# Corollary 2.5 from Atiyah-Macdonald small question

I'm having a doubt in the proof of Corollary 2.5 from Atiyah-Macdonald's Introduction to Commutative Algebra (page 21).

The proof is simple: apply Proposition 2.4 (basically a version of Cayley-Hamilton's theorem) with the identity module homomorphism: from $$\phi^n+a_1\phi^{n-1}+...+a_n \text{ and } \phi(x)=x,$$ we get $$x + a_1 x + a_2 x + ... + a_{n-1}x+a_n = x(1+a_1+...+a_{n-1})+a_n.$$

Proof follows by picking $$x=1+a_1+...+a_n$$ as if you could put in evidence the factor $$x$$ from the previous expression. But you can't. There is no $$x$$ multiplying the last $$a_n$$.

Can someone go through some trouble and explain what I'm missing? Thanks in advance.

• Sorry, what is the statement of corollary 2.5? – jgon Mar 5 '20 at 3:02
• "If M is a finitely generated A-module, I an ideal of A such that $I.M=M$, then there exists $x\in A$ such that $x-1 \in I$ and $xM=0$" – Marra Mar 5 '20 at 3:11

I'm having a bit of trouble parsing what you're saying. I did look up the version in A-M. in A-M, $$x$$ is an element of $$A$$, defined to be the sum of the coefficients of the characteristic polynomial of $$\phi=\textrm{id}_M$$.

In detail the proof goes as follows:

Let $$A$$ be a commutative ring, $$M$$ a finitely generated $$A$$ module. If $$I\subseteq A$$ is an ideal such that $$IM=M$$, then there exists $$x\in A$$, $$x\equiv 1 \pmod{I}$$ such that $$xM=0$$.

Proof

Let $$\phi=\textrm{id}_M$$. $$\phi(M)=M=IM$$, so $$\phi(M)\subseteq IM$$. Then by 2.4 (generalized Cayley-Hamilton kinda) $$\phi$$ satisfies a polynomial equation $$\phi^n + a_1\phi^{n-1}+\cdots + a_n=0,$$ with $$a_j\in I^j$$. Let $$x=1+a_1+\cdots + a_n$$. Certainly $$x\equiv 1\pmod{I}$$. Then if $$m\in M$$, $$xm = (1+a_1+\cdots a_n)m = m + a_1m + \cdots a_nm = \phi^n(m) + a_1\phi^{n-1}(m)+\cdots a_nm=0.$$ Hence $$x$$ annihilates $$M$$, as desired.

• Exactly. You first equation with $\phi$ should be equal to zero. The last element $a_n$ has no $x$ multiplying it. Now you need to compare this equation to your last one, but in the last one there is $m$ times $a_n$ . Why can you use the first to reduce that the second is zero if they have this difference? – Marra Mar 5 '20 at 9:40
• @Marra, please try to use $x$ to mean what it means in A-M and my answer. I suspect what you're asking about now has to do with $a_n$ being the constant term of the polynomial equation. The point then is the mapping into the endomorphism ring sends an element $a$ of $A$ to the endomorphism $\phi(m)=am$. Thus $a_n$ is acting as it should. I'll try to come back and edit my answer later, to add this explanation in and expand on it, but hopefully this is helpful. – jgon Mar 5 '20 at 15:50
• I think I got it. Each $a_n$ is actually the A-module homomorfism $x\mapsto a_n\cdot x$ is that right? – Marra Mar 6 '20 at 12:02
• @Marra Sorry for the slow reply, but yes, exactly. – jgon Mar 9 '20 at 3:51