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In this note I just want to record some facts about commutative algebra about which I often just think “that’s something to do with Nakayama’s lemma, which is a thing” and try to do them justice. In particular, I want to prove the following very nice fact.

Theorem: Let $A$ be a local ring (one with a unique maximal ideal $m$). Then if the finitely-generated $A$-module $M$ has the property that $M/mM = (x_1+mM ,x_2 + mM,...,x_n + mM)$ as a finite dimensional $A/mA$ vector space, it follows that $M = (x_1,...,x_n)$. In other words, we can lift a basis of $M/mM$ to get a generating set of $M$.

First, some remarks about why this isn’t trivial and why all the conditions are necessary. The hypothesis of locality is essential. Indeed, if there are two maximal ideals $m_1,m_2$, then $A/m_2$ is an $A$-module whose image under projection by $m_1$ is the zero vector space. This has as a basis the empty set, which certainly doesn’t lift to a generating set of the whole module. In general, having loads of maximal ideals floating about will often result in such ‘cancellation’ behaviour. The hypothesis that $M$ is finitely generated is also essential. Indeed, an important aspect of the proof will involve taking polynomial equations whose degree is the dimension of the space.

So how are we going to prove it? Well, the module $N = (x_1,...,x_n)$ is certainly a submodule of $M$, and in fact we can write $M = N + mM$ (since the $x_i$ generate the quotient module, $N \rightarrow M \rightarrow M/mM$ is surjective). The essential ingredient is therefore to show that $mM$ is substantially smaller than M. In fact, we want it to be so much smaller that in fact $M/N = mM/N$ is trivial. What we have just written leads us to the following plausible conjecture.

Plausible but Optimistic Conjecture: If $M$ is a finitely generated $A$-module with $aM = M$, for some proper ideal $a$, then $M=0$. Or, to put it another way, multiplication by an ideal of $A$ must always make a nonzero module smaller.

This is plausible because multiplication by ideals feels like it should usually make things smaller. However, it is optimistic because it is obviously false outside local rings (multiplying a cyclic group of order 3 by 2 doesn’t make it smaller). Our hope is that with a single maximal ideal governing behaviour, $A$-modules will be unable to ‘dodge’ having to get smaller in such a way. What nice properties do local rings have that will facilitate a proof of non-dodgingness?
Why local rings are lovely: Take $a$ any proper ideal, and consider the coset $1+a$. This certainly has no intersection with $m$, and therefore every element of $1+a$ is a unit!

Contrast with the horrible old integers, where 2+1 = 3, which isn’t a unit. So yes, maybe our conjecture is true. In fact, given the loveliness of local rings, it will now suffice to find an element $x \in 1+a$ which kills $M$ (since then $1=x^{-1}x$ will kill $M$, proving that $M = 0$).

But we have an obvious way to get such an element, using the fact that $M$ is finitely generated. What does the condition $M = aM$ mean? Well, take a bunch of generators $e_1,...,e_s$ for $M$, and write out equations for each of them in terms of elements of $aM$:

$e_i = \sum_j a_{ij} e_j.$

Then, multiplying by the adjugate, we deduce that $x := \text{det}(\delta_{ij} - a_{ij}) = 0$ as an endomorphism of the module, and this determinant expands to a give precisely a number in $1+a$ which therefore annihilates $M$. So our plausible conjecture was true, and we have proved the theorem.

The content of this post is actually part of a more general story, centred around Nakayama’s lemma, which is just our ‘plausible conjecture’ but with the locality condition replaced by the (what feels like almost as stringent) condition that $a$ must be contained in the intersection of all maximal ideals of $A$. Chapter II of Atiyah and MacDonald is the obvious reference.

Probability and Measure Notes
I have uploaded my course notes for part II Probability and Measure. The course is one which occupies an odd place in the tripos, containing a huge amount of important analysis interspersed with probability theory in a way that I found slightly confusing. Anyway, the above notes put things in what is in my personal view a more intuitive order, though I have omitted details of many proofs, and I miss out some of the topics which I feel could maybe be usefully dropped from the course anyway.

I think Probability and Measure is possibly the hardest course in part II (certainly it will be once someone writes a canonical set of notes for Algebraic Geometry), and hope these notes are useful for giving a slightly different flow and a bit more motivation to complement Norris’s very thorough and content-rich but perhaps overly concise notes.

While idly browsing I came across a rather nice paper of Zhou and Markov (2009), which led to the first time I have ever learnt a proof of the famous and slightly random fact that $\pi$ is irrational.

