\(\newcommand{\bb}[1]{\mathbb{#1}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\F}{\mathbb{F}} \newcommand{\E}{\mathcal{E}} \newcommand{\H}{\mathcal{H}} \newcommand{\l}{\mathcal{l}} \newcommand{\iso}{\cong} \newcommand{\orb}[1]{\text{Orb}({#1})} \newcommand{\stab}[1]{\text{Stab}({#1})} \newcommand{\tleq}{\trianglelefteq} \newcommand{\tgeq}{\trianglerighteq} \newcommand{\tf}{\therefore} \newcommand{\qed}{ // Q.E.D } \newcommand{\sub}{\subseteq} \newcommand{\sup}{\supseteq}\) 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
- Chapter 2 - Irreducibility Criteria
- Chapter 3 - Field Extensions
- Chapter 4 - Splitting Fields
- Chapter 5 - Cyclotomic Extensions
- Chapter 6 - Finite Fields
- Chapter 7 - Galois Groups
- Chapter 8 - Normal + Seperable Extensions
- Chapter 9 - Fundamental Theorem of Galois Theory
- Chapter 10 - Galois Groups of Polynomials
- Chapter 11 - Solvability by Radicals
- Bonus Material - The Fundamental Theorem of Algebra
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.
- We say
$G$is a$p$-group,$p$prime, if$|G| = p^n, n \in \bb{N}$ - We say
$H \leq G$is$p$-subgroup if$H$is a$p$-group. - 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]
- Suppose a finite group
$G$acts on a finite set$X$, i.e.$\bullet: G \times X \to X$where$\forall x, e \cdot x = x$$\forall g, h \in X, \ g(hx) = (gh)x$
- For
$x \in X$,$\orb{X}$ - Orbit Stabilizer Theorem:
$\forall x \in X$
$$|G| = |\stab{X}| \cdot |\orb{X}|$$ - 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. - 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$) $|\orb{X}|=1 \iff \orb{X} = \{x\} \iff \forall g, gxg^{-1} = x \iff Z(G)$- 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$
*Why?**
$|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$.
$n_p \equiv 1 \mod{p}$$n_p \mid m$
Corollary
$n_p = [G : N_G(P)]$ 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
Corollary
$n_p = 1 \iff P \tleq G$, 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
$H_p$Sylow$p$-subgroup,$H_q$Sylow$q$-subgroup,$H_p \cap H_q = \{e\}$$|G| = pm, p \nmid m$, if$H_1 \neq H_2$are Sylow$p$-subgroups, then$H_1 \cap H_2$= {e}- Suppose
$H_p \tleq G$or$H_q \tleq G$,$\tf H_pH_q \leq G$and
$$|H_pH_q| = \frac{|H_p| \cdot |H_q|}{|H_p \cap H_q|} = |H_p| \cdot |H_q|$$ $|H_p \cup H_q| = |H_p| + |H_q| -1$
Exercise $|G| = pq$, $p < q$ are primes where $p \nmid (q-1)$. Prove $G$ is cyclic.
Proof $n_p \equiv 1 \mod{p}$, $n_p \mid q \implies n_p = 1$
Then $n_q \equiv 1 \mod{q}, n_q \mid p \implies n_q = 1$
So $H_p, H_q \tleq G$. We know that $H_p$ and $H_q$ are abelian (prime order). Let $a \in H_p, b \in H_q$, then $aba^{-1}b^{-1} \in H_p \cap H_q = \{e\}, \ \tf ab = ba$. So $H_pH_q$ is abelian! By the Fundamental Theorem of finite abelian groups,
$$H_p H_q \iso \Z_p \cdot \Z_q \equiv \Z_{pq}$$ and therefore cyclic.
Note: $H_p H_q \leq G$ where $|H_pH_q| = pq = |G|$, therefore $H_pH_q = G \ \qed$
Proposition $|G|=30$. Then there exists $H \tleq G$ such that $H \iso \Z_{15}$
Why?
$$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$$
Since $3\mid(5-1)$ from last time $H_3H_5$ is cyclic, $\tf H_3 H_5 \iso \Z_{15}$
Since $[G:H_3H_5]=2, H_3H_5 \tleq G \ \qed$.
Proposition
$|G|= 60$, if $n_5 > 1$, then $G$ is simple. Note $60 = 2^2 \times 3 \times 5$
Why?
$n_5 \equiv 1 \mod 5, n\mid 12 \implies n_5 = 6$
This gives us 6(5-1) = 24 elements of order 5.
Assume $G$ has a proper non-trivial $H \tleq G$.
Case 1: $5 \mid |H|$
Since $H \tleq G$ and $H$ contains a Sylow 5-subgroup of $G$, then $H$ contains ALL Sylow 5
$$\tf \ |H| \mid 60 \text{ and } |H| \geq 24+1$$
$\implies |H| = 30$
We know $\exists \ H_0 \tleq H$ such that $H_0 \iso \bb{Z}_{15}$
Again $H_0$ contains all Sylow 5-subgroups of $G$. Since $H_0$ is abelian, $n_5 = 1$. Contradiction!
