## Galois Theory, Part II Michaelmas 2013

• Lectures: Monday, Wednesday and Friday, 9:00 MR3
• Example sheets: (will appear here) 1 - 2 - 3 - 4
• Supplementary material:
• Summary of relevant material from Groups, Rings and Modules
• Summary of Zorn's Lemma and related material.
• Proof of transitivity of norm for a tower $$M/L/K$$ (the corresponding result for trace was proved in the lectures).

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### Notes on Lecture 1

I didn't have time to go into any detail about the (colourful) history of the solution of equations of degree 3 and 4 by radicals. A good place to explore this (and other topics in the history of mathematics) is the St. Andrew's History of Maths archive. Start with Tartaglia. For the solution itself, search for "Cubic equation" in Wikipedia.

### Notes on Lecture 11

This post is about various set-theoretic niceties which you certainly don't need to understand in order to follow the course. Definitely to be ignored unless these things bother you (in which case you should certainly go to Logic and Set Theory next term).

In the course of proving the existence of algebraic closure for a countable field $$K$$, I enumerated the polynomials over $$K$$ as $$\{ f_n \mid n\ge 1 \}$$, and constructed a sequence of fields $$K_n$$, where $$K_0 = K$$ and $$K_n$$ is a splitting field for $$f_n$$ over $$K_{n-1}$$. There are two issues I skirted over somewhat.

(i) I assumed that that $$K_{n-1}$$ was a subfield of $$K_n$$. If you don't have any problem with this, you probably don't want to read on, but if you are troubled, here is a simple illustration of the point. We usually define the complex numbers as the set of ordered pairs $$\CC = \{ (x,y) \mid x,y\in \RR \}$$ with addition defined coordinatewise and multiplication by $$(x,y)(x',y') = (xx'-yy',xy'+x'y)$$. Purists will notice that this doesn't actually contain $$\RR$$, but only contains $$\{ (x,0) \mid x\in \RR \}$$, which is a subfield isomorphic to $$\RR$$. To make $$\CC$$ actually contain $$\RR$$, we need to define it to be the disjoint union of $$\RR$$ and $$\{ (x,y) \mid y\ne 0 \}$$. In a similar way we can construct a splitting field as an extension which contains the base field as an actual subfield (not just an isomorphic copy). This is the way it is done in Lang's Algebra, for example.

Personally I prefer just to work with extensions of fields as homomorphisms $$K \hookrightarrow L$$. In this case, we have a sequence of extensions $$K=K_0 \hookrightarrow K_1 \hookrightarrow K_2 \dots$$. Let $$X$$ be the disjoint union of the fields $$K_n$$ and define an equivalence relation on $$X$$ such that if $$x \in K_m$$ and $$y\in K_n$$ with $$m \le n$$, then $$x \sim y$$ iff the composite embedding $$K_m \hookrightarrow K_n$$ maps $$x$$ to $$y$$. Define $$\overline K$$ to be the quotient of $$X$$ by this equivalence relation (i.e. the set of equivalence classes). (If each $$K_n$$ is actually a subset of $$K_{n+1}$$ then this is isomorphic to the union $$\bigcup K_n$$). This is then a field, and there are embeddings $$K_n \hookrightarrow \overline K$$ which are compatible with the embeddings $$K_n \hookrightarrow K_{n+1}$$. (Aside: this is a special case of a more general construction, the direct limit.)

(ii) You may have observed is that there are various choices involved in constructing a splitting field of an irreducible polynomial; after adjoining one root, the polynomial becomes reducible, and you have to choose which irreducible factor to adjoin a root of at the next next. So in constructing an infinite sequence of splitting fields, you have to make an infinite number of choices, and this appears to require the axiom of choice.

In fact this isn't necessary here. Let $$L$$ be a countable field, so we are given an injection $$i : L \hookrightarrow \NN$$. This can be used to give an injection $$j : L[X] \hookrightarrow \NN$$ (by your favourite way of showing that polynomials over a countable field are countable). This in turn determines, for every simple algebraic extension $$L(x)/L$$, an injection $$k : L(x) \hookrightarrow \NN$$ - let $$\phi : L[X] \to L(x)$$ be the $$L$$-homomorphism with $$\phi(X)=x$$, and define $$k(y) = \mathrm{min}\{ j(f) \mid f \in L[X],\ f(x)=y \}$$.

