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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 $$h(\alpha) = 0$$
$$\tf <f(x)> = < h(x)>$$$\impliesf(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 $$\alpha = \sqrt[3]{2}, \beta = \sqrt[3]{2} \zeta_3$$
–> primitive third root of 3 is $$\zeta_3$$
$$m(x)=x^3-2$$ (2-eisenstein)

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

Goal for Today $$K/F$$, Explore $$K$$ as an $$F$$-vector space.

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

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

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

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

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

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 $$\alpha$$] $$K/F$$

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 $$\sqrt[5]{3}, \sqrt[5]{3} \zeta_5, \sqrt[5]{3} \zeta_{5^2}, \sqrt[5]{3} \zeta_{5^3}, \sqrt[5]{3} \zeta_{5^4}$$ $$\tf K = \Q(\sqrt[5]{3}, \zeta_5)$$ 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 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 $$K = \Q(\sqrt[3]{2})$$. $$\varphi \in Gal(K/\Q)$$ $$\varphi(\sqrt[3]{2}) \in \{\sqrt[3]{2}, \sqrt[3]{2} \zeta_3, \sqrt[3]{2}\zeta_3^2\} \cap K \sub \R$$ $$\implies \varphi(\sqrt[3]{2}) = \sqrt[3]{2}$$ $$\implies Gal(K/\Q)=\{\text{id}\}$$ Recall 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: $$\alpha_1 = \sqrt[3]{2}, \alpha_2 = \sqrt[3]{2} \zeta_3, \alpha_3 = \sqrt[3]{2} \zeta_3^2$$ 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 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 $$K = \Q(\sqrt[3]{2}), F=\Q$$. 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: 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 $$\Q(\sqrt[3]{2})/ \Q$$ 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 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 $$\Q$$ $$K = \Q(\alpha, \zeta_3), \alpha = \sqrt[3]{2}$$ Since $$f(x)$$ is irreducible (2-Eis) and $$\Q$$ is perfect, $$K/\Q$$ is Galois and $$Gal(K/\Q) \leq S_3$$, Since $$|Gal(K/\Q)| = [K:\Q]=6$$, $$Gal(K/\Q)=S_3$$ ![[West LA - 1150 AM 3.png]] Corollary $$K/F$$ finite, Galois Then there are finitely many fields $$E$$ such that $$F \sub E \sub K$$ Why? $$Gal(K/F)$$ has finitely many subgroups. Investigation $$K/E/F$$, $$\varphi \in Gal(K/F)$$ what does it mean for $$\psi \in Gal(K/\varphi(E))$$ $$\iff \forall a \in E, \psi(\varphi(a)) = \varphi(a)$$ and of course, $$\psi$$ is a $$K$$-automorphism. $$\iff \forall a \in E, (\varphi^{-1} \circ \psi \circ \varphi)(a)=a$$ $$\iff \varphi^{-1} \circ \psi \circ \varphi \in Gal(K/E)$$ $$\iff \psi \in \varphi Gal(K/E) \varphi^{-1}$$ –> $$g H g^{-1}$$ moment 1. $$Gal(K/\varphi(E)) = \varphi Gal(K/E) \varphi^{-1}$$ 2. $$Gal(K/E) \tleq Gal(K/F) \iff \forall \varphi \in Gal(K/F), \varphi(E)=E$$ Theorem $$K/F$$ finite, Galois, $$K/E/F$$, TFAE: 1. $$E/F$$ is Galois 2. $$E/F$$ Normal 3. $$Gal(K/E) \tleq Gal(K/F)$$ 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 $$K=\Q(\sqrt[3]{2}, \zeta_8)$$. 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 $$\alpha \in K$$, 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(
\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 $$E_r=E(\zeta, \sigma_1(\alpha), \cdots, \sigma_r(\alpha))$$

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

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

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

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

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

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

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

Inductively, we get the same result when $$K/E$$ is radical.

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). $$\qed$$