Case 2: $5 \nmid |H|$
$|H| \in \{2, 3,4,6, 12\}$
$|H| = 12 = 2^2 \times 3$. HOMEWORK to show$n_2 = 1$or$n_3 = 1$
$\tf \ H$contains a Sylow 2-or-3 subgroup which is normal. Call it$K$.$|K| \in \{3, 4\}$- HOMEWORK, If
$|H|=6$, then$n_3=1$. Let$K \tleq$Sylow 3-subgroup of$H$
Note: by Sylow 2nd theorem, in either case$K$is normal in$G$(see Case 1 Argument)
By replacing$H$with$K$(if necessary), we may assume$|H|\in\{2,3,4\}$
$\bar{G} := G/H$, then$|\bar{G}| \in \{15, 20, 30\}$. HOMEWORK in any case above,$\exists \bar{P} \tleq \bar{G}, |\bar{P}| = 5$
So then $\bar{P} = P/H, \text{ where } H \leq P \tleq G$ (by the correspondence theorem)
$\tf P \tleq G \text{ s.t. } 5 = \frac{|P|}{|H|}$
$\implies 5 \mid |P|$. This contradicts that Case 1 is impossible. Therefore $G$ is simple $\qed$
Corollary: $A_5$ is simple.
Chapter 2 - Irreducibility Criteria
Motivation
$\F$ a field, $p[x] \in \F[x]$. Let $I$ be a non-zero proper ideal of $\F[x]$. Say $I = <p(x)>$.
Then $\F[x]/<p(x)>$ is a field $\iff$ $<p(x)>$ maximal $\iff p(x)$ irreducible.
Recall
$R$ integral domain (“ID”). Then $p(x) \in R[x]$ is irreducible $\iff$
$p(x) \neq 0$$p(x) \notin R[x]^\times = R^\times$- 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 $\iff$
$p(x) \neq 0$$p(x) \notin R^\times$$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 $f(a)$
Why?
$f(x) = (x-a)q(x)+r(x)$ 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.
Why?
$$f(a) = 0$$
$$f(x) =(x-a) q(x) + 0 = (x-a)q(x)$$
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.
Why?
$f(x) \text{ reducible} \iff \text{linear factor } \iff \text{root}$
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 $p$
If:
$\deg f(x) = \deg \bar{f}(x)$and$\bar{f}(x) \in \Z_p[x]$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 $Z_2[x]$
Proof
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:
$a_{n-1}, \cdots, a_1, a_0 \in P$$a_0 \notin P^2$
Then $f(x)$ is irreducible over $R$.
Recall [Pairwise Ideals] $R$ a ring, $I, J \sub R$ ideals
$$IJ = \{\sum a_i b_i : a_i \in I, b_i \in J\}$$
and $IJ$ itself is an ideal.
Proof. Suppose $f(x) = g(x) h(x)$ where $\deg g(x), \deg h(x) < \deg f(x)$.
Then $\bar{f}(x) = \bar{g}(x) \bar{h}(x) = x^n \in (R/P)$
Since $R/P$ is an integral domain ($\star$)
$\bar{g}(0) = \bar{h}(0) = 0, \tf g(0), h(0) \in P$
$\implies a_0 = f(0) = g(0)h(0) \in P^2$. This is a contradiction $\qed$
example $f(x, y) = y^2 + x^2 - 1 \in \Q[x, y]$. Claim: $f(x, y)$ is irreducible.
Consider $g(y) = y^2 + (x^2-1) \in \Q[x][y]$.
Note: $x^2-1 = (x-1)(x+1) \in <x-1>$. Since $x-1$ is irreducible in $\Q[x]$, $P$ is maximal (prime). Also $P^2 = <x-1>^2 = <(x-1)^2>$
Clearly $(x-1)^2 \nmid x^2-1$, so $x^2-1 \notin P^2$ and so $g(f)$ is irreducible by Eisenstein.
Corollary
$$f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in \Z[x]$$
$p \in \N$ prime. If $p$ divides $a_{n-1},\cdots, a_1, a_0$ but $p^2 \nmid a_0$ then $f(x)$ is irreducible over $\Q$.
Proof Let $P = <p>$
Exercises: Are these irreducible over $Q$?
$f(x) = x^7 + 21x^5 + 15x^2 + 9x+6$
Yes irreducible by 3-Eisenstein.$g(x) = x^3+2x+16$
Notice that$\bar{g}(x) = x^3+2x+1 \in \Z_3[x]$, so$\deg g(x) = \deg \bar{g}(x)$
$\bar{g}(0)=1, \bar{g}(1)=1, \bar{g}(2)=1$
Since$\deg \bar{g}(x) = 3$,$\bar{g}(x)$is irreducible. By the Mod-3 irreducibility test,$g(x)$is irred.$p(x) = x^4+5x^3+6x^2-1$
$\bar{p}(x) = x^4+x^3+1 \in \Z_{2}[x]$, with$\bar{p}(0)=1, \bar{p}(1)=1$
check: the only irreducible quadratic is$x^2+x+1$
Moreover$(x^2+x+1)^2 = x^4+x^2+1 \neq \bar{p}(x)$
$\implies$$\bar{p}(x)$is irreducible, and by the Mod-2 irred. test,$p(x)$is irred.$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 $K$.
Remark
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 $\Q(\sqrt{2}, \sqrt{3}) = \Q(\sqrt{2}+\sqrt{3})$
$\sqrt{2}+\sqrt{3} \in \Q(\sqrt{2}, \sqrt{3})$
By minimality,$Q(\sqrt{2}+\sqrt{3}) \sub Q(\sqrt{2}, \sqrt{3})$$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}$
where
$\psi(\bar{g}(x)) = \varphi(g(x)) = g(\alpha)$
is the isomorphism afforded by the FIT.
