0 \implies g(\zeta_n^{p_1 p_2})=0 \implies \cdots \implies g(\zeta_n^{k})=0\)

Claim If \(\zeta \in \C\) with \(g(\zeta)=0\) and \(p\) is prime with \(p \nmid n\), then \(g(\zeta^p)=0\).

Proof of Claim: Since \(g(\zeta)=0\), \(\phi_n(\zeta)=0\) (divisibility)
\(\tf \zeta\) is a primitive nth ROU \(\implies \zeta^p\) is a prim nth ROU, since \(p\nmid n\), \(\gcd(p,n)=1\)

\(\implies \phi_n(\zeta^p)=0\). For contradiction, suppose \(g(\zeta^p) \neq 0\). Hence, \(h(\zeta^p)=0\)

Note: by Gauss, \(h(x) \in \Z[x]\)
–> b/c both monic and \(\phi_n(x)\), by proof of GL, \(g(x), h(x) \in \Z[x]\)

Define \(f(x) = h(x^p)\) \(\implies f(\zeta)=0\) \(\implies g(x) | f(x)\) \(\implies f(x)=g(x)K(x), K(x) \in \Z[x]\) Gauss

$$h(x) = \sum b_j x^j$$ $$\implies f(x) = \sum b_j x^{pj}$$

In \(\Z_p[x]\),

\(\bar{f}(x) = \sum \bar{b_j} x^{pj} = \sum \bar{b_j}^p x^{pj} \text{ FLT } = (\sum \bar{b_j} x^j)^p = \bar{h}(x)^p\)

\(\tf \bar{h}(x)^p = \bar{f}(x) = \bar{g}(x)\bar{K}(x)\)

Let \(\bar{l}(x)\) be an irreducible factor of \(\bar{g}(x)\) in \(\Z_p[x]\)

\(\bar{l}(x) | \bar{h}(x)^p \implies \bar{l}(x) | \bar{h}(x)\)

Now, \(\bar{\phi}_n(x) = \bar{g}(x)\bar{h}(x)\)

\(\implies \bar{l}(x)^2 | \bar{\phi}_n(x)\) \(\implies \bar{l}(x)^2 | \underbrace{x^n-1}_{\Z_p[x]}\)

\(\implies x^n-\bar{1} = \bar{l}(x)^2 \bar{q}(x)\)
\(\implies \bar{n}x^{n-1} = \bar{l}(x)^2 \cdot \bar{q}'(x) + 2 \bar{l}(x) \bar{l}'(x) \bar{q}(x)\)
\(= \bar{l}(x) \cdot \text{[stuff]}\)

Note: \(p \nmid n \implies \bar{n} \neq \bar{0}\)

\(\tf \bar{l}(\bar{0}) = \bar{0}\)

Since \(\bar{l}(x) | x^n-\bar{1}\), \(\bar{0}^n-1 = \bar{0}\)  
\(\implies \)\bar{1}=\bar{0} \in \Z_p \implies p 1$. Contradiction!

For \(n \in \N\), \(\phi_n(x)\) is the minimal polynomial for \(\zeta_n\) over \(\Q\). In particular,
$$[\Q(\zeta_n):\Q] = \phi(n)$$

Examples: let \(K\) be the splitting field of \(f(x) = x^5-3\) over \(\Q\)

  1. Describe \(K\)
  2. Compute \([K : \Q]\)
  3. Find a basis for \(K / \Q\)

1) The complex roots of \(f(x)\) are \(\sqrt[5]{3}, \sqrt[5]{3} \zeta_5, \sqrt[5]{3} \zeta_{5^2}, \sqrt[5]{3} \zeta_{5^3}, \sqrt[5]{3} \zeta_{5^4}\)
\(\tf K = \Q(\sqrt[5]{3}, \zeta_5)\)

2) \([\Q(\sqrt[5]{3}):\Q] = \underbrace{\deg (x^5-3)}_{\text{3-Eis}} = 5\)
\([\Q(\zeta_5):\Q] = \phi(5)=4\)
Since \(\gcd(4, 5)=1\), \([K : \Q] = 5 \cdot 4 = 20\)

3) \([\Q(\zeta_5)(\sqrt[5]{3}):\Q(\zeta_5)] = 5\)
(Tower Theorem)

A basis for \(\Q(\zeta_5)(\sqrt[5]{3})/\Q(\zeta_5)\) is
\(B_1 = \{1, \sqrt[5]{3}, (\sqrt[5]{3})^2, (\sqrt[5]{3})^3, (\sqrt[5]{3})^4\}\)

A basis for \(\Q(\zeta_5)/\Q\) is
$$B_2 = \{1, \zeta_5, \zeta_5^2, \zeta_5^3\}$$

A basis for \(K/\Q\) is
$$\{(\sqrt[5]{3})^i \zeta_5^j : 0 \leq i \leq 4, 0 \leq j \leq 3\}$$ –> proof of the Tower Theorem

Chapter 6 - Finite Fields

Proposition Let \(F_q\) be a finite field.
Then \(F_{q^\ast}\) is cyclic.

Proof: \(F_{q^\ast}\) has \(q-1\) elements
So \(F_{q^\ast} \iso C_1 \times C_2 \times \cdots \times C_r\)
where \(C_i\) is cyclic. If \(i \neq j\) and \(d|\gcd(|C_i|, |C_j|)\) then the equation \(x^d =1\) has at most \(d\) solutions.
in \(F_{q^\ast}\), has exactly \(d\) solutions in \(C_i\) and in \(C_j\). So intotal, \(2d-1\) solutions in \(F_{q^\ast}\) So \(2d-1 \leq d \implies d \leq 1\)
so the product \(\Pi C_i\) is cyclic. \(\qed\).

Proposition If \([K:F_q] = d\), then \(|K| = q^d\)

\(K = \{a_1x_1+\cdots+a_dx_d : a_i \in F_q\}\)
\(\implies |K| = q^d\)
where \(\{x_1, \cdots, x_d\}\) is an \(F_q\) basis of \(K\) \(\qed\)

Proposition If \(K/F_q\) is finite, then \(K = F_q(\alpha)\) for some \(\alpha \in K\).

Proof: Let \(\alpha =\) generator of \(K^\ast\) \(\qed\)

The characteristic of \(F_q\) is some prime \(p\). So the image of the charac homomorphism \(\phi: \Z \to F_q\) is a field isomorphic to \(F_p = \Z / p \Z\). So \(F_q = F_p (\alpha)\) for some \(\alpha\) and \(|F_q| = q = p^n\) for some \(n \in \Z\)

Proposition \(F_q\) have \(q=p^n\) elements. Then \(F_q\) is a splitting field for \(x^{p^n}-x\) over \(\Z/p\Z = F_p\)
–> if we can prove this, any two finite fields are isomorphic

\(x^{p^n}-x\) splits in \(F_q\) because its roots satisfy \(x=0\) or \(x^{p^n-1}=1\), so every root of \(x^{p^n}-x\) lies in \(F_q\)
Conversely, the set of roots of \(x^{p^n}-x\) is a field \(F_q\), so
$$F_q = \{r_1, \cdots, r_{p^n}\} = F_p (r_1, \cdots, r_{p^n})$$ where \(r_1, \cdots, r_{p^n}\) are the roots of \(x^{p^n}-x\) \(\qed\)

Corollary So any two fields with \(p^n\) elements are isomorphic.
Also, for any prime \(p\) and positive integer \(n\), there is a field with \(p^n\) elements.
If \(K\) is a field with two subfields \(L_1, L_2\) of order \(p^n\), then \(L_1 = L_2\), because they are both the splitting field (set of roots of) \(x^{p^n}-x\) in \(K\)

Proposition The field \(F_{p^n}\) contains a subfield of order \(p^m\) iff \(m|n\)

Proof This follows from \(x^{p^m}-x\) divides \(x^{p^n}-x\) iff \(m|n\) \(\qed\)

Let \(K\) be any field of characteristic \(p > 0\). The Frobenius homomorphism
Frob: \(K \to K\) is defined by:
$$\text{Frob}(\alpha) = \alpha^p$$ Check:
\((\alpha+\beta)^p = \alpha^p + \binom{p}{1} \alpha^{p-1}\beta + \cdots + \binom{p}{p-1} \alpha \beta^{p-1}+\beta^p = \alpha^p + \beta^p\)

If \(K=F_p\), then \(\text{Frob} = \text{id}\)

If \(K=F_{p^2}\), then say \(\text{Frob}(\alpha) = \alpha\), then \(\alpha^p=\alpha\) so \(\alpha\) is a root of \(x^p-x\) so \(\alpha \in F_p\). Thus, Frob moves every element of \(F_{p^2} \to F_p\)

In general, the fixed set of Frob is always \(F_p\).

\(\text{Frob}^2 = \alpha^{p^2}\). Its fixed set is \(F_{p^2} \cap K\), any \(K\) characteristic \(p\)

In even more general, the fixed set of \(\text{Frob}^n\) is \(F_{p^n} \cap K\)

Note: If \(K\) is finite, then Frob\(: K \to K\) is isomorphism is an isomorphism, because it’s injective

If \(K\) is not finite, then sometimes Frob is an isomorphism, and sometimes it isn’t.

Example \(q=9\), \(F_q \iso F_3(i)\), \(i^2=-1\)
What is \(\text{Frob}(a+bi)\)?
\((a+bi)^3 = a^3 + (b_i)^3 = a^3-b^3i: a, b \in F_3\)
If \(q\) is odd, then \(F_{p^2} \iso F_p(\sqrt{d})\), so Frob(\(a+b\sqrt{d}\))=\((a-b\sqrt{d})\)


  1. \(F\) is a finite field, \(|F| = p^n\) where
    1. \(p = Char(F)\)
    2. \(n = [F:\Z_p]\)
  2. There is a unique (upto iso) field of order \(p^n\)
    It is the splitting field of \(f(x)=x^{p^n}-x\) over \(\Z_p\)
  3. \(\F_{p^n}\) has a unique subfield of order \(p^d\) for every \(d | n\). And, these are all the subfields of \(\F_{p^n}\)

Proof of (1): \(p = Char(F)\), \(F^\times = <\alpha>\), \(F = \Z_p(\alpha)\).
Let \(n = \deg_{\Z_p} (\alpha) = [F:\Z_p]\)
\(F=\Span_{\Z_p}\{1, \alpha, \cdots, \alpha^{n-1}\}\)
\(\implies |F|=p^n\)

Chapter 7 - Galois Groups

Context: Galois Theory is the study of the roots of polynomials and how they “interact”.

