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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). $\qed$