Investigation
$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\}$
- If
$h(x) \in F[x]$such that$h(\alpha)=0$, then$f(x) | h(x)$ - Let
$h(x) \in F[x]$be irreducible such that$h(\alpha) = 0$
$\tf <f(x)> = < h(x)>$$\implies$ $f(x) = c h(x)$, $c \in F^\times$ - 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)
$\deg_{\Q}(\alpha)=4$
$\tilde{m}(x) = x - \alpha$, $\deg_{\R} (\alpha)=1$
Example $a \in F$, $\deg_F (a)=1$ $(x-a)$
Remark
$K/F$. 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, $|F(\alpha)| = |F|^n$
Why?
$F(\alpha) = \Span_F \{1, \alpha, \cdots, \alpha^{n-1}\}$
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 $|F(\alpha)| = |F|^n$ $\qed$
Note:
$\dim_F F(\alpha) = \deg_F (\alpha)$
$F$ as $\Z_p$ btw for assignment
Proposition $K/F$
If $\alpha, \beta \in K$ have the same min poly over $F$ then $F(\alpha) \iso F(\beta)$
Why?
$F(\alpha) \iso F(\beta) \iso F[x]/<m(x)>$
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)
$\Q(\alpha) \iso\Q(\beta)$$\Q(\alpha) \neq \Q(\beta)$since$\Q(\alpha) \sub \R$and$\Q(\beta) \not\subseteq \R$$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, $[K:F] = [K:E][E:F]$
Proof:
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: $[\Q(\sqrt[3]{5}, i):\Q] = [\Q(\sqrt[3]{5})(i):\Q(\sqrt[3]{5}] \cdot [\Q(\sqrt[3]{5}):\Q]$
$p(x) = \text{ min poly } i \text{ over } \Q(\sqrt[3]{5})$
$q(x) = \text{ min poly } \sqrt[3]{5} \text{ over } \Q$
Note:
$q(x)=x^3-5$ (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 $\alpha$] $K/F$
- We say
$\alpha \in K$is algebraic over$F$iff there exists$0 \neq f(x) \in F[x]$s.t.$f(\alpha)=0$ - We say
$K/F$is algebraic iff$\alpha$is algebraic over$F$for all$\alpha \in K$
Proposition
If $K/F$ is finite, then $K/F$ is algebraic.
Why?
$[K:F] = n < \infty$. 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.
Proposition
$$K/F \text{ finite } \iff [K:F] < \infty$$
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 $\qed$
Proposition
$K/E, E/F$ are algebraic, then $K/F$ algebraic
Proof
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.
Why?
$\alpha, \beta \in L$, $\alpha \neq 0$
$\alpha-\beta, \alpha \beta, \alpha^{-1} \in F(\alpha, \beta)$
$[F(\alpha, \beta):F] < \infty$ and therefore algebraic
Example[Revisited]
$p_1 < p_2 < \cdots$ 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, $f(\bar{t}) = 0$ $\qed$
Remark
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
$K/F$$f(x)$splits over$K$- Whenever
$f(x)$splits over$F \sub L$, then$K \sub L$
Remark
$F, f(x) \in F[x]$ as before. Let $K/F$ be an ext. such that $f(x)$ splits over $K$
$f(x) = u(x-\alpha_1) \cdots u(x-\alpha_n) \in K[x]$
Then, $F(\alpha_1, \cdots, \alpha_n)$ is a splitting field for $f(x)$
Problem (picture drawn of different extensions $K$ and $E$ we could split over)
Goal $F(\alpha_1, \cdots, \alpha_n) \iso F(\beta_1, \cdots, \beta_n)$
i.e. splitting fields are unique up to isomorphism
Example $f(x)=x^4+x^2-6 \in \Q[x]$
$=(x^2+3)(x^2-2)$
“The” splitting field of $f(x)$ is $\Q(i \sqrt{3}, \sqrt{2})$
Remark
$F, F'$ are fields. $\varphi: F \to F'$ is an isomorphism
The natural map $\tilde{\varphi} : F[x] \to F'[x]$ is an isomorphism
We write $\tilde{\varphi} = \varphi$
Test 1: Chapter 1-3(Irreducibles, Irreducibility, Field Extensions)
- a)[5 marks] (Sylow 3), b)[5 marks] Sylow 2
- a)[5 Marks] (Irreducibility of a polynomial), b)
$\deg_F(\alpha)$ - a)[5 Marks] Proof
$F(\alpha_1, \cdots, \alpha_n)$, b) [5 Marks] Proof$[K:F]$
Test is out of 30 + 2/30 for showing up
Lemma [Isomorphism Extension Lemma]
$F, F'$ fields, $\varphi : F \to F'$ isomorphism. $f(x) \in F[x]$ irreducible. Let $\alpha$ be a root of $f(x)$ in some extension of $F$.
Let $\beta$ be a root of $\varphi(f(x))$ in some extension of $F'$
Then $\exists \ \psi: F(\alpha) \to F'(\beta)$ such that:
$\psi |_F = \varphi$$\psi(\alpha) = \beta$
Why?
$\rho_1 (g(\alpha)) = \overline{g(x)} = g(x) + <f(x)>$
$\rho_2 (\overline{h(x)}) = h(\beta)$
–> 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.