Recall \(K\) field
\(Aut(K) = \{\varphi: K \to K \text{ isomorphism }\}\) is the group of automorphisms under function composition.

Definition [Galois Group] \(K/F\)
\(Gal(K/F) = \{\varphi \in Aut(K): \varphi|_{F} = id\}\)
–> automorphisms of K that leave \(F\) alone
called the Galois Group of \(K/F\)
We call \(\varphi \in Gal(K/F)\) a Galois automorphism of \(K\).

\(Gal(K/F) \leq Aut(K)\)

Proposition \(K/F\)
\(f(x) = a_nx^n +a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \in F[x]\)
If \(\alpha \in K\) is a root of \(f(x)\) and \(\varphi \in Gal(K/F)\), then \(\varphi(\alpha)\) is a root of \(f(x)\).

\(a_n\alpha^n + \cdots + a_1 \alpha + a_0 = 0\)
\(\implies\) \(\varphi(a_n)\varphi(\alpha^n) + \cdots + \varphi(a_1)\varphi(\alpha)+\varphi(a_0)=\varphi(0)=0\)
\(\implies a_n \varphi(\alpha)^n + \cdots + a_1 \varphi(\alpha)+a_0 = 0\)

Corollary \(K/F\), \(\varphi \in Gal(K/F)\)
Suppose \(\alpha \in K\) is algebraic over \(F\).

Then \(\alpha, \varphi(\alpha)\) have the same minimal polynomial.

Remark \(K/\Q\)

  1. \(Gal(K/\Q) = Aut(K)\)
    \(\varphi \in Aut(K)\), \(\varphi(1)=1\), \(\varphi(\underbrace{1+1+\cdots+1}_n) = n\), i.e. \(\varphi(n)=n\). Then \(\varphi(-n)=-\varphi(n)=-n\)
    i.e. \(\varphi(x)=x\), \(\forall x \in \Z\) and
    \(\varphi(\frac{a}{b}) = \frac{\varphi(a)}{\varphi(b)} = \frac{a}{b}\)
    –> everything fixes

Remark \(K=F(\alpha_1, \cdots, \alpha_n)\)
\(\varphi \in Gal(K/F)\) is completely determined by \(\varphi(\alpha_i), 1 \leq i \leq n\)

Examples \(Gal(\C/\R)\), \(\C = \R(i)\)
For \(\varphi \in Gal(\C/\R)\), \(\varphi(i) = \pm i\) (root of \(x^2+1\))
So \(\varphi = \text{id}\) or \(\varphi =\) complex conjugation
\(Gal(\C/\R) = \Z_2\)

Example \(K=\Q(\sqrt{2})\). Take \(\varphi \in Gal(K/\Q)\)
\(\varphi(\sqrt{2}) = \pm \sqrt{2}\) (root of \(x^2-2\))
By the extension lemma, there exists an isomorphism \(\psi\) such that \(\psi : \Q(\sqrt{2}) \to \Q(\sqrt{2}): \sqrt{2} \to -\sqrt{2}\)
\(\tf Gal(K/\Q) = \{\text{id}, \psi\} = \Z_2\)

\(K = \Q(\sqrt{2}, \sqrt{3})\). \(\varphi \in Gal(K/\Q)\)
\(\varphi(\sqrt{2}) = \pm \sqrt{2}\), \(\varphi(\sqrt{3}) = \pm \sqrt{3}\)
–> can’t just send one root to any other root b/c not roots of the same minimal polynomial.

By extension lemma: do the diagram![[West LA - 1150 AM 1.png]]
\(\tf Gal(K/\Q(\sqrt{2})) = \{\varphi_1, \varphi_2, \varphi_3, \varphi_4\}\)
\(\varphi_1 = \text{id}\)
\(\varphi_2(\sqrt{2}) = \sqrt{2}, \varphi_2(\sqrt{3}) = -\sqrt{3}\)
\(\varphi_3(\sqrt{2}) = -\sqrt{2}, \varphi_3(\sqrt{3}) = \sqrt{3}\)
\(\varphi_4(\sqrt{2}) = -\sqrt{2}, \varphi_4(\sqrt{3}) = -\sqrt{3}\)

\(K = \Q(\sqrt[3]{2})\). \(\varphi \in Gal(K/\Q)\)
\(\varphi(\sqrt[3]{2}) \in \{\sqrt[3]{2}, \sqrt[3]{2} \zeta_3, \sqrt[3]{2}\zeta_3^2\} \cap K \sub \R\)
\(\implies \varphi(\sqrt[3]{2}) = \sqrt[3]{2}\)
\(\implies Gal(K/\Q)=\{\text{id}\}\)


  1. \(Gal(K/F) = \varphi \in Aut(K): \forall a \in F, \varphi(a)=0\)
  2. \(f(x) \in F[x], \alpha \in K, f(\alpha)=0 \ \forall \varphi \in Gal(K/F)\)

Definition [Seperable] We say that \(f(x) \in F[x]\) is seperable if \(f(x)\) has no repeated roots in its splitting field.

Definition [\(Gal(f(x))\)] Let \(f(x) \in F[x]\) be non-constant. \(Gal(f(x)) := Gal(K/F)\), \(K\) is the splitting field of \(f(x)\) over \(F\).


  1. \(f(x) \in F[x]\) seperable (non constant), \(K\) be the splitting field of \(f\). Roots \(\alpha_1, \cdots, \alpha_n \in K\) where \(n = \deg f(x)\)
    Let \(G = Gal(f(x)\)). Then \(G\) acts on \(\{\alpha_1, \cdots, \alpha_n\}\) via \(\varphi \cdot \alpha_i = \varphi(\alpha_i)\). We can say \(\varphi(\alpha_i) = \alpha_{\sigma(i)}\)

    Then \(G\) is isomorphic to a subgroup of \(S_n\) via \(\varphi \mapsto \sigma\)

  2. In addition, assume \(f(x)\) is irreducible.
    By the extension lemma, \(\forall i, j, \exists \varphi \in G\) such that \(\varphi(\alpha_i)=\alpha_j\)
    ![[West LA - 1150 AM.png]]
    i.e. the group action is transitive.
  3. \(|G| = |Stab(\alpha_i)| \cdot \underbrace{|Orb(\alpha_i)|}_{n}\)
    \(\implies |G| \mid n!\), \(n | |G|\)

Example \(f(x) = (x^2-2)(x^2-3) \in \Q[x]\). This polynomial is separable (just check roots). We compute \(Gal(f(x))\) we have that
$$Gal(f(x)) \iso \{e, (1 2), (3 4), (1 2)(3 4)\}$$

Example \(G = Gal(x^3-2)\) where \(x^3-2 \in \Q[x]\). The roots are:
$$\alpha_1 = \sqrt[3]{2}, \alpha_2 = \sqrt[3]{2} \zeta_3, \alpha_3 = \sqrt[3]{2} \zeta_3^2$$

Minimal polynomial for \(\zeta_3\) over \(\Q\) is
$$\phi_3(x) = x^2+x+1$$ By an argument of roots the minimal polynomial for \(\sqrt[3]{2}\) over \(\Q(\zeta_3)\) is \(x^3-2\).
Suppose not, then
$$[\Q(\sqrt[3]{2}):\Q] \mid [\Q(\zeta_3):\Q]$$ which means that \(3 \mid 2\). We know that \(G \leq S_3\) and \(3 \mid |G|\). So \(G = S_3\) or \(G=A_3 \iso Z_3\). Using the extension lemma we have that:
$$\varphi(\alpha_1) = \alpha_1$$ $$\varphi(\alpha_2) = \varphi(\sqrt[3]{2}) \varphi(\zeta_3) = \sqrt[3]{2} \zeta_3^2$$ $$\varphi(\alpha_3) = \varphi(\sqrt[3]{2})\varphi(\zeta_3)^2 = \alpha_2$$ Hence \(\varphi = (2 3)\) which is odd. So it must be the case that \(G=S_3\).

Example \(f(x) = x^4-4x^2+2 \in \Q[x]\). Let \(G =Gal(f(x))\). Using the quadratic formula, you can check roots are:
$$\alpha_1 = \sqrt{2+\sqrt{2}}, \alpha_2 = - \sqrt{2+\sqrt{2}}, \alpha_3 = \sqrt{2-\sqrt{2}}, \alpha_4 = -\sqrt{2-\sqrt{2}}$$

Note that \(\alpha_1\alpha_3 = \alpha_1^2-2\) so that \(\alpha_3 = \frac{\alpha_1^2-2}{\alpha_1}\). Since \(f(x)\) is irreducible, for all \(1 \leq i \leq 4, \exists \ \varphi_i \in G\) such that \(\varphi_i (\alpha_1) = \alpha_i\)
From before,
$$G=\{\alpha_1, \alpha_2, \alpha_3, \alpha_4\}$$ We have that
$$\varphi_2(\alpha_1) = \alpha_2$$ $$\varphi_2(\alpha_2)= -\varphi_2(\alpha_1) = -\alpha_2=\alpha_1$$ $$\varphi_2(\alpha_3) = \varphi_2 \left(\frac{\alpha_1^2-2}{\alpha_1}\right) = \alpha_4$$ $$\varphi_2(\alpha_4) = -\varphi_2(\alpha_3) = \alpha_3$$

Hence \(\varphi_2 = (1 2)(3 4)\). You can also deduce \(\varphi_3 = (1 3 2 4)\) and by group properties (inverses and closure) you get that \(\varphi_4 = (1 4 2 3)\). Hence \(G = \{e, (12)(34), (1324), (1 4 2 3)\} = <(1 3 2 4)> \iso \Z_4\)

Goal from here: Start to develop the theory of Galois Groups.

