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I have a nice solution to the following problem and I thought of writing a paper about it but beforehand, I wanted to ask the problem here to see if this is an easy problem and if you people can solve it easily. If its easy then I will not write a paper about my solution. I know of many people who did not manage to solve the problem so let's see if you can.

For every finitely generated $A$-module $N$ we define $Supp N = V(Ann(N)) \subset Spec(A)$.

$Spec(A)$ denotes the prime ideals of $A$.

$Ann(N)$ denotes all the members a of $A$ such that $aN=0$.

Let $A$ be a commutative noetherian ring. Let $M$ be a finitely generated $A$-module. Prove that for any prime ideal $p\in\operatorname{Supp}M$, $M$ has a quotient isomorphic to $A/p$.

Good Luck :)

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    $\begingroup$ To write down a paper for yourself is always a good idea. It helps you to fix notation / conventions, run through definitions (which can be different for the same object in different references), set down a coherent proof and store the material for the next 50000000 years :) $\endgroup$ – Avitus Aug 4 '14 at 15:11
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    $\begingroup$ @MartinBrandenburg: I must say, this is the one time that I've found your unparalleled capacity for condescension useful. OP: Published papers are for new results, and there is far more material in the literature than you can possibly know right now. Do not be discouraged though - the fact that you attempted to write down a proof (despite being flawed) indicates that you have at least some inclination towards commutative algebra. I urge you to continue in your studies, and learn all you can $\endgroup$ – zcn Aug 4 '14 at 17:18
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    $\begingroup$ @Student: You are certainly welcome to post your proof, either as an answer to this or as another question (I'm not sure which is more appropriate - in any case the flaws will be pointed out to you). A new proof can be enlightening, but typically only when the result is non-trivial, and previous proofs were non-elementary $\endgroup$ – zcn Aug 4 '14 at 17:36
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    $\begingroup$ By the way, you should use the @ to notify a user - @Avitus was not notified of your earlier comment, but now he should be $\endgroup$ – zcn Aug 4 '14 at 17:37
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    $\begingroup$ @Student: One more point - please use dollar signs to render the LaTeX correctly (I personally can't read the answer as it is now). If no one else has pointed out the flaws, then I will try sometime later $\endgroup$ – zcn Aug 4 '14 at 17:53
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The claim is not true. If $A$ is an integral domain which has a non-trivial Picard group, there is some invertible $A$-module $M$ which is not free. But then $M$ has no quotient isomorphic to $A$, since any epimorphism $M \to A$ is an isomorphism (using that $M$ is locally free of rank $1$).

However, we can prove the following: Let $M$ be a finitely generated $A$-module and let $\mathfrak{p} \in \mathrm{supp}(M)$ i.e. $M_\mathfrak{p} \neq 0$. I claim that $M$ admits some quotient $ \neq 0$ which embeds into $A/\mathfrak{p}$.

$M/\mathfrak{p}M$ is an $A/\mathfrak{p}$-module and satisfies $(M/\mathfrak{p}M)_\mathfrak{p} \neq 0$, since otherwise $M_\mathfrak{p}=0$ by Nakayama. Hence, we may replace $A$ by $A/\mathfrak{p}$ (and $M$ by $M/\mathfrak{p}M$) and therefore assume that $A$ is an integral domain and $M$ is a finitely generated $A$-module with $M \otimes Q(A) \neq 0$. It follows that the image of $M \to M \otimes Q(A)$ is non-zero. After replacing $M$ by the image, we may assume that $M$ is torsion-free and $M \neq 0$.

After choosing a basis of $M \otimes Q(A)$ over $Q(A)$, we find that $M$ embeds into $Q(A)^n$ for some $n \in \mathbb{N}$. One of the $n$ projections $M \to Q(A)$ must be non-zero, so we may replace $M$ by the image and assume that $M$ embeds into $Q(A)$ and $M \neq 0$. Choose a finite generating set of $M$ and let $q$ be the product of all the denominators. It follows that $M$ embeds into $q^{-1} A$. But this module is isomorphic to $A$. Hence, $M$ embeds into $A$.

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  • $\begingroup$ What does Q(A) denotes? fraction field of A? $\endgroup$ – Guy L. Aug 4 '14 at 17:20
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$\DeclareMathOperator{\Hom}{\operatorname{Hom}}$There is a well-known criterion for $\Hom$ to vanish:

Proposition: Let $A$ be a Noetherian ring, $M, N$ f.g. $A$-modules. Then $\Hom_A(M, N) = 0$ iff $\text{ann}_A(M)$ contains a nonzerodivisor on $N$.

Taking $N = A/p$ for $p \in V(\text{ann}_A(M))$ gives that $\Hom_A(M, A/p) \ne 0$, i.e. some nonzero quotient of $M$ embeds in $A/p$. As mentioned before, this is the best you can do: e.g. for $M := (2, 1 + \sqrt{-5}) \subseteq A := \mathbb{Z}[\sqrt{-5}]$, $0 \in \text{Supp}(M)$, but there is no surjection $M \twoheadrightarrow A$, since $M_p \cong A_p$ for all $p \in \text{Spec}(A)$, but $M \not \cong A$.