Using this, given any nonconstant polynomial $$f \in L[X]$$ we can construct a canonical (i.e. no choices involved) splitting field $$M$$ for $$f$$, together with an injective map $$M \hookrightarrow \NN$$, as follows: if $$f$$ is a product of linear factors take $$M=L$$. Otherwise, there is a unique irreducible non-linear factor $$g$$ of $$f$$ with $$j(g)$$ minimal. Construct the simple extension $$L' = L[X]/(g)$$. By the above, there is a canonical choice of maps $$i' : L' \hookrightarrow \NN$$ and $$j' : L'[X] \hookrightarrow \NN$$. Now apply the same process to $$f$$ over $$L'$$. After at most deg($$f$$) steps we end up with a splitting field $$M$$.

So there is a canonical way to write down each splitting field $$K_n$$ in the procedure described, given an enumeration of $$K$$, and the axiom of choice isn't required.

### Notes on Lecture 17

In this lecture I proved the useful fact that, for a monic separable polynomial $$f\in \ZZ[X]$$, its Galois group contains a conjugate (in $$S_n$$) of the Galois group of its reduction mod $$p$$.

The best way to prove this is using Algebraic Number Theory, which is the subject of next term's Number Fields course. I will explain how it works here for the benefit of the curious or impatient (who will need to fill in all the missing details!):

Let $$\{x_i\}$$ be the roots of $$f$$ in a splitting field $$L$$. Let $$R=\ZZ[x_1,\dots,x_n] \subset L$$. Algebraic number theory tells us that $$R$$ is a finitely generated $$\ZZ$$-module of rank $$N = [L:\QQ] = \#G$$, and that $$1/p \notin R$$. So $$pR \subset R$$ is a proper ideal, and by Zorn's Lemma$${}^*$$ there is a maximal ideal $$P\subset R$$ containing $$p$$. The quotient $$F = R/P$$ is then a finite field of characteristic $$p$$, and under the quotient homomorphism $$\phi : R \to F$$ the factorisation $$f=\prod(X-x_i)$$ is mapped to a factorisation $$\bar{f} = \prod (X-\bar{x}_i)$$, where $$\bar{x}_i = \phi(x_i)$$, and $$F = \FF_p(\bar{x}_i, \dots, \bar{x}_n)$$. Assume that $$\bar{f}$$ is separable, so that the $$\bar{x}_i$$ are distinct. So we have simultaneously numbered the roots of $$f$$ in $$L$$ and of $$\bar{f}$$ in $$F$$. We'll show that, with this numbering, $$H = \Gal(\bar{f}/\FF_p) \subset G = \Gal(f/\QQ)$$.

The Galois group $$G$$ maps $$R$$ to itself. Let $$H'$$ be the subgroup of $$G$$ consisting of those elements which map $$P$$ to itself. Then any $$\sigma\in H'$$ gives an automorphism of $$F = R/P$$ by $$\sigma\colon x+P \mapsto \sigma(x)+P$$, giving a homomorphism $$H' \to H$$, which is injective. We want to prove that it is a bijection.

Let $$\{ P_i | 1\le i\le s \}$$ be the set of maximal ideals of $$R$$ containing $$p$$, and let $$m_i = [ R/P_i : \FF_p ]$$. The Chinese Remainder Theorem says that

$R/pR \simeq R/P_1\times \cdots\times R/P_s.$

Because $$R$$ is a free $$\ZZ$$-module of rank $$N$$, $$\# R/pR = p^N$$ and so this isomorphism gives $$N = \sum m_i$$. Now suppose that the set of $$G$$-conjugates of $$P$$ is $$\{ P_1=\sigma_1(P), \dots, P_r = \sigma_r(P) \}$$. Then $$\sigma_i \colon R/P \to R/P_i$$ is an isomorphism, so $$m_i = [R/P:\FF_p]=\#H=m$$ say for $$1\le i\le r$$. In particular $$N \ge mr$$. On the other hand, $$rm \ge r\#H' = N$$ by Orbit-Stabiliser Theorem. So we have equality everywhere, which means that $$H = H'$$ as required. For good measure we have also shown that the $$P_i$$ are all conjugate under $$G$$, and all the $$m_i$$ are equal.

( $${}^*$$This is overkill: the index of $$pR$$ in $$R$$ is $$p^N$$, and therefore the (finite) ring $$R/pR$$ has maximal ideals without using Zorn's Lemma.)