$a \in F$.
\(\begin{aligned} \psi(a) &= \rho_2(\sigma(\rho_1(a))) \\ &=\rho_2(\sigma(\bar{a})) \\ &= \rho_2(\overline{\varphi(a)}) \\ &= \varphi(a) \end{aligned}\)- \[\begin{aligned} \psi(\alpha) &= \rho_2(\sigma(\rho_1(\alpha))) \\ &= \rho_2 (\sigma(\bar{x})) \\ &= \rho_2 (\overline{\varphi({x})}) \\ &= \rho_2(\bar{x}) \\ &= \beta \end{aligned}\]
Corollary
$F$ 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? $\varphi: F \to F \ \text{ id }$
Lemma
$F$ 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
$\bar{F}/F$algebraic extension- 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.
Why?
$f(x) \in \bar{F}[x]$ 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.
Proof
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.
$x^p-1 = (x-1)(\underbrace{x^{p-1}+x^{p-2} + \cdots + x + 1}_{\phi_p (x)}$
$\phi_p(\zeta_p)=0$
$\phi_p (x)$ minimal polynimial for $\zeta_p$
$[\Q(\zeta_p):\Q]=p-1$
Remark
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
$\zeta^n=1$$\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
Investigation
-
$\{z \in \C: z^n =1\}$
$= \cup_{d|n} \{z \in \C: z \text{prim dth rou} \}$ -
$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)}$
$=\frac{(x-1)(x^2+x+1)(x+1)(x^2-x+1)}{(x-1)(x+1)(x^2+x+1)}$
$= x^2-x+1$
Goal
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$
Proposition
$\phi_n(x)$ is irreducible over $\Q$.
Proof
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
Say,
$$h(x) = \sum b_j x^j$$
$$\implies f(x) = \sum b_j x^{pj}$$
In $\Z_p[x]$,
$\bar{f}(x) = \sum \bar{b_j} x^{pj} = \sum \bar{b_j}^p x^{pj} \text{ FLT } = (\sum \bar{b_j} x^j)^p = \bar{h}(x)^p$
$\tf \bar{h}(x)^p = \bar{f}(x) = \bar{g}(x)\bar{K}(x)$
Let $\bar{l}(x)$ be an irreducible factor of $\bar{g}(x)$ in $\Z_p[x]$
$\bar{l}(x) | \bar{h}(x)^p \implies \bar{l}(x) | \bar{h}(x)$
Now, $\bar{\phi}_n(x) = \bar{g}(x)\bar{h}(x)$
$\implies \bar{l}(x)^2 | \bar{\phi}_n(x)$ $\implies \bar{l}(x)^2 | \underbrace{x^n-1}_{\Z_p[x]}$
$\implies x^n-\bar{1} = \bar{l}(x)^2 \bar{q}(x)$
$\implies \bar{n}x^{n-1} = \bar{l}(x)^2 \cdot \bar{q}'(x) + 2 \bar{l}(x) \bar{l}'(x) \bar{q}(x)$
$= \bar{l}(x) \cdot \text{[stuff]}$
Note: $p \nmid n \implies \bar{n} \neq \bar{0}$
$\tf \bar{l}(\bar{0}) = \bar{0}$
Since $\bar{l}(x) | x^n-\bar{1}$, $\bar{0}^n-1 = \bar{0}$ |
|
$\implies $\bar{1}=\bar{0} \in \Z_p \implies p |
1$. Contradiction! |
Corollary
For $n \in \N$, $\phi_n(x)$ is the minimal polynomial for $\zeta_n$ over $\Q$. In particular,
$$[\Q(\zeta_n):\Q] = \phi(n)$$
Examples: let $K$ be the splitting field of $f(x) = x^5-3$ over $\Q$
- Describe
$K$ - Compute
$[K : \Q]$ - 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 $\gcd(4, 5)=1$, $[K : \Q] = 5 \cdot 4 = 20$
(A4)
3) $[\Q(\zeta_5)(\sqrt[5]{3}):\Q(\zeta_5)] = 5$
$[\Q(\zeta_5):\Q]=4$
(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 $|K| = q^d$
Proof:
$K = \{a_1x_1+\cdots+a_dx_d : a_i \in F_q\}$
$\implies |K| = q^d$
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
Proof:
$x^{p^n}-x$ 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 $\text{Frob}(a+bi)$?
$(a+bi)^3 = a^3 + (b_i)^3 = a^3-b^3i: a, b \in F_3$
$=a-bi$
If $q$ is odd, then $F_{p^2} \iso F_p(\sqrt{d})$, so Frob($a+b\sqrt{d}$)=$(a-b\sqrt{d})$
Recall
$F$is a finite field,$|F| = p^n$where$p = Char(F)$$n = [F:\Z_p]$
- 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$ $\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$.
Remark
$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)$.
Why?
$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$
$Gal(K/\Q) = Aut(K)$
Why?
$\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$
Example
$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}$
Example
$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}\}$
Recall
$Gal(K/F) = \varphi \in Aut(K): \forall a \in F, \varphi(a)=0$$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 $F$.
Investigation
-
$f(x) \in F[x]$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$\varphi \mapsto \sigma$ - 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. $|G| = |Stab(\alpha_i)| \cdot \underbrace{|Orb(\alpha_i)|}_{n}$
$\implies |G| \mid n!$,$n | |G|$
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 $x^3-2$.
