Topics: Algebraically closed fields, Maps from simple extensions, Splitting fields, Multiple roots, Groups of automorphisms of fields, Separable, normal, and Galois extensions, The fundamental theorem of Galois theory
[Homeworks]
Tuesday, 16 May, 2017
(also download from here)

We began by reviewing several of the most important basic things from Module 1, for example:

Minimum polynomial of an element of an extension field.

Degree of a field extension, multiplicativity of degrees along a tower of fields.

Construction of a field extension containing a root of a given polynomial of positive degree.

$[F(\alpha):F] = \deg f$, where $f$ is the min poly for $\alpha$ over $F$.

If $\alpha$ and $\beta$ have same min poly over $F$, $\exists!\ F$iso
$F(\alpha) \to F(\beta)$ taking $\alpha$ to $\beta$. Picture to illustrate.

We defined what it means to say that a polynomial splits in $K$.

We did some examples to motivate the notion of a splitting field:

Find min poly of $\sqrt{2+\sqrt2}$ over $\Q$. (Splitting field has degree $4$.)

Splitting field of $X^32$ over $\Q$ has degree $6$. Picture to illustrate.

We defined splitting field $K$ of $f(X)$ over $F$ and proved that $[K:F] \le (\deg f)!$ . The question arose as to whether the degree is always a divisor of $d!$ . On the spot, we said that we did not know whether or not it is true in general, but that we would see later that that is true in characteristic $0$ (using Galois Theory).
Wednesday, 17 May, 2017
(also download from here)

Basic stuff on maps of simple algebraic extensions: ``Roots must map to roots." Also, if the polynomial is irreducible, a root can map to any root. Vague discussion of the basic ideas of Galois Theory: You don't necessarily get n! different maps, e.g., with $\alpha := \sqrt{2+\sqrt2}$, with roots $\pm \alpha$ and $\pm\beta$, where $\beta = \sqrt{2\sqrt2}$, if $\alpha \mapsto \beta$, then $\alpha \mapsto \beta$ is forced.

We proved that splitting field is unique up to $F$iso.

Basic stuff on algebraically closed and (relative) algebraic closure.

Proposition 1.44; mentioned Gilmer's theorem: If $K/F$ is algebraic and every $f(X)\in F[X]$ has a root in $K$, then $K = F^a$.

Set of algebraic elements over $F$ forms a subfield.
Thursday, 18 May, 2017
(also download from here)

Definition of
multiple roots
. Trivial examples, e.g., $X^22X+1$, irreducible example $X^2a$ in characteristic $2$. Criterion for multiple roots in terms of derivative. Permanence of GCD
under extensions. No multiple roots $\iff$ $f$ and $f'$ are relatively prime. Importance of char$(F) = 0$
. Milne, Proposition 2.12.

Field homomorphisms are onetoone. Def of automorphism. $\text{Aut(}K/f)$ as a group. When $K/F$ is finite, every $F$hom $K \to K$ is an automorphism. Mentioned without proof that $\#\text{Aut}\mathbb{C} = 2^{2^{\aleph_0}}$. Returned to results on # of autos/isos in Section 2.
Tuesday, 23 May, 2017
(also download from here)
 Finishing up Section 2

Finish up Proposition 2.7 (the case of an extension that is not simple).

Do Milne, Proposition 2.13, define separable polynomial, and warn students that some books define separability in terms of the irreducible factors. (For us, $X^2\in \mathbb{Q}[X]$ is not separable, but in other books all polynomials are separable if the field has characteristic $0$.)

Define perfect field and do Milne, Proposition 2.16 (characterization of perfect fields).

Continuing with Section 3

Define em fixed field $E^G$. Point out that every automorphism of a field fixes the prime field; so, for example, $\Aut(\mathbb{Q}(\root 4\of 2)/\mathbb{Q})$ $= \Aut(\mathbb{Q}(\root 4\of 2)$.

Examples : $\Aut(\mathbb{Q}(\root3\of2)) = \{1\}$. $\Aut(\mathbb{Q}(\sqrt{2+\sqrt2}\, ))$ is cyclic of order $4$. (Reminder that there are only two groups of order $4$, up to $\cong$ )

State and prove Artin's Theorem (3.4): $[E:E^G] \le G$ for $G$ finite. (Proof to be finished next time.)
Wednesday, 24 May, 2017
(also download from here)

We finished the proof of Artin's Theorem (which should probably be called a lemma). The lemma (Theorem 3.4) is the technical crux of the argument, but the real theorem is Corollary 3.5:
For a finite group $G$ of automorphisms of a field $E$ we have $G= \Aut(E/E^G)$. Also, we pointed out that the inequality in Theorem 3.4 is actually an equality: $[E:E^G] = G$\,.

We did Proposition 3.2.

