# Inductive proof of a version of Nakayama's lemma

I have already asked this at math.stackexchange, but since no one answered there after my edit, I decided to try here, although it might be a non-research level question.

The following version of Nakayama's lemma is from Matsumura's Commutative Ring Theory:

Let $M$ be a finitely generated $A$-module, $I\subseteq A$ an ideal s.t. $IM=M$. Then there exists an $a\in A$ with $a\equiv 1\pmod{I}$ and such that $aM=0$. In particular, if $I\subseteq\operatorname{rad}(A)$ we have $M=0$.

After the proof of this via a generalized Cayley-Hamilton, he mentions that the result 'can easily be proved [..] by induction on the number of generators of $M$.' I wonder: how? I tried doing it similarly to the inductive proof of the 'in particular' part, but it didn't work out for me (see MSE for more information on what I think I was doing wrong).

Wouldn't I need to be able to find an $N\subseteq M$, $IN=N$, with fewer generators than $M$ in a somewhat obvious way to use the induction hypothesis? How could I do this?

• You need to do it the other way around. Let $N$ be generated by the first generator of $M$, then $M/N$ needs fewer generators than $M$ and satisfies $I.(M/N)=M/N$, and you can take it from there. – Neil Strickland May 21 '12 at 8:20
• @Neil: What should we do with some $a \equiv 1$ mod $I$ satisfying $a M \subseteq N$? @Rand: I think that Matsumura only refers to the particular case that $I \subseteq jac(R)$, this can be done by induction. The general form is equivalent to Cayley-Hamilton and no reasonable induction is possible. – Martin Brandenburg May 21 '12 at 8:37
• I believe Martin is right and said so here: mathoverflow.net/questions/41836/… – Phillip Williams May 21 '12 at 22:09
• Hey InvisiblePanda, I saw your name in chat.SE and couldn't resist contacting you! I think we have a book series in common :-) The name of this room is pretty funny. – Rand al'Thor Dec 29 '14 at 21:22

After a little work I think I found an inductive proof:

Let $$M$$ be finitely generated by $$n$$ elements. We perform induction on $$n$$. If $$n=1$$, $$M=\langle m\rangle$$ and there is some $$x\in I$$ such that $$xm=m$$. Thus $$1-x$$ annihilates $$M$$.

Assume by induction that we proved the claim for some $$n-1$$ and suppose $$M = \langle m_1,\ldots,m_n \rangle.$$ Consider $$N = M / \langle m_n \rangle$$. It is generated by $$n-1$$ elements and satisfies $$N=IN$$, so there exists $$x \in I$$ such that $$1+x$$ annihilates $$N$$. Therefore $$(1+x)M \subset \langle m_n \rangle.$$ Since $$m_n \in IM$$, $$(1+x)m_n \in (1+x)IM = I(1+x)M \subset I\langle m_n \rangle.$$ Take $$y\in I$$ such that $$(1+x)m_n = ym_n$$. Then $$(1+x-y)m_n = 0$$ and $$(1+x)(1+x-y) \in 1+I$$ annihilates all $$M$$.

In fact, one can use the same method to replace the determinant trick entirely. Unfortunately the resulting polynomial has high degree.

Claim: Let $$M$$ be finitely generated and $$\varphi:M \rightarrow M$$ an $$R$$-module homomorphism satisfying $$\varphi(M) \subset IM$$ where $$I$$ is an ideal (possibly all $$R$$). Then $$\sum_{i=0}^{k-1} a_i \varphi^i + \varphi^k = 0$$ on $$M$$ for some $$a_0,\ldots,a_{k-1} \in I$$.

Proof: This is really the same as the previous proof, just replacing $$\mathrm{id}_M$$ by $$\varphi$$ in some places and keeping track of what this entails.

The case $$M=\langle m \rangle$$ is clear: $$\varphi(m) = xm$$ for some $$x \in I$$, so $$\varphi - x = 0$$ on $$M$$.

Assume by induction that we proved the claim for some $$n-1$$ and suppose $$M = \langle m_1,\ldots,m_n \rangle.$$ Consider $$N = M / \langle \{\varphi^i (m_n)\}_{i\ge 0} \rangle$$. It is generated by $$n-1$$ elements and satisfies $$\varphi(N) \subset IN$$. Hence there exists some polynomial $$p(x)$$ (monic, with all but the top coefficient in $$I$$) satisfying $$p(\varphi) = 0$$ on $$N$$. Therefore $$p(\varphi)(M) \subset \langle \{\varphi^i (m_n)\}_{ i\ge 0} \rangle.$$

In fact it suffices to take $$\langle \{\varphi^i (m_n)\}_{0\le i \le k} \rangle,$$ where $$k$$ is some natural number (take a maximum over the powers needed for the elements $$p(\varphi)(m_i),\ \ i \le n$$.)

Since $$\varphi^{k+1}(m_n) \in IM$$, $$(p(\varphi))(\varphi^{k+1}(m_n)) \in p(\varphi)(IM) = I\cdot p(\varphi)(M) \subset I \langle \{\varphi^i (m_n)\}_{0\le i \le k} \rangle.$$

Take $$y_0,\ldots,y_k \in I$$ such that $$q(\varphi) = \sum_{i=0}^k y_i \varphi^i$$ satisfies $$p(\varphi)\circ\varphi^{k+1}(m_n) = q(\varphi)(m_n),$$ and then since $$p(\varphi)\circ\varphi^{k+1} - q(\varphi)$$ is monic with all but the top coefficient in $$I$$, $$p(\varphi)\circ(p(\varphi)\circ\varphi^{k+1} - q(\varphi))$$ gives a polynomial of the required form: composition of polynomials of an endomorphism = multiplication of polynomials. So the induction step gives $$(p(z)^2 \cdot z^{k+1} - q(z)p(z))\restriction_{z=\varphi}$$ where $$y \in I.$$

(edit: thanks to Yakov Varshavsky for pointing out an error near the last step, where I mistakenly thought $$q$$ could be taken to be a single monomial.)

• @Martin-Brandenburg, others: (In response to a comment on the question above, which I don't have the reputation to reply to directly.) It's worth noting that Cayley-Hamilton itself is an elementary computation, so it should not be too surprising that a simpler computation is possible for this use-case (Cayley-Hamilton has significant advantages in general, like a degree bound and an explicit description of the polynomial.) This proof is mainly useful for showing determinants are not somehow necessary for the basic theory, e.g. a book like A-M could have been written entirely without them. – gyashfe Nov 10 '19 at 8:48