Why?
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
$\varphi$is a homomorphism$\forall a \in F, \varphi(a)=a$
Remark Let $\varphi: K \to E$ be an F-map. Then:
$\varphi$is injective$(\ker \varphi = \{0\})$- For all
$u, v \in K$,$\varphi(u+v)=\varphi(u)+\varphi(v)$ $\forall \alpha \in F, u \in K$,$\varphi(\alpha u)=\alpha\varphi(u)$
This implies that$\varphi$is a linear transformation!. Moreover:
Remark
If $[K:F] < \infty$ and $E=K$ then $\varphi \in Gal(K/F)$
Why?
$K$ 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
$$|Gal(K/F)|=[K:F]$$
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.
Proof:
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 $a_k=0$, $k= 1, \cdots, n$
$\iff k a_k = pm_k a_k, m_k \in \N, k=1, \cdots, n$
$\iff f(x)=a_nx^{pm_n} + \cdots + a_1x^{m_1}+a_0$
$\qed$
Corollary: If $F$ is finite, $F$ is perfect.
Recall
$$|Gal(K/F)| \leq [K:F]$$
$$F \text{ perfect } \iff (K/F \text{ alg } \implies K/F \text{ sep})$$
–> 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.
Proof
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.
$\tf \forall i, \exists b_i \in F, a_i = b_i^p$
Hence,
$f(x)=b_{n}^p x^{p_n} + \cdots + b_1^p x^p + b_0^p$
$=(b_n x^n + \cdots + b_1x^ + b_0)^p$. 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.
Theorem
$f(x) \in F[x]$ is seperable and non-constant.
Let $K$ be the splitting field of $f(x)$ over $F$. Then,
$|Gal(K/F)| = |Gal (f(x))|=[K:F]$
Remark
$K/E/F$. $Gal(K/E) \leq Gal(K/F)$
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,
$|Gal(K/E)|=[K:E]$
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 $\alpha_1 \in E$
$\implies \varphi_j^{-1} \circ \varphi_i \notin Gal(K/E)$
$\implies \varphi_i Gal(K/E)\neq \varphi_j Gal(K/E)$
Hence,
$$\frac{Gal(K/F)}{Gal(K/E)} \geq m$$
$\implies Gal(K/F) \geq [K:E] \cdot m = \frac{n}{m} \cdot m = n$
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 $K^\times = <\alpha>$, $\alpha \in K$
$K=F(\alpha)$
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.
Consider
$$S = \left\{\frac{\alpha_i-\alpha_1}{\beta_1-\beta_j} : i \neq 1, j \neq 1\right\}$$
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 $j\neq 1$
$h(x)=x-\beta \in F(\gamma)[x]$
$\implies \beta \in F(\gamma)$, $\implies \alpha \in F(\gamma)$ $\implies F(\alpha, \beta) \sub F(\gamma)$
Recall
$$Gal(\Q(\sqrt[3]{2})/\Q) = \{1\}$$
$\varphi(\sqrt[3]{2}) = \{\sqrt[3]{2}, \underbrace{\sqrt[3]{2}\zeta_3}_{\notin \Q(\sqrt[3]{2}}, \underbrace{\sqrt[3]{2}\zeta_3^2}_{\notin \Q(\sqrt[3]{2}}\}$
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:
$K/F$is normal- For every
$L/K$and$F$-map$\varphi: L \to L$,$\varphi |_K \in Gal(K/F)$
$F$map is a homo that fixes$F$ - If
$\alpha \in K$then all$F$-conjugates of$\alpha$belong to$K$ - If
$\alpha \in K$then its min poly over$F$splits over$K$
Proof
(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
However,
$$\beta = \varphi(\alpha) = \varphi |_K (\alpha) \in K$$
(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.
Remark
$F$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:
$K$is the splitting field of a separable polynomial in$F[x]$$|Gal(K/F)| = [K:F]$$Fix(Gal(K/F)) = F$$K/F$Galois
Proof:
(1) $\implies$ (2) done. Proved before
(2) $\implies$ (3). Assume $|Gal(K/F)|=[K:F]$.Let $E = Fix(Gal(K/F))$.
Note: $F \sub E \sub K$ by definition, and
$Gal(K/E) \leq Gal(K/F)$
Let $\varphi \in Gal(K/F)$ and let $a \in E$. So $\varphi(a)=a$
$\implies Gal(K/E)=Gal(K/F)$
$\tf [K:F] = |Gal(K/F)=|Gal(K/E)| \leq [K:E] \leq [K:F]$
$\implies [E:F]=1$
(3) $\implies$(4). Suppose $Fix(G)=F$ where $G = Gal(K/F)$
Let $\alpha \in K$ and let $p(x)$ be the minimal poly for $\alpha$ over $F$. We show $p(x)$ splits (normal) as a product of distinct linear factors (separable) over $K$.
Let $$\Delta = \{\varphi(\alpha) : \varphi \in G\} \sub \{\text{roots of $p(x)$} \} \cap K$$
Let $\alpha = \alpha_1, \alpha_2, \cdots, \alpha_n$ be the distinct elements of $\Delta$
Consider
$$h(x)=(x-\alpha_1) \cdots (x-\alpha_n) \in K[x]$$
Clearly $h(x)|p(x)$ in $K[x]$.
