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.