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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. $|a|=p$

*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 $G$.

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

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

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 $H \iso \Z_{15}$

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: $5 \nmid |H|$
$|H| \in \{2, 3,4,6, 12\}$

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

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 $\iff$

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 $f(a)$

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:

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 $Z_2[x]$

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:

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.

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 $Q$?

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 $K$.

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 $\Q(\sqrt{2}, \sqrt{3}) = \Q(\sqrt{2}+\sqrt{3})$

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 $F(\alpha) = \psi (F[x]/<f(x)>)$
$=\overline{c_0+c_1\alpha+\cdots+c_{n-1}\alpha^{n-1}: c_i \in F}$
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: $<f(x)> = \{h(x) : h(\alpha) = 0\}$

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)

$\deg_{\Q}(\alpha)=4$
$\tilde{m}(x) = x - \alpha$, $\deg_{\R} (\alpha)=1$
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 $\alpha = \sqrt{2}, \beta = \sqrt{2} \zeta_3$
–> 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, $[K:F] = [K:E][E:F]$

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

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

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

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

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

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 $\alpha \in K$

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

1. $K/F$
2. $f(x)$ splits over $K$
3. 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)
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. $\psi(\alpha) = \beta$

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.

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}

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

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.

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

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

Investigation

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

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 $\Q$ 1. Describe $K$ 2. Compute $[K : \Q]$ 3. Find a basis for $K / \Q$ 1) The complex roots of $f(x)$ are $\sqrt{3}, \sqrt{3} \zeta_5, \sqrt{3} \zeta_{5^2}, \sqrt{3} \zeta_{5^3}, \sqrt{3} \zeta_{5^4}$ $\tf K = \Q(\sqrt{3}, \zeta_5)$ 2) $[\Q(\sqrt{3}):\Q] = \underbrace{\deg (x^5-3)}_{\text{3-Eis}} = 5$ $[\Q(\zeta_5):\Q] = \phi(5)=4$ Since $\gcd(4, 5)=1$, $[K : \Q] = 5 \cdot 4 = 20$ (A4) 3) $[\Q(\zeta_5)(\sqrt{3}):\Q(\zeta_5)] = 5$ $[\Q(\zeta_5):\Q]=4$ (Tower Theorem) A basis for $\Q(\zeta_5)(\sqrt{3})/\Q(\zeta_5)$ is $$B_1 = \{1, \sqrt{3}, (\sqrt{3})^2, (\sqrt{3})^3, (\sqrt{3})^4\}$$ 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 $$\{(\sqrt{3})^i \zeta_5^j : 0 \leq i \leq 4, 0 \leq j \leq 3\}$$ –> 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 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 $K$. 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 $K/\Q$ 1. $Gal(K/\Q) = Aut(K)$ 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 $K = \Q(\sqrt{2})$. $\varphi \in Gal(K/\Q)$ $\varphi(\sqrt{2}) \in \{\sqrt{2}, \sqrt{2} \zeta_3, \sqrt{2}\zeta_3^2\} \cap K \sub \R$ $\implies \varphi(\sqrt{2}) = \sqrt{2}$ $\implies Gal(K/\Q)=\{\text{id}\}$ Recall 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 $F$. Investigation 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: $$\alpha_1 = \sqrt{2}, \alpha_2 = \sqrt{2} \zeta_3, \alpha_3 = \sqrt{2} \zeta_3^2$$ Minimal polynomial for $\zeta_3$ over $\Q$ is $$\phi_3(x) = x^2+x+1$$ By an argument of roots the minimal polynomial for $\sqrt{2}$ over $\Q(\zeta_3)$ is $x^3-2$. Why? Suppose not, then $$[\Q(\sqrt{2}):\Q] \mid [\Q(\zeta_3):\Q]$$ 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: $$\varphi(\alpha_1) = \alpha_1$$ $$\varphi(\alpha_2) = \varphi(\sqrt{2}) \varphi(\zeta_3) = \sqrt{2} \zeta_3^2$$ $$\varphi(\alpha_3) = \varphi(\sqrt{2})\varphi(\zeta_3)^2 = \alpha_2$$ 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: 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 $K = \Q(\sqrt{2}), F=\Q$. 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 $$Gal(\Q(\sqrt{2})/\Q) = \{1\}$$ $\varphi(\sqrt{2}) = \{\sqrt{2}, \underbrace{\sqrt{2}\zeta_3}_{\notin \Q(\sqrt{2}}, \underbrace{\sqrt{2}\zeta_3^2}_{\notin \Q(\sqrt{2}}\}$ 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 $\sqrt{2}$ are $\sqrt{2} \zeta_3, \sqrt{2}\zeta_3^2$ 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 $K$ 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 $\Q(\sqrt{2})/ \Q$ is NOT normal, because $\sqrt{2}\zeta_3 \in \Q(\sqrt{2})$ –> 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 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 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: 1. $[K:F] = |H|$ 2. $K/F$ is Galois 3. $Gal(K/F)=H$ 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: 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 $\qed$ 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 $\Q$ $K = \Q(\alpha, \zeta_3), \alpha = \sqrt{2}$ 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. $Gal(K/E) \tleq Gal(K/F)$ 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$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 Recall $f(x) \in F[x]$ 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: $$disc(f(x))=\Pi_{i < j} (\alpha_i-\alpha_j)^2$$ Remarks: 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. $f(x)=x^2+bx+c = (x-\alpha_1)(x-\alpha_2)$ $disc(f(x))=(\alpha_1-\alpha_2)^2 = \alpha_1^2+\alpha_2^2-2\alpha_1\alpha_2 = b^2-4c$ 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: 1. $G \sub A_n$ 2. $\forall \varphi \in G, \varphi(d)=d$ 3. $d \in F$ 4. $discf(x)$ is a square in $F$. 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: 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. $x^2+(b-u) \in F[x] = (x-(\alpha_1+\alpha_2))(x-(\alpha_3+\alpha_4))$ 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 $Res f(x)=x^3-bx^2-4dx+4bd-c^2$ $G=Gal(f(x))$, $m = |Gal Res f(x)|$ $\begin{array}{c | c c c c c} G & S_4 & A_4 & D_4 & V & \Z_4 \\ \hline G \cap V & V & V & V & V & \Z_2 \\ \hline m & 6 & 3 & 2 & 1 & 2 \end{array}$ 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 $L=sf Res f(x)$ 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: 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 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 $K=\Q(\sqrt{2}, \zeta_8)$. Clearly, $K \sup \Q(\sqrt{2}) \sup \Q$ 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 $\alpha \in K$, 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)$[\underbrace{a_2(\sigma_2(\beta))-\sigma_1(
\beta))}_{\neq 0}\sigma_2(\alpha)+\cdots+a_n(\sigma_n(\beta)-\sigma_1(\beta))\sigma_n(\alpha)]=0$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: 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. $\phi_n(x) \in FixGal(F(\alpha)/F)[x] = F[x]$ 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(