We defined
separable
, normal
, and Galois
extensions and gave several examples to illustrate. Informally, Galois
means that there are enough automorphisms to make the fixed field as small as possible. To get enough automorphisms, you need enough roots, and separable
and normal
help provide these roots.

These ideas become formalized in Theorem 3.10, which we stated and proved.
The part of the argument (in proving (b) $\implies$ (c)) involving symmetric polynomials might have been a little hard going for people not familiar with these things. Another approach might be to point out that $G$ acts on $E[X]$ by taking $X$ to $X$. The elements $\alpha_1,\dots,\alpha_m$ are the distinct images of $\alpha$ under the action of $G$, and the action of $G$ just permutes these elements. Therefore $g(X) = (X\alpha_1)\cdots(X\alpha_m)$ is fixed by the action of $G$ and hence is in $F[X]$. Milne is a bit careless in the proof that (c) $\implies$ (a). It could happen that $f_1 = f_2$, for example, and this would prevent the product $f$ from being separable. Easy remedy: Just define $f$ as the product of the {\em distinct} polynomials among the $f_i$. Then, for $i\ne j$, the polynomials $f_i$ and $f_j$, being distinct monic irreducible polynomials, are relatively prime, and hence by
permanence of gcd
cannot have a common root in any extension.

We did Corollary 3.13: If $E \supset M \subset F$ is a tower of fields and $E/F$ is Galois, then so is $E/M$. (However, the example $\mathbb{Q}(\root3\of2) \supset \mathbb{Q}(\root3\of2) \supset \mathbb{Q}$ shows that $M/F$ is not necessarily Galois.)

We gave a brief
preview
of FTGT (the fundamental theorem) and mentioned that the terminology normal
for field extensions is not an accident; it has to do with normality of subgroups. Next time: FTGT and illustrations.

Confession: The solution we gave for Problem 9 in Homework #1 was not quite in line with the instructions, since we had extensions $\mathbb{Q}(\alpha_1,\alpha_2)$ and $\mathbb{Q}(\alpha_1,\alpha_3)$. The repetition of $\alpha1$ was definitely bad form. Mahdav Sharma pointed out that $X^82$ works: The eight roots are $\zeta^j\alpha\,, 0 \le j \le 7$, where $\alpha =\root8\of2$ and $\zeta = \frac{1}{\sqrt2}(1+i)$. Now take these four roots: $\zeta\alpha$, $\zeta\alpha = \zeta^4\alpha$, $\alpha$, $i\alpha= \zeta^2\alpha$. Then $[\mathbb{Q}(\zeta\alpha, \zeta\alpha):\mathbb{Q}]$ $= 8$, while $[\mathbb{Q}(\alpha, i\alpha) = 16$. Probably there are examples with $\deg f = 6$. Are there examples with $\deg f = 4$? (We don't know.)
Thursday, 25 May, 2017
(also download from here)

First, there was a typo in the notes on Class #5: Twelve lines from the bottom,
$\mathbb{Q}(\root3\of2) \supset $ $\mathbb{Q}(\root3\of2) \supset $ $\mathbb{Q}$
should have been $\mathbb{Q}(\root3\of2, \omega) \supset $ $\mathbb{Q}(\root3\of2) \supset$ $\mathbb{Q}$
.

Probably there should be another equivalent condition for
Galois
in Theorem 3.10, namely, (e) $\Aut(E/F) = $ $[E:F] < $ $\infty$. To see that this is equivalent, note that $\Aut(E/F) = $ $\Aut(E/E^G)$, where $G= \Aut(E/F)$. Since $G = [E:E^G]$ (the important point that should have been part of Corollary 3.5), we see that (e) is equivalent to (d).

In Corollary 3.12 there's the same problem encountered in Theorem 3.10: You have to take the distinct $f_i$.

I skipped Remark 3.14 because I wanted to be sure to have time for FTGT.

Neither Sylvia nor I could make any sense out of Remark 3.11 (b), so we told the students to ignore it. Unfortunately the proof of FTGT refers to 3.11 (b) twice, but actually Corollary 3.5 is what is needed. I added a little to FTGT, namely, that the intermediate field $M$ is normal (Galois) over $F$ if and only if it's stable under the action of the Galois group. (Of course, this is what Milne proves, but it seems worthwhile to make it explicit.)

I drew lots of sidebyside pictures of field extensions and upside down subgroup lattices, and explained how joins correspond to intersections. (Reminded that for subgroups $H_1$ and $H_2$ the product $H_1H_2$ is not necessarily a subgroup, so I used
$\langle H_1H_2\rangle$
for the join. I did not get to 3.18 or 3.19 or 3.20.

I illustrated FTGT with splitting fields of $(X^22)(X^23)$ and $X^32$ over $\mathbb{Q}$ (with sidebyside pictures of intermediate fields and subgroups). I told them to go over Example 3.21 and draw these pictures.