Definition [\(F\)-Map] Let \(K/F\) and \(E/F\) be field extensions. We say that \(\varphi K \to E\) is an \(F\)-Map iff

  1. \(\varphi\) is a homomorphism
  2. \(\forall a \in F, \varphi(a)=a\)

Remark Let \(\varphi: K \to E\) be an F-map. Then:

  1. \(\varphi\) is injective \((\ker \varphi = \{0\})\)
  2. For all \(u, v \in K\), \(\varphi(u+v)=\varphi(u)+\varphi(v)\)
  3. \(\forall \alpha \in F, u \in K\), \(\varphi(\alpha u)=\alpha\varphi(u)\)
    This implies that \(\varphi\) is a linear transformation!. Moreover:

If \([K:F] < \infty\) and \(E=K\) then \(\varphi \in Gal(K/F)\)
\(K\) is a finite dimensional \(F\) vector space and hence \(\varphi\) is injective iff \(\varphi\) is surjective.

Lemma Let \(K/F\) and \(E/F\) be field exensions and \([K:F]\) be finite. Then the number of \(F\)-maps \(\varphi: K \to E\) is at most \([K:F]\)

Proof: We can write \(K = F (\alpha_1, \cdots, \alpha_n)\). We proceed by induction on \(n\). Suppose \(K=F(\alpha_1)\). An \(F\)-map is completely determined by \(\varphi(\alpha_1)\). But \(\alpha_1, \varphi(\alpha_1)\) have the same minimal polynomial. The number of choices of for \(\varphi(\alpha_1)\) is at most:
\(\deg_F(\alpha_1) = [F(\alpha_1):F] = [K:F]\)

Proceeding inductively, assume \(K = F(\alpha_1, \cdots, \alpha_n), n > 1\). Let \(L=F(\alpha_1, \cdots, \alpha_{n-1})\) and let \(\varphi: K \to E\) be an \(F\)-map. Note: \(\varphi |_L\) is an \(F\)-map. Since \(\varphi\) is completely determined by \(\varphi |_L\) and \(\varphi(\alpha_n)\). there are at most:
$$\underbrace{[L:F]}_{(\text{IH})} \cdot \underbrace{\deg_L (\alpha_n)}_{[\underbrace{L(\alpha_n)}_K:L]} = [K:F]$$ \(\qed\)

Lemma Suppose that \(K/F\) extension and \([K:F]\) is finite. Then:
$$|Gal(K/F)| \leq [K:F]$$ Why? \(\varphi \in Gal(K/F) \iff \varphi: K \to K\) \(F\)-map.

Example Suppose \(K = \Q(\sqrt[3]{2}), F=\Q\). Then
$$|Gal(K/F) < 3 = [K:F]$$ Example \(K = \Z_2(t)\) and \(F = \Z_2(t^2)\). Then \([K:F]=2\). Let \(\varphi \in Gal(K/F)\). Then, \(\varphi(t)\) is a root of \(x^2-t=(x-t)^2 (char(F)=2)\). And hence \(\varphi(t)=t\) so that \(\varphi=\text{id}\) and hence \(|Gal(K/F)|=1\)

Remark We are interested when

Definition [Separable Element] Let \(K/F\) be an extension. We say that an algebraic \(\alpha \in K\) is seperable over \(F\) iff \(m_{\alpha} \in F[x]\) is separable.

Definition [Separable Extension] \(K/F\). An algebraic extension \(K/F\) is seperable iff \(\alpha \in K\) is separable over \(F\) for all \(\alpha \in K\)

Definition [Perfect] \(K/F\). \(F\) is perfect iff \(K/F\) is separable for all algebraic extensions \(K/F\).

Example \(\Z_p(t^p)\) is NOT perfect.

Recall Let \(f(x) \in F[x]\) be irreducible. Then \(f(x)\) is separable iff \(f'(x) \neq 0\)

Proposition Every field where \(char(F)=0\) is perfect.

Proposition Let \(F\) be a field with \(char(F)=p>0\). Let \(f(x) \in F[x]\) be irreducible. Then \(f(x)\) is not seperable if and only if \(f(x)=g(x^p)\) for some \(g(x) \in F[x]\)

Why? \(f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\), \(f(x)\) not seperable \(\iff f'(x) =0\) \(\iff ka_k=0\), \(k=1, \cdots, n\)
\(\iff k=0\) or \(a_k=0\), \(k= 1, \cdots, n\)
\(\iff k a_k = pm_k a_k, m_k \in \N, k=1, \cdots, n\)
\(\iff f(x)=a_nx^{pm_n} + \cdots + a_1x^{m_1}+a_0\)

Corollary: If \(F\) is finite, \(F\) is perfect.

$$|Gal(K/F)| \leq [K:F]$$ $$F \text{ perfect } \iff (K/F \text{ alg } \implies K/F \text{ sep})$$ –> F is perfect iff every irreducible polynomial has no repeated roots. Remember that every irreducible is a minimal, take the kronecker extension

Proposition Every finite field is perfect.
Suppose \(F\) is finite with \(char(F)=p>0\).
Suppose \(f(x) \in F[x]\) is irreducible but not seperable.
Then, \(f(x)=g(x^p)\) where \(g(x) \in F[x]\). Say \(g(x) = a_nx^n + \cdots + a_1 x + a_0\)
\(\implies f(x) = a_n x^{p_n} + \cdots + a_1 x^p + a_0\)

Now, \(\varphi: F \to F: \varphi(x)=x^p\) is an injective homomorphism.
Since \(F\) is finite, \(\varphi\) is surjective.
\(\tf \forall i, \exists b_i \in F, a_i = b_i^p\)
\(f(x)=b_{n}^p x^{p_n} + \cdots + b_1^p x^p + b_0^p\)
\(=(b_n x^n + \cdots + b_1x^ + b_0)^p\). A contradiction! \(\qed\)
–> Contradicted!

Example \(F=\Z_2(t)\)
\(f(x)=x^2-t \in F[x]\) irreducible (because no roots degree 2)
Let \(K\) be the splitting field of \(f(x)\) over \(F\).
Let \(\alpha \in K\) s.t. \(f(\alpha)=0\).
\(\implies \alpha^2 = t\)
\(f(x)=x^2-t=x^2-\alpha^2=(x-\alpha)^2 \in K[x]\)
\(\tf K/F\) is not seperable and \(F\) is not perfect.

\(f(x) \in F[x]\) is seperable and non-constant.
Let \(K\) be the splitting field of \(f(x)\) over \(F\). Then,
\(|Gal(K/F)| = |Gal (f(x))|=[K:F]\)

\(K/E/F\). \(Gal(K/E) \leq Gal(K/F)\)

Proof of Theorem
We proceed by induction on \([K:F]\).
If \([K:F]=1\), then \(1 \leq |Gal(K/F)| \leq [K:F] = 1\)
Proceeding inductively, assume \([K:F] = n > 1\).

Therefore \(\exists\) \(p(x) \in F[x]\) irred. such that \(p(x)|f(x)\) and \(\deg p(x) = m > 1\).

Say the roots of \(p(x)\) are \(\alpha_1, \cdots, \alpha_m \in K \setminus F\)

Note: \(\alpha_i \neq \alpha_j\)
Moreover, \(K\) is the splitting field of \(f(x)\) over \(\underbrace{F(\alpha_1)}_{E}\)

We have \([K:E]=\frac{[K:F]}{[E:F]} = \frac{n}{m} < n\)

By induction,

Since \(p(x)\) is irreducible, \(\forall 1 \leq i \leq m\)
\(\exists\) \(\varphi_i \in Gal(K/F)\) such that \(\varphi_i(\alpha_1) = \alpha_i\)
–> isomorphism extension lemma is back!

For \(i \neq j\), \(\alpha_i \neq \alpha_j\) \(\implies \varphi_i(\alpha_1) \neq \varphi_j(\alpha_1)\).
\(\implies (\varphi_j^{-1} \circ \varphi_i)(\alpha_1) \neq \alpha_1\) but \(\alpha_1 \in E\)

\(\implies \varphi_j^{-1} \circ \varphi_i \notin Gal(K/E)\)
\(\implies \varphi_i Gal(K/E)\neq \varphi_j Gal(K/E)\)
$$\frac{Gal(K/F)}{Gal(K/E)} \geq m$$ \(\implies Gal(K/F) \geq [K:E] \cdot m = \frac{n}{m} \cdot m = n\)

Chapter 8 - Normal + Seperable Extensions

Goal: We show that for \([K:F] < \infty\), \(K\) is often the splitting field of a sep poly over \(F\)

Definition [Simple Extensions] \(K/F\)
We say \(K/F\) is simple iff \(\exists\) \(\alpha \in K, K = F(\alpha)\)
We call \(\alpha \in K\) a primitive element for \(K\) over \(F\)

Theorem [Primitive Element Theorem]
If \(K/F\) is finite + seperable, then \(K/F\) is simple.

Corollary \([K:F]\) finite, \(F\) perfect, then \(K=F(\alpha)\) for some \(\alpha \in K\)

Proof of Theorem
Case 1: \(F\) is finite
Since \(K/F\) is finite, \(K\) is finite. From before \(K^\times = <\alpha>\), \(\alpha \in K\)

Case 2: \(F\) is infinite
Assume \(K/F\) is finite and seperable. Say \(K=F(\alpha_1, \cdots, \alpha_n)\)
By induction, we may assume \(n=2\) and \(K=F(\alpha, \beta)\)
Let \(p(x)\) the min poly for \(\alpha\) over \(F\). and \(q(x)\) be the min poly for \(\beta\) over \(F\)
–> (both seperable since \(K/F\) seperable)

Let \(L\) be the splitting field of \(p(x)q(x)\) over \(K\)
Say the roots of \(p(x)\) and \(q(x)\) are \(\alpha = \alpha_1, \cdot,s \alpha_n \in L\) and \(\beta = \beta_1, \cdots, \beta_m \in L\) respectively.

$$S = \left\{\frac{\alpha_i-\alpha_1}{\beta_1-\beta_j} : i \neq 1, j \neq 1\right\}$$

Since \(S\) is finite, and \(F\) is infinite, there exists \(u \neq 0\) in \(F\) such that \(u \notin S\)
Let \(\gamma = \alpha + u \beta\).

