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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?

Thanks in advance!

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  • $\begingroup$ 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. $\endgroup$ – Neil Strickland May 21 '12 at 8:20
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    $\begingroup$ @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. $\endgroup$ – Martin Brandenburg May 21 '12 at 8:37
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    $\begingroup$ I believe Martin is right and said so here: mathoverflow.net/questions/41836/… $\endgroup$ – Phillip Williams May 21 '12 at 22:09
  • $\begingroup$ 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. $\endgroup$ – Rand al'Thor Dec 29 '14 at 21:22
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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.)

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  • $\begingroup$ @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. $\endgroup$ – gyashfe Nov 10 '19 at 8:48

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