The idea is to construct a sequence $A_n$ of expressions involving integrals, which are on the one hand (integrating by parts) given by a recurrence relation which guarantees that they are positive integers, but which on the other hand must clearly tend to zero, and of course this gives a contradiction. The rabbit from the analyst’s hat seems to be to consider the functions $f_n(x) = (\pi x - x^2)^n$ and then integrate through by a sine function. This then very quickly just works. Ok, let’s do it.

Suppose $\pi = \frac{a}{b}$, and consider $A_n = \frac{1}{n!} \int_{0}^\pi (a x - b x^2)^n \sin x \rightarrow 0$. However, we will also show that the $A_n$ are positive integers. Indeed, $A_0 = 2, A_1 = 4b$ and for $n\geq 2$, integrating by parts gives that

$A_n = b(4n-2) A_{n-1} - a^2 A_{n-2} \in \mathbb{N}$.

This is secretly a post to experiment with Luca Trevisan‘s amazing LaTeX-to-Wordpress tool, and maybe do a token amount of revision in the process. Enjoy, and please point out if I could have achieved anything more efficiently. Note that though this proof is logically self-contained, it requires the reader to already be pretty familiar with the subject, so is probably quite unfriendly to others (sorry).

We roughly follow Teruyoshi Yoshida‘s `minimal notes’ to prove the fundamental theorem as efficiently as possible using basic linear algebra. I have probably weakened the generality of his results a bit but hopefully made the presentation slightly friendlier along the way. We pass via two important propositions (the tower laws and Dedekind’s lemma) to a proof that whenever ${F/K}$ is Galois, and ${F/L}$ a subextension, ${F/L}$ is Galois. We then construct the final important algebraic ingredient, which gives us an all-important upper bound on the degree of an extension over a fixed field.

If ${F/L}$, ${L/K}$ finite, then ${F/K}$ is finite and

$\displaystyle [F:L] [L:K] = [F:K].$

Also, for any ${E/K}$,

$\displaystyle |\text{Hom}_K(F, E)| = \coprod_{\sigma \in \text{Hom}_K(L,E)} |\text{Hom}_L(F, E_\sigma)|.$

Proof: If ${e_1,...,e_n}$ a basis for ${L/K}$, and ${f_1,...,f_m}$ a basis for ${F/K}$ then ${e_1f_1, e_1f_2, ....,e_nf_m}$ is a basis for ${M/K}$, proving the first part. The second is a simple check that the sets on both sides contain each other, by restricting a map to ${L}$ and then viewing as an appropriate ${L}$ embedding of ${F}$ (the other direction being trivial). $\Box$

If ${\sigma_1,...,\sigma_n}$ are distinct elements of ${\text{Hom}_K(F,E)}$, they are linearly independent considered as elements of ${E^F}$. In particular, since they are all linear maps of ${K}$-vector spaces ${F \rightarrow E}$, we have

$\displaystyle |\text{Hom}_K(F,E)| \leq [F:K].$

Proof: Suppose not, so there exists a dependence relation between the ${\sigma_i}$. Let ${\sum_{i=1}^k \lambda_i \sigma_i = 0}$ be such a relation with ${k}$ minimal. Since they are distinct morphisms, we can pick ${x \in F}$ with ${\sigma_k(x) \not= \sigma_1(x)}$. Applying the dependence relation twice and subtracting, for any ${y \in F}$,

$\displaystyle 0 = \sum_{i=1}^k \lambda_i (\sigma_i(xy) - \sigma_k(x)\sigma_i(y)) = (\sum_{i=1}^{k-1} \lambda_i (\sigma_i(x) - \sigma_k(x))\sigma_i)y.$

So we have found a dependence relation with ${k}$ smaller, a condtradiction. $\Box$

If ${F/K}$ is Galois (in the sense that ${|\text{Aut}_K(F)| = [F:K]}$) then for any intermediate field ${F/L/K}$, ${F/L}$ is Galois.

Proof: Using the tower law for homomorphisms,

$\displaystyle [F:K] = |\text{Aut}_K(F)| = \coprod_{\sigma \in \text{Hom}_K(L,F)} |\text{Hom}_L(F, F_\sigma)|.$

So since there are at most ${[L:K]}$ ${K}$-homomorphisms ${L \rightarrow F}$, and at most ${[F:L]}$ ${L}$-homomorphisms in each factor, the tower law for fields implies we have equality everywhere, and in particular that (taking ${\sigma: L \rightarrow F}$ as the inclusion map) ${|\text{Hom}_L(F, F)| = [F:L]}$. $\Box$

We have still not yet talked about fixed fields. This will be an important part of the fundamental theorem of Galois theory, which is about subfields fixed by subgroups of the automorphism group of an extension. Indeed, one way to define a Galois extension ${F/K}$ is as one with the property that ${K}$ is precisely the set of elements of ${F}$ fixed by ${\text{Aut}_K(F)}$. Let us see why this is true.