Claim: \(K = F(\alpha, \beta) = F(\gamma)\)
By minimality \(F(\gamma) \sub F(\alpha, \beta)\)

New stuff: WWTS \(F(\alpha, \beta) \sub F(\gamma)\)
Let \(h(x)\) be the min poly for \(\beta\) over \(F(\gamma)\)
–> We want \(\deg h(x) = 1\)

Note: \(h(x) | q(x)\). The roots of \(h(x)\) are a subcollection of the \(\beta_j\)’s.

Moreover, if \(k(x)=p(\gamma-ux) \in F(\gamma)[x]\)
\(\implies k(\beta)=p(\gamma-u\beta) = p(\alpha)=0\)
\(\implies h(x) | k(x)\)

For \(j \neq 1\), \(k(\beta_j) = 0 \iff p(\gamma-u\beta_j)=0\)
\(\iff \gamma - u\beta_j = \alpha_i (i \neq 1, \text{ since otherwise } (\gamma-u\beta_j=\gamma-u\beta_1, \beta_1=B_j))\)
\(\iff \alpha_1+u\beta_1-u\beta_j=\alpha_i\)
\(\iff u = \frac{\alpha_i-\alpha_1}{\beta_1-\beta_j}\)
By choice of \(u\), \(K(\beta_j) \neq 0\) for \(j \neq 1\)
\(\implies h(\beta_j)\neq 0\) for \(j\neq 1\)
\(h(x)=x-\beta \in F(\gamma)[x]\)
\(\implies \beta \in F(\gamma)\), \(\implies \alpha \in F(\gamma)\) \(\implies F(\alpha, \beta) \sub F(\gamma)\)

$$Gal(\Q(\sqrt[3]{2})/\Q) = \{1\}$$ \(\varphi(\sqrt[3]{2}) = \{\sqrt[3]{2}, \underbrace{\sqrt[3]{2}\zeta_3}_{\notin \Q(\sqrt[3]{2}}, \underbrace{\sqrt[3]{2}\zeta_3^2}_{\notin \Q(\sqrt[3]{2}}\}\)

Definition [Normal Extensions] \([K:F] < \infty\)
We say \(K/F\) is normal iff \(K\) is the splitting field of a non-constant \(f(x) \in F[x]\)

Definition [\(F\)-conjugates] \(K/F\), \(\alpha \in K\) is algebraic over \(F\)
Let \(f(x)\) be the min poly for \(\alpha\) over \(F\).
The roots of \(f(x)\) in its splitting field are called the \(F\)-conjugates or just conjugates of \(\alpha\)
Example the \(\Q\)-conjugates of \(\sqrt[3]{2}\) are \(\sqrt[3]{2} \zeta_3, \sqrt[3]{2}\zeta_3^2\)

Theorem [Normality Theorem] \([K:F] < \infty\). TFAE:

  1. \(K/F\) is normal
  2. For every \(L/K\) and \(F\)-map \(\varphi: L \to L\), \(\varphi |_K \in Gal(K/F)\)
    \(F\) map is a homo that fixes \(F\)
  3. If \(\alpha \in K\) then all \(F\)-conjugates of \(\alpha\) belong to \(K\)
  4. If \(\alpha \in K\) then its min poly over \(F\) splits over \(K\)

(1) \(\implies\) (2) Assume \(K\) is the splitting field of \(f(x) \in F[x]\). Say the roots of \(f(x)\) are \(\alpha_1, \alpha_2, \cdots, \alpha_n \in K\).
Note: \(K=F(\alpha_1, \cdots, \alpha_n)\)
Let \(L/K\) be an extension, and let \(\varphi: L \to L\) be an \(F\)-map (homo that fixes \(F\))
For every \(\alpha_i\), \(\varphi(\alpha_i)=\alpha_j\) for some \(j\). \(\tf \varphi |_K : K \to K\) is an injective homomorphism. (injective linera transformation)
Since \([K:F] < \infty\), \(\varphi\) is surjective. i.e. \(\varphi \in Gal(K/F)\)

(2) \(\implies\) (3)
Assume (2). Let \(\alpha \in K\) and let \(p(x)\) be its min poly over \(F\)
Since \([K:F] < \infty\), there exists \(\alpha_1, \cdots, \alpha_n\) such that \(K=(\alpha_1, \cdots, \alpha_n)\)

Let \(h_i(x) \in F[x]\) be the min poly for each \(\alpha_i\). Consider \(f(x)=p(x)h_1(x)\cdots h_n(x)\)

Let \(L\) be the splitting field for \(f(x)\) over \(F\). Then, \(L \sup K \sup F\)

Let \(\beta \in L\) be a root of \(p(x)\). Since \(p(x)\) irreducible over \(F\), \(\exists \varphi \in Gal(L/F)\) such that \(\varphi(\alpha)=\beta\)
–> extension lemma

$$\beta = \varphi(\alpha) = \varphi |_K (\alpha) \in K$$

(3) \(\iff\) (4)

(4) \(\implies\) (1)
Assume 4. As before, let \(K=F(\alpha_1, \cdots, \alpha_n)\)
Let \(h_i(x)\) be the min poly for \(\alpha_i\) over \(F\). By (4), \(K\) is the splitting field for \(f(x)= h_1(x)h_2(x) \cdots h_n(x)\)

Example \(\Q(\sqrt[3]{2})/ \Q\) is NOT normal, because \(\sqrt[3]{2}\zeta_3 \in \Q(\sqrt[3]{2})\)
–> easiest way to show something is not normal is to show it is not conjugate closed

Example \(\Q(\zeta_n)/\Q\). Normal b/c it is the splitting field for \(\phi_n(x)\) over \(\Q\)

Example \(\F_{p^n}/\F_p\). Normal b/c splitting field for \(x^{p^n}-x\)

Example \(\Z_p(t)/\Z_p(t^p)\). Normal b/c splitting field for \(x^p-t^p\)

–> to show normality, just show it’s the splitting field of something.

Definition [Galois Extensions] \([K:F]<\infty\)
We say \(K/F\) is Galois iff \(K/F\) is normal and seperable.


  1. \(F\) is perfect, then \(K/F\) is Galois for iff \(K/F\) is normal

Definition [Fixed Field] \(K\) field, \(G \leq Aut(K)\)
We call \(Fix(G) = \{a \in K: \forall \varphi \in G, \varphi(a)=a\}\) the fixed field of \(G\)

Homework: \(Fix(G)\) is a subfield of \(K\).

Theorem [Characterization Theorem] \([K:F] < \infty\), TFAE:

  1. \(K\) is the splitting field of a separable polynomial in \(F[x]\)
  2. \(|Gal(K/F)| = [K:F]\)
  3. \(Fix(Gal(K/F)) = F\)
  4. \(K/F\) Galois

(1) \(\implies\) (2) done. Proved before

(2) \(\implies\) (3). Assume \(|Gal(K/F)|=[K:F]\).Let \(E = Fix(Gal(K/F))\).
Note: \(F \sub E \sub K\) by definition, and
\(Gal(K/E) \leq Gal(K/F)\)

Let \(\varphi \in Gal(K/F)\) and let \(a \in E\). So \(\varphi(a)=a\)
\(\implies Gal(K/E)=Gal(K/F)\)
\(\tf [K:F] = |Gal(K/F)=|Gal(K/E)| \leq [K:E] \leq [K:F]\)
\(\implies [E:F]=1\)

(3) \(\implies\)(4). Suppose \(Fix(G)=F\) where \(G = Gal(K/F)\)
Let \(\alpha \in K\) and let \(p(x)\) be the minimal poly for \(\alpha\) over \(F\). We show \(p(x)\) splits (normal) as a product of distinct linear factors (separable) over \(K\).

Let $$\Delta = \{\varphi(\alpha) : \varphi \in G\} \sub \{\text{roots of (p(x))} \} \cap K$$

Let \(\alpha = \alpha_1, \alpha_2, \cdots, \alpha_n\) be the distinct elements of \(\Delta\)
$$h(x)=(x-\alpha_1) \cdots (x-\alpha_n) \in K[x]$$

Clearly \(h(x)|p(x)\) in \(K[x]\).

For \(\varphi \in G\), \(\varphi(h(x)) = h(x)\)
\(\implies h(x) \in Fix(G)[x]\)
\(\implies h(x) \in F[x]\)
\(\implies p(x) | h(x)\) \(\implies p(x) =h(x)\)

\(\tf K/F\) is normal+seperable.

(4) \(\implies\) (1). Assume \(K/F\) Galois.
By the PET, \(\exists \in K\) such that \(K=F(\alpha)\). Easily, \(K\) is the splitting field of the minimal polynomial \(p(x)\) for \(\alpha\) over \(F\). \(\qed\)
–> Galois extensions are the splitting field of an irreducible polynomial.

Chapter 9 - Fundamental Theorem of Galois Theory

Theorem [Artin’s Theorem]
If \(H\) is a finite subgroup of \(Aut(K)\) and \(F=Fix(H)\).

  1. \([K:F] = |H|\)
  2. \(K/F\) is Galois
  3. \(Gal(K/F)=H\)

First, \(H \leq Gal(K/F)\).
\(\tf |H| \leq |Gal(K/F)| \leq [K:F]\). It suffices to prove
$$[K:F] \leq |H|$$

Let \(\beta_1, \cdots, \beta_n \in K^\times\) s.t. \(n > m\) where \(m=|H|\)

Claim: distinct \(\{\beta_1, \cdots, \beta_n\}\) is linearly dependent.

Proof of Claim Consider the system:
$$\varphi(\beta_1)x_1+\varphi(\beta_2)x_2 + \cdots+\varphi(\beta_n)x_n=0$$ where \(\varphi \in H\)

Since there are \(m\) equations and \(n > m\) unknowns, this system has a non-trivial solution \((x_1, x_2, \cdots, x_n) \in K^n\)

Note: Fix \(\psi \in H\) and let \(\varphi \in H\) be arbitrary. Then:
$$\varphi(\beta_1)\psi(x_1) + \cdots \varphi(\beta_1)\psi(x_n)$$ $$=\psi(\underbrace{\psi^{-1}\circ \varphi}_{\in H}(\beta_1)x_1 + \cdots+\underbrace{\psi^{-1}\circ \varphi(\beta_n)}_{\in H}x_n)$$ $$=\psi(0)=0$$

Let \((x_1, \cdots, x_n) \in K^n\) be a non-trivial solution with a minimal amount of non-zero entries. By reordering we may assume:
$$(x_1, \cdots, x_n)=(\underbrace{x_1, \cdots, x_r}_{\neq 0}, 0, 0, \cdots, 0)$$

Note: If \(r=1\), \(\underbrace{\varphi(\beta_1)}_{\neq 0}x_1 = 0 \implies x_1=0\). Contradiction, \(\tf r > 1\)

Since \((1, \frac{x_2}{x_1}, \cdots, \frac{x_r}{x_1}, 0, \cdots, 0)\) is a solution, we may assume \(x_1=1\).