For $\varphi \in G$, $\varphi(h(x)) = h(x)$
$\implies h(x) \in Fix(G)[x]$
$\implies h(x) \in F[x]$
$\implies p(x) | h(x)$ $\implies p(x) =h(x)$
$\tf K/F$ is normal+seperable.
(4) $\implies$ (1). Assume $K/F$ Galois.
By the PET, $\exists \in K$ such that $K=F(\alpha)$. Easily, $K$ is the splitting field of the minimal polynomial $p(x)$ for $\alpha$ over $F$. $\qed$
–> Galois extensions are the splitting field of an irreducible polynomial.
Chapter 9 - Fundamental Theorem of Galois Theory
Theorem [Artin’s Theorem]
If $H$ is a finite subgroup of $Aut(K)$ and $F=Fix(H)$.
Then:
$[K:F] = |H|$$K/F$is Galois$Gal(K/F)=H$
Proof:
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
$Gal(K/\cdot): \E \to \H : E \mapsto Gal(K/E)$
$Fix: \H \to \E : H \mapsto Fix H$
Note:
$E_1, E_2 \in \E$,$E_1 \sub E_2$.
$\implies Gal(K/E_2) \sub Gal(K/E_1)$$H_1, H_2 \in \H$,$H_1 \sub H_2$
$\implies FixH_2 \sub Fix H_1$
i.e. the Galois correspondences are inclusion-reversing.
Theorem [Fundamental Theorem of Galois Theory]
Let $K/F$ be a finite Galois extension.
- For
$E \in \E$,$Fix Gal(K/E)=E$,$|Gal(K/E)|=[K:E]$
-> fix is a left inverse of Gal - For
$H \in \H$,
$Gal(K/Fix H)=H$,$[K:FixH]=H$
i.e. the Galois correspondences are inverses of each other.
Big Picture:
![[West LA - 1150 AM 2.png]]
Proof of Theorem
By A7, $K/E$ is Galois.
$\tf FixGal(K/E)=E$ (characterization theorem)
$|Gal(K/E)| = [K:E]$
(2) Follows from Artin
$\qed$
Corollary $K/F$ finite Galois
If $H_1 \sub H_2$ are in $\H$, then $[H_2:H_1] = [Fix H_1 : H_2]$
if $E_1 \sub E_2$ are in $\E$, then
$[E_2:E_1] = [Gal(K/E_1):Gal(K/E_2)]$
Why?
$[FixH_1:FixH_2] = \frac{[K:FixH_2]}{[K:FixH_1]} \stackrel{Artin}{=} \frac{|Gal(K/FixH_2)|}{|Gal(K/FixH_1)|} \stackrel{Artin}{=} \frac{H_2}{H_1} = [H_2:H_1]$
Next,
$$|Gal(K/E_1)|/|Gal(K/E_2)| \stackrel{FT}{=} \frac{[K:E_1]}{[K:E_2]} = [E_2:E_1]$$ $\qed$
Example $K= \text{ s.f. } f(x)=x^3-2$ over $\Q$
$K = \Q(\alpha, \zeta_3), \alpha = \sqrt[3]{2}$
Since $f(x)$ is irreducible (2-Eis) and $\Q$ is perfect, $K/\Q$ is Galois and $Gal(K/\Q) \leq S_3$, Since $|Gal(K/\Q)| = [K:\Q]=6$, $Gal(K/\Q)=S_3$
![[West LA - 1150 AM 3.png]]
Corollary $K/F$ finite, Galois
Then there are finitely many fields $E$ such that $F \sub E \sub K$
Why? $Gal(K/F)$ has finitely many subgroups.
Investigation $K/E/F$, $\varphi \in Gal(K/F)$
what does it mean for $\psi \in Gal(K/\varphi(E))$
$\iff \forall a \in E, \psi(\varphi(a)) = \varphi(a)$ and of course, $\psi$ is a $K$-automorphism.
$\iff \forall a \in E, (\varphi^{-1} \circ \psi \circ \varphi)(a)=a$
$\iff \varphi^{-1} \circ \psi \circ \varphi \in Gal(K/E)$
$\iff \psi \in \varphi Gal(K/E) \varphi^{-1}$
–> $g H g^{-1}$ moment
$Gal(K/\varphi(E)) = \varphi Gal(K/E) \varphi^{-1}$$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:
$E/F$is Galois$E/F$Normal$Gal(K/E) \tleq Gal(K/F)$
Proof:
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, $Gal(K/F) / Gal(K/E) \iso Gal(E/F)$
Why?
$\psi : Gal(K/F) \to Gal(E/F)$
$\psi(\varphi)=\varphi |_E$ (Normality Theorem)
$\ker \psi = Gal(K/E)$
Surjectivity:
$$|Gal(K/F)/Gal(K/E)| = \frac{[K:F]}{[K:E]} \stackrel{Tower Thm}{=} [E:F] = |Gal(E/F)|$$
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 $\psi: \Z_n^\times \to Gal(K/\Q) :
\psi(k)= \varphi_k, \varphi_k(\zeta_n)=\zeta_n^k$
$\varphi_{ab}(\zeta_n) = \zeta_n^{ab}=(\zeta_n^b)^a = \varphi_a(\varphi_b(\zeta_n)) \implies \varphi_{ab}=\varphi_a \circ \varphi_b$
$\implies \psi(ab)=\psi(a) \circ \psi(b)$$\ker \psi = \{1\}$$|\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
Recall
$f(x) \in F[x]$ irred, sep. If $G = Gal(f(x))$:
$G$is a transitive subgroup of$S_n$,$n=\deg f(x)$.$n \mid |G|$(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:
$$disc(f(x))=\Pi_{i < j} (\alpha_i-\alpha_j)^2$$
Remarks:
$disc(f(x))=0$$\iff$$f(x)$is NOT separable$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$$f(x)=x^2+bx+c = (x-\alpha_1)(x-\alpha_2)$
$disc(f(x))=(\alpha_1-\alpha_2)^2 = \alpha_1^2+\alpha_2^2-2\alpha_1\alpha_2 = b^2-4c$
Investigation
$char(F)\neq2$. $f(x) \in F[x]$ separable. $K$ = splitting field of $f(x)$. $f(x)=(x-\alpha_1) \cdots (x-\alpha_n) \sub K[x]$ distinct roots.