If ${F/K}$ a finite field extension, ${H \leq \text{Aut}_K(F)}$, then

$\displaystyle [F:F^H] = |H|.$

Proof:It is clear that ${[F:F^H] \geq |H|}$ from proposition 1.2. To do the other direction, let us use a similar algebraic method to the proof of proposition 1.2. Suppose ${n = [F:F^H] > |H| = m}$, so let ${\{\sigma_1,...,\sigma_m \} = H}$, and ${x_1,...,x_n}$ be a basis for ${F/F^H}$, and consider the system of linear equations (in variables ${c_1,...,c_n}$ taking values in ${F}$)

$\displaystyle c_1 \sigma_i(x_1) + c_2 \sigma_i(x_2) + ... + c_n \sigma_i(x_n) = 0\ \ \ \ \ (i=1,2,...,m).$

There are fewer equations than unknowns, so there must exist nontrivial solutions to the system. Let us take such a solution with minimal ${k}$

$\displaystyle c_1 \sigma_i(x_1) + c_2 \sigma_i(x_2) + ... + c_k \sigma_i(x_k) = 0\ \ \ \ \ \forall i.$

Dividing through by ${c_k}$ and noting that since the ${x_i}$ are a basis, some ${c_l}$ must lie outside ${F^H}$, take any ${\sigma \in H}$ not fixing ${c_l}$. Since ${H}$ is a group, premultiplying by ${\sigma}$ permutes its elements. Therefore, taking ${\sigma}$ of each of the above equations and subtracting them from another corresponding equation will cancel the term in the ${k}$th position and fail to cancel a term in the ${l}$th position, giving a nontrivial solution with smaller ${k}$. $\Box$

We have now assembled enough tools to prove the fundamental theorem of Galois theory.

Let ${F/K}$ be Galois. The mappings ${L \mapsto \text{Aut}_L(F)}$ and ${H \mapsto F^H}$ are mutually inverse.

Proof: Obviously ${F^{\text{Aut}_L(F)} \supset L}$ and ${\text{Aut}_{F^H}(F) \supset H}$, and we shall establish both are equalities by dimension counting. Since ${F/L}$ is Galois by theorem 1.3, proposition 1.4 gives ${[F: F^{\text{Aut}_L(F)}] = |Aut_L(F)| = [F:L]}$. Since ${F/F^H}$ is Galois by theorem 1.3, proposition 1.4 gives ${|Aut_{F^H}(F)| = [F:F^H] = |H|}$. $\Box$

The fundamental theorem of algebra (polynomials have roots in the complex numbers) is a classic result which one has very little idea of how to do at school, and at the end of an undergraduate course has about four different ways to prove, maybe using Liouville’s theorem, facts about maps between Riemann spheres (compactness implies surjective), maybe using Rouche’s theorem and facts about winding numbers, which amounts to the intuitive topological proof involving thinking about small circles and big circles and how many times they orbit the origin.

While reading around, I stumbled across the first almost purely algebraic proof I’ve ever seen in Lang’s epic “Algebra” book. I’d say purely algebraic, but we are going to need the fact that every odd degree polynomial has a solution in the reals, which is an analytic fact, but a much more obvious one than Liouville’s theorem. However, from this we’ll be able to get to the fundamental theorem of algebra just using basic Galois theory and group theory, and the easily verifiable fact that every quadratic equation with complex coefficients has a complex root. Indeed, once one knows that these are the ingredients, the proof is just “throw them all into a pan.”

Here we go. Consider an arbitrary finite extension $E/\mathbb{C}$. Since the reals have characteristic zero, E is separable over $\mathbb{R}$ and we may therefore take $F/\mathbb{R}$ Galois and containing $E$. Consider the Sylow 2-group of the Galois group, and note by Galois theory that it fixes a field of odd degree over $\mathbb{R}$. Since there are no nontrivial odd algebraic extensions of $\mathbb{R}$ (there are no odd degree irreducible polynomials except the linear ones), this must in fact be the entire Galois group. Consider the subgroup corresponding to $\text{Gal}(F/\mathbb{R}(i))$. If nontrivial, this is a 2-group, so has a subgroup of index 2, and hence by Galois theory $\mathbb{R}(i)$ has a degree 2 extension. But this contradicts the fact that all quadratic equations with complex coefficients have complex roots, so $\text{Gal}(F/\mathbb{R}(i))$ is trivial, hence $E =\mathbb{C}$, so the complex numbers have no nontrivial algebraic extensions.