At this point, our minimal, non-trivial solution is \((1, x_2, \cdots, x_r, 0, \cdots, 0)\)

Subclaim \(x_2, \cdots, x_r \in F = Fix(H)\).
Suppose not. Then, WLOG, say \(\psi \in H\) such that \(\psi (x_2) \neq x_2\).
We have the following solutions to our system:
\((1, x_2, \cdots, x_r, 0, \cdots, 0)\)
\((1, \psi(x_2), \cdots, \psi(x_r), 0, \cdots, 0)\)
Subtracting the two above:
\((0, \underbrace{x_2-\psi(x_2)}_{\neq 0}, \cdots, x_r-\psi(x_r), 0, \cdots, 0)\)
This contradicts minimality (of number of zeros)
\(\tf x_2, x_3, \cdots, x_r \in F\)

For \(\varphi=1 \in H\), we have:
\(\beta_1+\beta_2x_2 + \cdots+\beta_rx_r = 0\)
\(\implies \beta_1 + x_2\beta_2 + \cdots + x_r \beta_r=0\)
\(\tf \{\beta_1, \cdots, \beta_n\}\) is \(F\) linearly dependent.
\(\qed\) Artin’s Theorem Proved

Notation: \(K/F\)
\(\mathcal{E} = \{E: K/E/F \text{ tower }\}\)
\(\mathcal{H} = \{H : H \leq Gal(K/F\}\)

Galois Correspondences

\(Gal(K/\cdot): \E \to \H : E \mapsto Gal(K/E)\)
\(Fix: \H \to \E : H \mapsto Fix H\)


  1. \(E_1, E_2 \in \E\), \(E_1 \sub E_2\).
    \(\implies Gal(K/E_2) \sub Gal(K/E_1)\)
  2. \(H_1, H_2 \in \H\), \(H_1 \sub H_2\)
    \(\implies FixH_2 \sub Fix H_1\)

i.e. the Galois correspondences are inclusion-reversing.

Theorem [Fundamental Theorem of Galois Theory]
Let \(K/F\) be a finite Galois extension.

  1. For \(E \in \E\), \(Fix Gal(K/E)=E\), \(|Gal(K/E)|=[K:E]\)
    -> fix is a left inverse of Gal
  2. For \(H \in \H\),
    \(Gal(K/Fix H)=H\), \([K:FixH]=H\)
    i.e. the Galois correspondences are inverses of each other.

Big Picture:
![[West LA - 1150 AM 2.png]]

Proof of Theorem

By A7, \(K/E\) is Galois.
\(\tf FixGal(K/E)=E\) (characterization theorem)
\(|Gal(K/E)| = [K:E]\)

(2) Follows from Artin

Corollary \(K/F\) finite Galois
If \(H_1 \sub H_2\) are in \(\H\), then \([H_2:H_1] = [Fix H_1 : H_2]\)
if \(E_1 \sub E_2\) are in \(\E\), then
\([E_2:E_1] = [Gal(K/E_1):Gal(K/E_2)]\)

\([FixH_1:FixH_2] = \frac{[K:FixH_2]}{[K:FixH_1]} \stackrel{Artin}{=} \frac{|Gal(K/FixH_2)|}{|Gal(K/FixH_1)|} \stackrel{Artin}{=} \frac{H_2}{H_1} = [H_2:H_1]\)

$$|Gal(K/E_1)|/|Gal(K/E_2)| \stackrel{FT}{=} \frac{[K:E_1]}{[K:E_2]} = [E_2:E_1]$$ \(\qed\)

Example \(K= \text{ s.f. } f(x)=x^3-2\) over \(\Q\)
\(K = \Q(\alpha, \zeta_3), \alpha = \sqrt[3]{2}\)

Since \(f(x)\) is irreducible (2-Eis) and \(\Q\) is perfect, \(K/\Q\) is Galois and \(Gal(K/\Q) \leq S_3\), Since \(|Gal(K/\Q)| = [K:\Q]=6\), \(Gal(K/\Q)=S_3\)

![[West LA - 1150 AM 3.png]]

Corollary \(K/F\) finite, Galois
Then there are finitely many fields \(E\) such that \(F \sub E \sub K\)
Why? \(Gal(K/F)\) has finitely many subgroups.

Investigation \(K/E/F\), \(\varphi \in Gal(K/F)\)
what does it mean for \(\psi \in Gal(K/\varphi(E))\)
\(\iff \forall a \in E, \psi(\varphi(a)) = \varphi(a)\) and of course, \(\psi\) is a \(K\)-automorphism.

\(\iff \forall a \in E, (\varphi^{-1} \circ \psi \circ \varphi)(a)=a\)

\(\iff \varphi^{-1} \circ \psi \circ \varphi \in Gal(K/E)\)
\(\iff \psi \in \varphi Gal(K/E) \varphi^{-1}\)
–> \(g H g^{-1}\) moment

  1. \(Gal(K/\varphi(E)) = \varphi Gal(K/E) \varphi^{-1}\)
  2. \(Gal(K/E) \tleq Gal(K/F) \iff \forall \varphi \in Gal(K/F), \varphi(E)=E\)

Theorem \(K/F\) finite, Galois, \(K/E/F\), TFAE:

  1. \(E/F\) is Galois
  2. \(E/F\) Normal
  3. \(Gal(K/E) \tleq Gal(K/F)\)

By A7, \(E/F\) is separable. \(\tf (1)\) and \((2)\) are equivalent.

Assume \(E/F\) is normal. Let \(\varphi \in Gal(K/F)\). We must show that \(\varphi(E)=E\)

By the Normality Theorem, \(\varphi |_E \in Gal(E/F)\)

\(\tf \varphi(E)=E\)

\((\impliedby)\), assume for all \(\varphi \in Gal(K/F)\), \(\varphi(E)=E\).

Let \(\alpha \in E\) and let \(p(x)\) be its minimal polynomial over \(F\). Assume \(\beta\) is a root of \(p(x)\) in \(K\)

There exists \(\varphi \in Gal(K/F)\) such that \(\varphi(\alpha)=\beta\)
–> by the extension lemma.

\(\tf \beta = \varphi(\alpha) \in \varphi(E)=E\),

and \(p(x)\) splits over \(E\). Hence, \(E/F\) is normal.

Proposition \(K/F\) finite, Galois, \(K/E/F\), \(E/F\) Galois

Then, \(Gal(K/F) / Gal(K/E) \iso Gal(E/F)\)

\(\psi : Gal(K/F) \to Gal(E/F)\)
\(\psi(\varphi)=\varphi |_E\) (Normality Theorem)

\(\ker \psi = Gal(K/E)\)

$$|Gal(K/F)/Gal(K/E)| = \frac{[K:F]}{[K:E]} \stackrel{Tower Thm}{=} [E:F] = |Gal(E/F)|$$

We conclude by computing two famous Galois Groups::

Example \(K=\Q(\zeta_n)\), \(F=\Q\)

Since \(K\) is the splitting field of the separable polynomial \(\phi_n(x)\), \(K/\Q\) is Galois.

Consider $\psi: \Z_n^\times \to Gal(K/\Q) :
\psi(k)= \varphi_k, \varphi_k(\zeta_n)=\zeta_n^k$

  1. \(\varphi_{ab}(\zeta_n) = \zeta_n^{ab}=(\zeta_n^b)^a = \varphi_a(\varphi_b(\zeta_n)) \implies \varphi_{ab}=\varphi_a \circ \varphi_b\)
    \(\implies \psi(ab)=\psi(a) \circ \psi(b)\)
  2. \(\ker \psi = \{1\}\)
  3. \(|\Z_n^\times|=|Gal(K/\Q)|=\phi(n)\)

Example \(K=\F_{p^n}, F=\F_p\)
Since \(K\) is the splitting field of the separable polynomial \(x^{p^n}-x\), \(K/F\) is Galois.

Consider the Frobenius automorphism: \(\varphi: \F_{p^n} \to \F_{p^n} : \varphi(a)=a^p\)

Note: \(\varphi \in Gal(K/F)\)

Let \(j = |\varphi|\). \(|Gal(K/F)|=[K:F]=n\) So \(\tf j \leq n\)
For all \(x \in \F_{p^n}\), \(\varphi^j(x)=x\)
\(\iff x^{p^j}=x \iff x^{p^j}-x=0\)
\(\tf p^n \leq p^j \implies n \leq j\), \(\implies n=j\)

\(\tf Gal(K/F) = <\varphi> \iso \Z_n\)

Chapter 10 - Galois Groups of Polynomials

\(f(x) \in F[x]\) irred, sep. If \(G = Gal(f(x))\):

  1. \(G\) is a transitive subgroup of \(S_n\), \(n=\deg f(x)\).
  2. \(n \mid |G|\) (orbit-stabilizer, every orbit will have size \(n\))
    In particular, if \(n=2\), \(G = S_2 \iso \Z_2\)

Definition \(f(x) \in F[x]\) monic, non-constant.
\(K\) = splitting field of \(f(x)\) over \(F\).
\(f(x)=(x-\alpha_1)\cdots(x-\alpha_n) \in K[x]\)
The discriminant of \(f(x)\) is:
$$disc(f(x))=\Pi_{i < j} (\alpha_i-\alpha_j)^2$$


  1. \(disc(f(x))=0\) \(\iff\) \(f(x)\) is NOT separable
  2. \(f(x)\) separable. \(\forall \varphi \in Gal(f(x)) = Gal(K/F)\)
    \(\varphi(disc(f(x)))=disc(f(x))\) (just changes up the ordering) (regardless of sep)
    \(\implies disc(f(x)) \in FixGal(K/F)=F\)
  3. \(f(x)=x^2+bx+c = (x-\alpha_1)(x-\alpha_2)\)
    \(disc(f(x))=(\alpha_1-\alpha_2)^2 = \alpha_1^2+\alpha_2^2-2\alpha_1\alpha_2 = b^2-4c\)

\(char(F)\neq2\). \(f(x) \in F[x]\) separable. \(K\) = splitting field of \(f(x)\). \(f(x)=(x-\alpha_1) \cdots (x-\alpha_n) \sub K[x]\) distinct roots.
Let \(d = \Pi_{i < j} (\alpha_i-\alpha_j)\) so that \(d^2=disc(f(x))\)
Let \(G = Gal(f(x))=Gal(K/F)\)
For all \(\varphi \in G\), \(\varphi(d)^2=d^2 \implies \varphi(d)\) root of \(x^2-d^2 \in F[x]\)
\(\implies \varphi(d)=\pm d\). \(char 2\), don’t want 1=-1

Fact: \(\varphi(d)=d \iff \varphi \in A_n\)

\(\tf\) TFAE:

  1. \(G \sub A_n\)
  2. \(\forall \varphi \in G, \varphi(d)=d\)
  3. \(d \in F\)
  4. \(discf(x)\) is a square in \(F\).