Let $d = \Pi_{i < j} (\alpha_i-\alpha_j)$ so that $d^2=disc(f(x))$
Let $G = Gal(f(x))=Gal(K/F)$
For all $\varphi \in G$, $\varphi(d)^2=d^2 \implies \varphi(d)$ root of $x^2-d^2 \in F[x]$
$\implies \varphi(d)=\pm d$. $char 2$, don’t want 1=-1
Fact: $\varphi(d)=d \iff \varphi \in A_n$
$\tf$ TFAE:
$G \sub A_n$$\forall \varphi \in G, \varphi(d)=d$$d \in F$$discf(x)$is a square in$F$.
Cubics
$f(x) \in F[x]$ monic irred., sep. $\deg f(x)=3$
$f(x)=x^3+\alpha x^2+\beta x + \gamma$
Assume $charF \neq 2, 3$
$g(x):=f(x-\alpha/3)$
$=\underbrace{x^3+ bx+c}_{\textbf{depressed cubic}}$
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, $Gal(f(x))=G=$
\(\begin{cases}
A_3 & \text{ if `$disc(f(x)$` square in `$F$`} \\
S_3 & \text{ otherwise }
\end{cases}\)
Example
$f(x)=x^3-3x+1 \in \Q[x]$. $f(x)$ is irreducible by Mod-2 test.
Since $\Q$ is perfect, $f(x)$ is separable. $disc(f(x))=-4(-3)^2-27(1)^2=4(27)-27=27(4-1)=27(3)=3^4=9^2$
$Gal(f(x))=A_3$
Quartics
Let $f(x)\in F[x]$ be a separable, irreducible, monic, quartic.
Say
$$f(x)=x^4+\alpha x^3+\beta x^2+\gamma x + \delta$$
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: $S_4, A_4, D_4, V\iso \Z_2 \times \Z_2, \Z_4$
Remark
$V = \{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)\}$
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 $L=F(u, v, w)$.
$\tf F \sub L \sub K$
Note:
$Gal(K/F)=Gal(f(x)), K/F$galois.$L/F$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$Gal(Res(f(x))=Gal(L/F) \iso Gal(K/F) / Gal(K/L)=G/(G \cap V)$
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 :
$x^2-ux+d \in F[x] = (x-\alpha_1\alpha_2)(x-\alpha_3\alpha_4)$$x^2+(b-u) \in F[x] = (x-(\alpha_1+\alpha_2))(x-(\alpha_3+\alpha_4))$
Prop
$G = <\sigma>\iso \Z_4$ $\iff$ (1) and (2) split over $L$.
Proof:
$(\implies)$ Suppose $Gal(f(x)) = <\sigma>$. Then $Gal(K/L)=<\sigma> \cap V = <\sigma^2>$
But $\alpha_1 \alpha_2, \alpha_3 \alpha_4, \alpha_1+\alpha_2, \alpha_3+\alpha_4 \in Fix<\sigma^2>=Fix(K/L)=L$
($\impliedby$) Suppose $\alpha_1 \alpha_2, \alpha_3 \alpha_4, \alpha_1+\alpha_2, \alpha_3+\alpha_4 \in L$
Since $\alpha_1, \alpha_2 \in L(\alpha_1)$
and $(\alpha_1-\alpha_2)(\alpha_3-\alpha_4)=v-w \in L$
$\implies \alpha_3-\alpha_4 \in L(\alpha_1) \implies \alpha_3, \alpha_4 \in L(\alpha_1)$, $charF\neq 2$
$\tf K=F(\alpha_1, \cdots, \alpha_4)=L(\alpha_1)$
Since $\alpha_1$ is a root of $x^2-(\alpha_1+\alpha_2)x+\alpha_1\alpha_2 \in L[x]$
$[K:L]=[L(\alpha_1):L] \leq 2$
However, $[L:F]=m=2$
$\implies [K:F] \leq 4$
But, $[K:F]=|G| \geq 4$
$\implies [K:F]=4 \implies |G|=4$ $\implies G \iso \Z_4$ $\qed$
Recall $char F \neq 2$
$f(x)=x^4+bx^2+cx+d$, irreducible separable
$Res f(x)=x^3-bx^2-4dx+4bd-c^2$
$G=Gal(f(x))$, $m = |Gal Res f(x)|$
if $m=2$, and $u \in F$ (root of $Res$), then $G=\Z_4$ $\iff$
(1) $x^2-ux+d$, (2) $x^2+(b-u)$ split over $L=sf Res f(x)$
Example
$f(x)=x^4-2x-2 \in \Q[x]$. By 2-Eis, $f(x)$ is irreducible. Since $\Q$ is perfect, $f(x)$ separable.
