Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would appreciate help on what should be an easy concept in the proof of a corollary leading up to Nakayama's Lemma.

This link to mathoverflow.com (in the green highlighted section) gives the development as presented in "Atiyah and Macdonald" as well as "Reid."

http://mathoverflow.net/questions/41836/elementary-proof-of-nakayamas-lemma

My question pertains to the second corollary (as in "Reid"):

If $M$ is a finite $A$-module and $M = IM$ then there exists an $x \in A$ such that $x \equiv 1$ mod $I$ and $xM = 0$.

I understand the use of $\phi = id_M$ in the relation of maps to get: $1 + a_1 + \dots + a_n = 0$ with $a_i \in I$. And $x$ is set equal to this, giving $xM = 0$.

Also this satisfies $x \equiv 1$ mod ($I$).

Here is my question:

How can $x \in A$ be = $0$ and be $\equiv 1$ (mod $I$)?

Thanks for straightening out what must be an error in my math understanding.

share|improve this question

2 Answers 2

up vote 4 down vote accepted

Let us review the previous proposition in Atiyah - Macdonald used in the proof of the Corollary above:

Proposition (Cayley - Hamilton): Let $M$ be a finitely generated $A$ - module, $I$ an ideal of $A$ and $\phi$ an $A$ - module endomorphism of $M$ such that $\phi(M) \subseteq IM$. Then $\phi$ satisfies an equation of the form $$\phi^n + a_1 \phi^{n-1} + \ldots + a_n = 0$$ where $a_i \in I$.

Now we want to apply this to prove your corollary above. What we are doing is this: Choose $x = 1 + a_1 + \ldots + a_n$. It is clear that $x \equiv 1 \mod I$ because $x - 1 = a_1 + \ldots + a_n = $ sum of guys in $I$. Now why is it that $xM= 0$? Well it is not because $x = 0$ but because $x$ is in the annihilator of $M$. This does not mean that $x = 0$. So how do we conclude for this $x$ that $xM = 0$? Well take any element in here, say $xm$ for $m\in M$. Now $\phi$ is the identity map so $\phi(xm) = x\phi(m) = xm$. However on the other hand

$$\begin{eqnarray*} \phi(xm)&=& \phi( (1 + a_1 +\ldots + a_n)m) \\ &=& (1+ a_1+ \ldots + a_n)\phi(m) \\ &=& \phi(m) + a_1\phi(m) + \ldots + a_nm\\ &=& \phi^n (m) + a_1\phi^{n-1}(m) + \ldots + a_n(m) \\ &=& (\phi^n + a_1\phi^{n-1} + \ldots + a_n)(m)\\ &=& 0\cdot m\\ &=& 0 \end{eqnarray*}$$

from which we conclude that $xm = 0$. But this was any element in $xM$ and so $xM= 0$.

share|improve this answer
    
Thanks - I appreciate it. –  Andrew Oct 28 '12 at 0:22

Ben's answer shows the correct way to conclude the result, but I thought it could help to explain where your reasoning goes astray.

When you have the equation of the form $\phi^n + a_1 \phi^{n-1} + \cdots + a_n \text{id}_M=0,$ that is an equation in the ring of endomorphisms of $M.$ It says the map $\phi^n + a_1 \phi^{n-1} + \cdots + a_n \text{id}_M : M \to M$ is the same as the map $0 : M \to M.$

You tried to plug in $\phi=\text{id}_M$ and also evaluate the map at $1$ to get $1+a_1+\cdots+a_n=0.$ You then set $x=1+a_1+\cdots+a_n$ and wondered how $x=0$ and $x\equiv 1 \pmod I$ at the same time. The mistake was that all these operators are maps from $M$ to $M,$ so actually you can only plug in $1_M$ to get $1_M + a_1 1_M + \cdots + a_n 1_M=0$ as an equality in $M.$ This not actually an element of $A$! So it doesn't fit the conditions of how we need to pick our $x,$ and in fact $xM$ is not even a defined object.

What Reid and Atiyah-Macdonald do is pick $x=1_R + a_1 + \cdots + a_n,$ which is indeed an element of $A,$ and the proof proceeds as in Ben's answer.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.