\(f(x) \in F[x]\) monic irred., sep. \(\deg f(x)=3\)
\(f(x)=x^3+\alpha x^2+\beta x + \gamma\)
Assume \(charF \neq 2, 3\)
\(=\underbrace{x^3+ bx+c}_{\textbf{depressed cubic}}\)

Important Note
\(g(t)=0 \iff t=s+\frac{\alpha}{3}\), \(f(s)=0\)
\(\tf Gal(f(x))=Gal (g(x))\)

WLOG, \(f(x)=x^3+bx+c\)

Fact: \(disc(f(x))=-4b^3-27c^2\).
Let \(G=Gal(f(x))\). Then, \(G \leq S_3\) and \(3 \mid |G|\) \(\implies G=A_3\) or \(S_3\)

From before, \(Gal(f(x))=G=\)
\(\begin{cases} A_3 & \text{ if `\(disc(f(x)\)` square in `\(F\)`} \\ S_3 & \text{ otherwise } \end{cases}\)

\(f(x)=x^3-3x+1 \in \Q[x]\). \(f(x)\) is irreducible by Mod-2 test.
Since \(\Q\) is perfect, \(f(x)\) is separable. \(disc(f(x))=-4(-3)^2-27(1)^2=4(27)-27=27(4-1)=27(3)=3^4=9^2\)

Let \(f(x)\in F[x]\) be a separable, irreducible, monic, quartic.
$$f(x)=x^4+\alpha x^3+\beta x^2+\gamma x + \delta$$

Assume \(Char(F)\neq2\). By making the substitution \(x \mapsto x-\frac{\alpha}{4}\), we may assume that
\(\underbrace{f(x)=x^4+bx^2+cx+d}_{\textbf{depressed quartic}}\)

We know \(G=Gal(f(x))\) is a transitive subgroup of \(S_4\) with \(4 \mid |G|\).

The options are: \(S_4, A_4, D_4, V\iso \Z_2 \times \Z_2, \Z_4\)

\(V = \{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)\}\)
Let \(u = \alpha_1 \alpha_2 + \alpha_3 \alpha_4\), \(v=\alpha_1 \alpha_3+\alpha_2 \alpha_4\), \(w=\alpha_1\alpha_4+\alpha_2\alpha_3\)
where \(\alpha_1, \alpha_2, \alpha_3, \alpha_4\) are the roots of \(f(x)\)

Let \(K\) be the splitting field of \(f(x)\) over \(F\). i.e. \(K=F(\alpha_1, \alpha_2, \alpha_3, \alpha_4)\)
and let \(L=F(u, v, w)\).

\(\tf F \sub L \sub K\)

  1. \(Gal(K/F)=Gal(f(x)), K/F\) galois.
  2. \(L/F\) Galois….
    \(Res f(x) := (x-u)(x-v)(x-w) = x^3-bx^2-4dx+4bd-c^2 \in F[x]\)
    This is called the resolvent cubic
  3. \(Gal(Res(f(x))=Gal(L/F) \iso Gal(K/F) / Gal(K/L)=G/(G \cap V)\)
    Let \(m=|Gal(Res(f(x))| = \frac{|G|}{|G \cap V|}\)

#### Insert table
For \(m \in \{1, 3, 6\}\), \(G\) is uniquely determined.

Assume \(m=2\), i.e. \(Gal (Resf(x))\iso \Z_2\). Say \(u \in F\), \(v, w \notin F\).

Note: \(L=F(v, w)\). Moreover, \(G \iso D_4\) or \(\Z_4\)
Both \(D_4\) and \(\Z_4\) contain a 4-cycle which fix
$$u=\alpha_1\alpha_2+\alpha_3\alpha_4$$ Why? \(F=FixGal(K/F)=Fix G\)

\(\tf \sigma = (1 3 2 4) \in G\)
\(\implies \sigma^2 = (1 2)(3 4) \in G\)

Consider :

  1. \(x^2-ux+d \in F[x] = (x-\alpha_1\alpha_2)(x-\alpha_3\alpha_4)\)
  2. \(x^2+(b-u) \in F[x] = (x-(\alpha_1+\alpha_2))(x-(\alpha_3+\alpha_4))\)

\(G = <\sigma>\iso \Z_4\) \(\iff\) (1) and (2) split over \(L\).

\((\implies)\) Suppose \(Gal(f(x)) = <\sigma>\). Then \(Gal(K/L)=<\sigma> \cap V = <\sigma^2>\)
But \(\alpha_1 \alpha_2, \alpha_3 \alpha_4, \alpha_1+\alpha_2, \alpha_3+\alpha_4 \in Fix<\sigma^2>=Fix(K/L)=L\)

(\(\impliedby\)) Suppose \(\alpha_1 \alpha_2, \alpha_3 \alpha_4, \alpha_1+\alpha_2, \alpha_3+\alpha_4 \in L\)
Since \(\alpha_1, \alpha_2 \in L(\alpha_1)\)
and \((\alpha_1-\alpha_2)(\alpha_3-\alpha_4)=v-w \in L\)
\(\implies \alpha_3-\alpha_4 \in L(\alpha_1) \implies \alpha_3, \alpha_4 \in L(\alpha_1)\), \(charF\neq 2\)

\(\tf K=F(\alpha_1, \cdots, \alpha_4)=L(\alpha_1)\)

Since \(\alpha_1\) is a root of \(x^2-(\alpha_1+\alpha_2)x+\alpha_1\alpha_2 \in L[x]\)

\([K:L]=[L(\alpha_1):L] \leq 2\)

However, \([L:F]=m=2\)
\(\implies [K:F] \leq 4\)

But, \([K:F]=|G| \geq 4\)
\(\implies [K:F]=4 \implies |G|=4\) \(\implies G \iso \Z_4\) \(\qed\)

Recall \(char F \neq 2\)
\(f(x)=x^4+bx^2+cx+d\), irreducible separable
\(Res f(x)=x^3-bx^2-4dx+4bd-c^2\)
\(G=Gal(f(x))\), \(m = |Gal Res f(x)|\)

\[\begin{array}{c | c c c c c} G & S_4 & A_4 & D_4 & V & \Z_4 \\ \hline G \cap V & V & V & V & V & \Z_2 \\ \hline m & 6 & 3 & 2 & 1 & 2 \end{array}\]

if \(m=2\), and \(u \in F\) (root of \(Res\)), then \(G=\Z_4\) \(\iff\)
(1) \(x^2-ux+d\), (2) \(x^2+(b-u)\) split over \(L=sf Res f(x)\)

\(f(x)=x^4-2x-2 \in \Q[x]\). By 2-Eis, \(f(x)\) is irreducible. Since \(\Q\) is perfect, \(f(x)\) separable.

\([b=0, c=-2, d=-2]\). \(Resf(x)=x^3+8x-4\)
[HW: No rational roots \(\implies\) irreducible ]

\(disc(Res(f(x))=-4(8)^3-27(-4)^2 < 0\), therefore \(discResf(x)\) is not a square in \(\Q\), and so \(Gal(Res(f(x))) = S_3\)

\(\tf m=6\) so \(Gal(f(x))=S_4\)

Example \(f(x)=x^4+5x+5 \in \Q[x]\)
by \(5\)-Eis, \(f(x)\) is irreducible. Since \(\Q\) is perfect, \(f(x)\) is separable.
[\(b=0, c=5, d=5\)]
Note \(5\) is a root, through poly long division,
\(Res f(x)=(x-5)(\underbrace{x^2+5x+5}_{\text{irred, 5-eis}})\)
\(\tf m=2\), so let \(u=5\).
roots of quadratic: $$\frac{-5 \pm \sqrt{25-20}}{2} = \frac{-5\pm \sqrt{5}}{2}$$ if \(L\) is the spliting field of \(Res f(x)\), \(L=\Q(\sqrt{5})\)

Then, we consider:

  1. \(x^2-5x+5\), roots: \(\frac{5\pm\sqrt{5}}{2} \in L\)
  2. \(x^2-5\), roots are: \(\pm \sqrt{5} \in L\)

\(\tf\) (1) and (2) split over \(L\), and so \(G=\Z_4\). //

Chapter 11 - Solvability by Radicals


Definition[Solvable Groups]
A group \(G\) is solvable if there exists:
$$\{e\} = H_0 \tleq H_1 \tleq H_2 \tleq \cdots \tleq H_n=G$$ such that \(H_{i+1}/H_{i}\) is abelian.

Example \(\{e\} \tleq <r> \tleq D_4\) solvable

Example \(S_4 \tgeq A_4 \tgeq V \tgeq \{e\}\)

Remark if \(G\) is simple then \(G\) is solvable \(\iff\) \(G\) is abelian (\(G \iso \Z_p)\)

Example \(A_5\) is not solvable –> simple, non-abelian.

If \(G\) is solvable and \(H \leq G\), then \(H\) is solvable.