$[b=0, c=-2, d=-2]$. $Resf(x)=x^3+8x-4$
[HW: No rational roots $\implies$ irreducible ]
$disc(Res(f(x))=-4(8)^3-27(-4)^2 < 0$, therefore $discResf(x)$ is not a square in $\Q$, and so $Gal(Res(f(x))) = S_3$
$\tf m=6$ so $Gal(f(x))=S_4$
Example $f(x)=x^4+5x+5 \in \Q[x]$
by $5$-Eis, $f(x)$ is irreducible. Since $\Q$ is perfect, $f(x)$ is separable.
[$b=0, c=5, d=5$]
$Resf(x)=x^3-20x-25$
Note $5$ is a root, through poly long division,
$Res f(x)=(x-5)(\underbrace{x^2+5x+5}_{\text{irred, 5-eis}})$
$\tf m=2$, so let $u=5$.
roots of quadratic: $$\frac{-5 \pm \sqrt{25-20}}{2} = \frac{-5\pm \sqrt{5}}{2}$$
if $L$ is the spliting field of $Res f(x)$, $L=\Q(\sqrt{5})$
Then, we consider:
$x^2-5x+5$, roots:$\frac{5\pm\sqrt{5}}{2} \in L$$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
Recall
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.
Recall
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.
Why?
$\implies$ 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 $N$, $H_0=k_m$
$\qed$
Recall
$G$ 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.
Investigation
$A, B, C$ 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.
Proof
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 $K$.
Proof
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 $\alpha \in K$,
$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$$a_1\sigma_1(\alpha)\sigma_1(\beta)+a_2\sigma_2(\alpha)\sigma_1(\beta)+\cdots+\sigma_n(\alpha)\sigma_1(\beta)=0$
subtract (1) and (2)
$[\underbrace{a_2(\sigma_2(\beta))-\sigma_1(
\beta))}_{\neq 0}\sigma_2(\alpha)+\cdots+a_n(\sigma_n(\beta)-\sigma_1(\beta))\sigma_n(\alpha)]=0$
hence $\{\sigma_2, \cdots, \sigma_n\}$ linearly dependent. Contradiction, $\qed$
Proposition Assume $F$ contains a primitive $n$th root of unity, and $K/F$ is cyclic of degree $n$. Then, $K$ is a simple radical extension of $F$.
Proof Let $G=Gal(K/F)$, so that $|G|=[K:F]=n$. Say $G=<\sigma>$
For any $\alpha \in K^\times$, let $g(\alpha)=\alpha+\zeta \sigma(\alpha)+\zeta^2\sigma^2(\alpha)+\cdots+\zeta^{n-1}\sigma^{n-1}(\alpha)$
where $\zeta \in F$ is a primitive $n$th root of unity.
Note:
- Since
$G$is LI over$K$,$\forall \alpha \neq 0$,$g(\alpha)=0$ $\sigma(g(\alpha))=\sigma(\alpha)+\zeta\sigma^2(\alpha)+\cdots+\zeta^{n-1}\alpha=\zeta^{-1}g(\alpha)$$\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.
$\phi_n(x) \in FixGal(F(\alpha)/F)[x] = F[x]$
Lemma
$[K:F] < \infty$, $K/E/F$
$K/E$ 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.
Proof
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.
Well,
$L = K(\text{ roots of f }) = K(\alpha, \text{ other roots}) = E(\alpha) \text{ other roots} = E(\text{roots of f})$
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 $E \sub \E(\zeta) \sub E(\zeta, \sigma_1(\alpha)) \sub E(\zeta, \sigma_1(\alpha), \sigma_2(\alpha)) \sub \cdots$
$\sub E(\zeta, \sigma_1(\alpha), \sigma_2(\alpha), \cdots, \sigma_r(
\alpha))=L$
$\implies$ $L/E$ radical.
Claim 3: $Gal(L/E)$ solvable.
Let $G_i = Gal(L/E_i)$ where $E_0 = E(\zeta)$ and $E_r=E(\zeta, \sigma_1(\alpha), \cdots, \sigma_r(\alpha))$
$\implies \{1\} \leq G_r \leq G_{r-1} \leq \cdots G_2 \leq G_1 \leq G_0 \leq \underbrace{Gal(L/E)}_{G'}$
-
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$Gal(L/E) / Gal(L/E(\zeta)) = G'/G_0 \iso \Z_n^\times \text{ abelian }$ -
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.
Corollary
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.
Proof
$(\implies)$ 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
$Gal(E/F) \iso \underbrace{Gal(L/F) / Gal(L/E)}_{\text{solvable}}$
$\qed$
Investigation
$S_5$ , $H = <(1 2), (1 2 3 4 5)>$
$(1 2 3 4 5)(1 2)(5 4 3 2 1) = (2 3) \in H$
$(1 2 3 4 5)(2 3)(5 4 3 2 1) = (3 4) \in H$
$(4 5), (5 1) \in H$
By instead conjugating by powers of $(1 2 3 4 5)$ (e.g. $(1 3 5 2 4)$):
$\forall \ \tau$ transposition, $\tau \in H$
$\tf H =S_5$
Remark:
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.
Proof
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). $\qed$