Proposition\(G\) solvable, \(N \tleq G\). Then \(G/N\) is solvable.
Why? \(\{e\} = H_0 \tleq H_1 \tleq \cdots \tleq H_n=G\)
\(\bar{\{e\}} = \bar{H_0} \tleq \bar{H_1} \tleq \cdots \tleq \bar{H_n}=\bar{G}\)
\(\bar{H_i}=H_i N / N\)

[3rd Iso Theorem] \(\overline{H_{i+1}}/\overline{H_i} \iso H_{i+1}/H_i\) abelian.

Proposition \(N \tleq G\)
Then, \(G\) is solvable \(\iff\) \(N\) and \(G/N\) are solvable.

\(\implies\) Done
\(\impliedby\) \(\overline{\{e\}}=\bar{H_0} = ..\)
\(\overline{H_i}=H_i/N\), \(N \sub H_i\), \(\overline{H_{i+1}}/\overline{H_i}\) abelian

Note: \(H_0 = N\)
$$\{e\} = K_0 \tleq K_1 \tleq \cdots \tleq K_m=N=H_0$$ $$\tleq H_1 \tleq H_2 \cdots \tleq G$$ Note:
\(H_{i+1}/H_{i}\iso\overline{H_{i+1}}/\overline{H_i}\) [3rd Iso Theorem]

\(K_{i+1}/K_i\) abelian by solvability of \(N\), \(H_0=k_m\)

\(G\) solvable \(\iff\) \(N, G/N\) solvable.

Example \(|G|=p^n\)

\(Z(G) \neq \{e\}\) solvable. b/c abelian groups solvable
\(G / Z(G)\) solvable (induction)
\(\implies G\) solvable.

\(A, B, C\) groups, \(A \tleq B \tleq C\), \(\underbrace{C/A \text{ abelian}}_{A \tleq C}\).

Let \(b_1, b_2 \in B \sub C\). \(\tf\) \((b_1A)(b_2A)=(b_2A)(b_1A)\)
\(\implies B/A\) abelian.

Take \(c_1, c_2 \in C\), we investigate Do \(c_1 B, c_2 B\) commute?

We know \((c_1A)(c_2 A)=(c_2 A)(c_1 A)\) \(\implies\) \(c_1c_2A=c_2c_1A\) \(\implies\) \(c_1^{-1}c_2^{-1}c_1c_2A=A\)
\(\implies c_1^{-1}c_2^{-1}c_1c_2 \in A \sub B\)

\(\implies (c_1B)(c_2B)=(c_2B)(c_1B)\) \(\implies C/B\) is abelian.

Suppose \(A \tleq C\) and there does not exist \(B\) such that \(A \tleq B \tleq C\).
–> no \(B\) in between \(A\) and \(C\)
and assume \(C/A\) abelian. \(\tf\) \(C/A\) simple, abelian.
\(\implies C/A \iso \Z_p\)

Suppose \(G\) is finite and solvable. By refining the chain as much as possible:
$$\{e\} =H_0 \tleq H_1 \tleq \cdots \tleq H_m=G$$ \(H_{i+1}/H_i\) cyclic, prime order.

Solvability by Radicals

Idea: Solving a polynomial by radicals means (informally) expressing its roots using arithmetic + radicals (nth roots).
In 1824, Abel proved that \(f(x)\) is solvable by radicals when \(\deg f(x) \leq 4\) and \(char \neq 2, 3\). He proved there exists quintics are not solvable by radicals.

Big assumption from now on: All fields have \(char=0\)

Definition [Simple Radical Extensions] We say \(K/F\) is a simple radical extension iff \(\exists \alpha \in K\), \(\exists n \in \N\) such that \(K=F(\alpha), \alpha^n \in F\)
–> the one thing you adjoined is an n-th root of something in the base field

Definition [Radical Tower] A radical tower over \(F\) is a tower of fields \(K_m / K_{m-1} \cdots / K_1/K_0=F\) such that \(K_{i+1}/K_i\) simple radical.

Definition [Radical Extension] We say \(K/F\) is radical if there exists a radical tower from \(F\) to \(K\) (i.e. \(K_m=K\))

Definition [Solvable by Radicals] We say \(f(x) \in F[x]\) is solvable by radicals if its splitting field is contained in a radical extension of \(F\).

Example \(K=\Q(\sqrt[3]{2}, \zeta_8)\). Clearly,
\(K \sup \Q(\sqrt[3]{2}) \sup \Q\)
So \(K/\Q\) is radical.

Example \(\Q(\sqrt{2+\sqrt{2}}) \sup \Q(\sqrt{2}) \sup \Q\)

Definition [Cyclic Extensions] We say \(K/F\) is cyclic iff \(K/F\) finite, Galois and \(Gal(K/F)\) is cyclic.

Proposition Suppose \(F\) contains a primitive nth root of unity, \(\zeta\).

If \(K=F(\alpha), \alpha^n \in F\), then \(K/F\) is cyclic.

The roots of \(f(x)=(x^n-\alpha^n)\) are \(\alpha, \zeta \alpha, \cdots, \zeta^{n-1}\alpha \in K\)

\(\tf\) \(K\) is the splitting field of the separable polynomial \(f(x)\). Hence, \(K/F\) is Galois,

For all \(\varphi \in Gal(K/F)\), there exists a unique \(0 \leq i \leq n-1\) such that \(\varphi(\alpha)=\zeta^i \alpha\)

Consider \(\psi: Gal(K/F) \to \Z_n\) given by \(\psi(\varphi)=i\) as above.

Claim \(\psi\) is an injective group homomorphism.

Take \(\varphi_1, \varphi_2 \in Gal(K/F)\) such that \(\varphi_1(\alpha)=\zeta^i \alpha\) and \(\varphi_2(\alpha)=\zeta^j \alpha\)

$\tf (\varphi_1 \circ \varphi_2)(\alpha) = \varphi_1(\zeta^j \alpha) = \zeta^j \varphi(\alpha)
\(\zeta^{i+j} \alpha \implies \psi(\varphi_1 \circ \varphi_2) = i+j=\psi(\alpha_1)+\psi(\alpha_2)\)

Now, \(\varphi \in \ker \varphi\) \(\iff \psi(\varphi)=0 \iff \varphi(\alpha)=\alpha \iff \varphi = \text{ id }\)

\(\tf \psi\) is injective. Hence, \(Gal(K/F)\) is isomorphic to a subgroup of \(Z_n\) and so is cyclic.

Definition \(\{\sigma_1, \sigma_2, \cdots, \sigma_n\} \in Aut(K)\)
We say \(\{\sigma_1, \sigma_2, \cdots, \sigma_n\}\) is linearly independent over \(K\) iff \(a_1 \sigma_1 + \cdots + a_n \sigma_n = 0 \implies a_1 = a_2 = \cdots = a_n = 0, (a_i \in K)\)

Lemma \([K:F]<\infty\), then \(G=Gal(K/F)\) is linearly independent over \(K\).

Let \(\{\sigma_1, \cdots, \sigma_n\} \sub G\) be a minimal linearly dependent set. –> if you threw out a \(\sigma_i\), we get a lin ind set.

This means \(\exists a_i \in K^\times\) such that \(a_1 \sigma_1 + \cdots + a_n \sigma_n = 0\)
–> by the minimality assumption, if u had \(a_1=0\), you could have thrown out \(a_1\).

Since \(a_1 \neq 0\) and \(\sigma_1 \neq 0\), \(n > 1\).

Since \(n \geq 2\), \(\exists \beta \in K\) such that \(\sigma_1(\beta)\neq\sigma_2(\beta)\)

For all \(\alpha \in K\),

  1. \(a_1\sigma_1(\alpha)\sigma_1(\beta)+a_2\sigma_2(\alpha)\sigma_2(\beta) +\cdots+a_n\sigma_n(\alpha)\sigma_n(\beta)=0\)
  2. \(a_1\sigma_1(\alpha)\sigma_1(\beta)+a_2\sigma_2(\alpha)\sigma_1(\beta)+\cdots+\sigma_n(\alpha)\sigma_1(\beta)=0\)
    subtract (1) and (2)
    \beta))}_{\neq 0}\sigma_2(\alpha)+\cdots+a_n(\sigma_n(\beta)-\sigma_1(\beta))\sigma_n(\alpha)]=0$

hence \(\{\sigma_2, \cdots, \sigma_n\}\) linearly dependent. Contradiction, \(\qed\)

Proposition Assume \(F\) contains a primitive \(n\)th root of unity, and \(K/F\) is cyclic of degree \(n\). Then, \(K\) is a simple radical extension of \(F\).

Proof Let \(G=Gal(K/F)\), so that \(|G|=[K:F]=n\). Say \(G=<\sigma>\)

For any \(\alpha \in K^\times\), let \(g(\alpha)=\alpha+\zeta \sigma(\alpha)+\zeta^2\sigma^2(\alpha)+\cdots+\zeta^{n-1}\sigma^{n-1}(\alpha)\)
where \(\zeta \in F\) is a primitive \(n\)th root of unity.


  1. Since \(G\) is LI over \(K\), \(\forall \alpha \neq 0\), \(g(\alpha)=0\)
  2. \(\sigma(g(\alpha))=\sigma(\alpha)+\zeta\sigma^2(\alpha)+\cdots+\zeta^{n-1}\alpha=\zeta^{-1}g(\alpha)\)
  3. \(\sigma(g(\alpha)^n)=\sigma(g(\alpha))^n=[\zeta^{-1}g(\alpha)]^n=g(\alpha)^n\)
    \(\tf g(\alpha) \notin F\) and \(g(\alpha)^n \in F\), \(F=FixG\)

Fix \(\alpha \in K^\times\).

For \(1 \leq i \leq n-1\), \(\sigma^i(g(\alpha))=\underbrace{\zeta^{-i}}_{\neq 1}g(\alpha) \neq g(\alpha)\)

If \(\{1\} \neq H \leq G\), then \(g(\alpha) \notin FixH\). (Why? \(H=<\sigma^i>\))

\(\tf F \subset E \subseteq K\) and \(g(\alpha) \in E\), then \(E=K\).

\(K=F(g(\alpha))\) and \(g(\alpha)^n \in F\), \(\qed\)

Remark \(F\) field.
\(W_n = \underbrace{\{z \in \bar{F}: z^n=1 \}}_{\text{ finite }} \leq \bar{F}^\times\)
From before, \(W_n\) is cyclic.

We say \(\alpha \in \bar{F}^\times\) is a primitive \(n\)th root of unity iff \(W_n=<\alpha>\)

Let \(\phi_n(x)=\Pi_{\text {prim nth root } \alpha}(x-\alpha) \in \)bar{F}\([x]\)

For a primitive \(n\)th ROU, \(\alpha\), \(F(\alpha)\) is the s.f. of \(x^n-1\)

Hence, \(F(\alpha)/F\) is normal = Galois.

\(\phi_n(x) \in FixGal(F(\alpha)/F)[x] = F[x]\)

\([K:F] < \infty\), \(K/E/F\)
\(K/E\) simple radical, \(E/F\) Galois. There exists \(L/K\) such that \(L/F\) Galois and \(L/E\) is radical.

Moreover, \(Gal(L/E)\) is solvable.

Suppose \(K=E(\alpha)\) where \(\alpha^n = \beta \in E\), and \(G=Gal(E/F)=\{\sigma_1, \sigma_2, \cdots, \sigma_r\}\)

Consider \(f(x)= \phi_n(x) \Pi_{i=1}^r (x^n - \sigma_i(\beta))\)

Let \(L\) be the splitting field of \(f(x)\) over \(K\)

Note: \(f(x) \in Fix G[x] = F[x]\) since \(E/F\) Galois.

Claim 1: \(L/F\) Galois.

\(L = K(\text{ roots of f }) = K(\alpha, \text{ other roots}) = E(\alpha) \text{ other roots} = E(\text{roots of f})\)
and so \(L\) is the splitting field \(f(x)\) over \(E\).
Since \(E/F\) Galois, \(E\) is the splitting field of some \(h(x) \in F[x]\) over \(F\).

Hence \(L\) is the splitting field of \(f(x)h(x)\) over \(F\)

Since \(char(F)=0\), \(L/F\) Galois. //

Claim 2: \(L/E\) is radical. Let \(\zeta\) be any root of \(\phi_n(x)\) in \(L\).
By the extension lemma, extend each \(\sigma_i\) to \(\sigma_i \in Gal(L/F)\)
Say \(\sigma_1 = \text{ id }\).

Since \(\sigma_i(\alpha)^n = \sigma_i(\beta)\), \(\sigma_i(\alpha)\) is a root of \(f(x)\)

\(\implies \sigma_i(\alpha)=\zeta^j \text{ or } \zeta^j \sigma_l(\alpha)\)

Therefore \(E \sub \E(\zeta) \sub E(\zeta, \sigma_1(\alpha)) \sub E(\zeta, \sigma_1(\alpha), \sigma_2(\alpha)) \sub \cdots\)
$\sub E(\zeta, \sigma_1(\alpha), \sigma_2(\alpha), \cdots, \sigma_r(

\(\implies\) \(L/E\) radical.

Claim 3: \(Gal(L/E)\) solvable.

Let \(G_i = Gal(L/E_i)\) where \(E_0 = E(\zeta)\) and \(E_r=E(\zeta, \sigma_1(\alpha), \cdots, \sigma_r(\alpha))\)

\(\implies \{1\} \leq G_r \leq G_{r-1} \leq \cdots G_2 \leq G_1 \leq G_0 \leq \underbrace{Gal(L/E)}_{G'}\)

  1. We have \(E(\zeta)/E\) is Galois (s.f. of \(\phi_n(x)\)) and \(Gal(E(\zeta)/E) \iso \Z_n^\times\)
    \(\underbrace{Gal(L/E(\zeta))}_{G_0} \tleq \underbrace{Gal(L/E)}_{G'}\)
    and \(Gal(L/E) / Gal(L/E(\zeta)) = G'/G_0 \iso \Z_n^\times \text{ abelian }\)

  2. We have \(E_{i+1} = E_i(\sigma_i(\alpha))\), \(\zeta \in E_i\)

    and so \(E_{i+1}/E\) is simple radical, and hence cyclic. (prev propositions)

    \(Gal(L/E_{i+1}) \tleq Gal(L/E_i)\)
    \(G_{i+1} \tleq G_i\) , and \(G_i/G_{i+1}\) cyclic.

    Hence \(Gal(L/E)\) is solvable. \(\qed\)

Proposition [Best of Both Worlds] \([K:F] < \infty, K/E/F\)
If \(K/E\) is simple radical, \(E/F\) Galois, then \(\exists L/K\) such that \(L/E\) radical, \(L/F\) Galois, \(Gal(L/E)\) solvable.

Inductively, we get the same result when \(K/E\) is radical.

If \(K/F\) is radical, then there exists an extension \(L/K\) such that \(L/F\) is radical and Galois and the \(Gal(L/F)\) is solvable.
Proof \(E=F\). \(\qed\)

Theorem [Galois’ Theorem]
Let \(f(x) \in F[x]\) be non-constant. Then \(f(x)\) is solvable by radicals over \(F\) iff \(Gal(f(x))\) is solvable.

\((\implies)\) By deleting repeated irreducible factors, we may assume \(f(x)\) is separable.

Suppose \(f(x)\) is solvable by radicals. Let \(E\) be the splitting field of \(f(x)\) over \(F\) and let \(K/F\) be radical such that \(E \sub K\).

By the corollary, \(\exists \ L/F\) radical and Galois, such that \(Gal(L/F)\) is solvable.

Since \(E/F\) is normal, \(Gal(L/E) \tleq Gal(L/F)\) and
\(Gal(E/F) \iso \underbrace{Gal(L/F) / Gal(L/E)}_{\text{solvable}}\)


\(S_5\) , \(H = <(1 2), (1 2 3 4 5)>\)
\((1 2 3 4 5)(1 2)(5 4 3 2 1) = (2 3) \in H\)
\((1 2 3 4 5)(2 3)(5 4 3 2 1) = (3 4) \in H\)
\((4 5), (5 1) \in H\)

By instead conjugating by powers of \((1 2 3 4 5)\) (e.g. \((1 3 5 2 4)\)):
\(\forall \ \tau\) transposition, \(\tau \in H\)

\(\tf H =S_5\)

In general, if \(p\) is prime, \(\tau \in S_p\) is a transposition, and \(\sigma \in S_p\) is a \(p\)-cycle, then \(<\tau, \sigma> = S_p\)
–> why \(p\) prime? Because otherwise, powers of \(p\)-cycles aren’t \(p\)-cycles.

Proposition Let \(f(x) \in \Q[x]\) be irreducible with prime degree \(p\). If \(f(x)\) contains exactly \(2\) non-real roots, then \(Gal(f(x))=S_p\)

Why? \(H=Gal(f(x))\)

\(|H|=[K:\Q]\), \(K\) = splitting field of \(f(x)\)
\(=[K:\Q(\alpha)]\underbrace{[\Q(\alpha):\Q]}_{p}\), \(f(\alpha)=0\)
\(\implies p \in |H|\)
\(\implies \exists\) p-cycle \(\sigma \in H\)

Consider \(\varphi : \C \to \C\), \(\varphi(z)=\bar{z}\)
By the normality theorem, \(\varphi |_K \in Gal(f(x))\)

By assumption, \(\varphi \sim \tau\), \(\tau\) is a transposition.

\(\tf\) \(\tau, \sigma \in H\),\(\implies H=S_p\)

Example \(f(x)=x^5+2x^3-24x-2 \in \Q[x]\)

\(f(x)\) is irreducible by \(2\)-Eis.

Claim: \(f(x)\) is not solvable by radicals.

\(f(-100) < 0\)
\(f(-1) > 0\)
\(f(1) < 0\)
\(f(100) > 0\)

by the Intermediate Value Theorem, \(f(x)\) has at least 3 real roots.

Let \(\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5\) be the roots of \(f(x)\).
\(\sum \alpha_i = -[x^4]f(x)=0\)
\(\sum \alpha_{i \leq j} \alpha_i \alpha_j = [x^3]f(x) = 2\)

\(\sum \alpha_i^2 = (\sum \alpha_i)^2 - 2 \sum\limits_{i < j} \alpha_i \alpha_j = -4\)

Therefore, not all roots of \(f(x)\) are real.

By the conjugate root theorem (M135), \(f(z)=0, f(\bar{z})=0\), \(f(x)\) has exactly two non-real roots.

Hence, \(Gal(f(x))=S_5\). Since \(S_5\) is not solvable (contains \(A_5 \leq S_5\) ), \(f(x)\) is not solvable by radicals.

Bonus Material - The Fundamental Theorem of Algebra

Theorem [Fundamental Theorem of Algebra]
\(\C\) is algebraically closed.

Suppose there exists \(L/\C\) algebraic, and \(\alpha \in L \setminus \C\). Since \(\C/\R\) is algebraic, \(L/\R\) is algebraic.

Let \(f(x)\) be the minimal polynomial of \(\alpha\) over \(\R\).

Consider the splitting field \(K\) of \(f(x)\) over \(\C\).

Then \(K\) is the splitting field of \(f(x)(x^2+1)\) over \(\R\).

\(\implies K/\R\) is Galois. Let \(G = Gal(K/\R)\), and say \(|G|=2^{j} m\) where \(2 \nmid m\).

Since \([C:\R]=2 |G|\), \(j \geq 1\).

For a Sylow-2 subgroup of \(G\), call it \(H\), consider \(E=Fix H\).

We know: \([K:E] = |H|=2^j\). \(\implies [E:\R]=m\)

For \(\beta \in E\), \(\deg (\beta) \mid m\), and so by Calculus, \(\implies \deg(\beta)=1\)
–> because odd degree polynomials always have a real root

\(\tf\) \(m=1\), \(\implies |G|=2^j\)

Let \(G' = Gal(K/\C)\) so that \(G' \leq G\).

So by good ol’ Lagrange’s Theorem, \(|G'|=2^{\l}\)

Consider \(N \leq G'\) with \(|N|=2^{l-1}\)

Since \([G':N]=2\), \(N \tleq G'\). For \(E'=Fix N\), \([K:E']=|N|=2^{l-1}\)

\(\tf [E':\C]=2\)

This contradicts the quadratic formula. All degree 2 elements of \(E'\) can’t have a minimal polynomial (b/c there aint no irreducible quadratics). \(\qed\)