(If $0 \to K \to M \to A \to 0$ is exact, then so is $0 \to K_p \to M_p \to A_p \to 0$ for any $p \in \text{Spec}(A)$, but then $A_p/K_p \cong M_p/K_p \cong A_p \implies K_p = 0$, since no nontrivial quotient of a domain is isomorphic to the domain).

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This was my answer to the question (which turns out to be wrong). Will be glad to hear any interesting comments :)

My solution, in a couple of steps:

1)

If $M$ is generated by 1 member m then if $\exists s\in A-p$ such that $sm=0$ then $\forall a\in A : sam=0$ and thus $sM=0$ and thus $s\in Ann(M)$ in contradiction to the fact that $p \supset Ann(M)$ while $s\in A-p$. Thus I showed that $Ann(m)\subset p$. Define $f:Am=M->A/p$ by $f(am)=a+p$. If $am=bm$ then $(a-b)m=0$ and thus $a-b \in Ann(m)\subset p$. Thus $a+p=b+p$. Thus $f$ is well defined. Now, for any $a\in A : f(am)=p <-> a\in p$ and thus $ker(f)=pm$ and thus by first isomorphism theorem it holds that: $M/pm \cong Am/pm \cong A/p$ as desired.

2)

Define $W$={$L\subset M$ : L has a quotient isomorphic to $A/p$}. By 1) $W$ is non empty. $A$ is a noetherian ring thus $W$ has a maximal member which I denote by $M'$. If $M=M'$ then we are done by the definition of $W$. So suppose that $M\neq M'$.

3)

If $(M/M')_p\neq 0$ then there is $m\in M-M'$ such that $(Am+M'/M')_p\neq 0$. Now by 1) it follows that $Am+M'/M'$ has a quotient isomorphic to $A/p$ (because $Am+M'/M'$ is generated by 1 member $m+M'$). Thus there is some $K\subset Am+M'/M'$ such that $(Am+M'/M')/K \cong A/p$. But there is $K'$ such that $M' \subset K' \subset Am+M'$ and $K'/M'=K$, and it holds by the third isomorphism theorem that: $Am+M'/K'\cong (Am+M'/M')/(K'/M') =(Am+M'/M')/K\cong A/p$. This means that $Am+M'$ has a quotient isomorphic to $A/p$. But $m\in M-M'$ and thus $M'\neq Am+M'$ which is a contradiction to the maximality of $M'$ in $W$. So I proved that $(M/M')_p\neq 0$ can't hold.

4)

Suppose that $(M/M')_p = 0$. $M/M'$ finitely generated by some $t1,...,tk$ because $M$ finitely generated. There is $s_i \in A/p$ for any $1\leq i\leq k$ such that $s_iti=0$. Thus $(\prod_{1\leq i\leq k}s_i)M/M'=0$. Thus $\prod_{1\leq i\leq k}s_i \in Ann(M/M') \subset p$. But p is prime and thus $\prod_{1\leq i\leq k}s_i \in A-p$ because $s_i \in A/p$ for any $1\leq i\leq k$. This is a contradiction. Thus $p\not\in Supp(M/M')$. Thus $p\not\supset Ann(M/M')$ and thus there is some $s\in A-p$ such that $s\in Ann(M/M')$ which gives $sM\in M'$.

5)

Look on the chain: $M \supset M' \supset sM \supset sM' \supset s^2M \supset s^2M'$... Using the morphism m->sm and the first isomorphism theorem we have that: $M/Ann(s^k) \cong s^{k}M$ where $Ann(s^k)={m\in M : s^{k}m=0}$. Thus we get the chain: $M/Ann(s) \supset M/Ann(s^2)$... and the chain $Ann(s) \subset Ann(s^2) \subset$ ... The last one is an ascending chain of ideals in $A$ and since $A$ is noetherian it follows that there is a $k$ such that $Ann(s^k)=Ann(s^{k+1})$ and thus $M/Ann(s^k)=M/Ann(s^{k+1})$ and thus as stated before (using $m->sm$ and first isomorphism theorem) I get that $s^{k}M=s^{k+1}M$. Thus I got that: $s^{k}M=s^{k}M'=s^{k+1}M$, denote this result by (*).

6)

I will now prove that if $M'$ is finitely generated with a quotient isomorphic to $A/p$, then for any $s\in A-p$ it holds that $sM'$ has a quotient isomorphic to $A/p$. Proof: By assumption, $M'$ has a quotient isomorphic to $A/p$. Thus there is some $K \subset M'$ such that $M'/K \cong A/p$.
I will prove first that $K \supset Ann_M'(s)={m'\in M' : sm'=0}$. For any $m' \in Ann_M'(s)$ holds $m' \in K <-> m'+K=K$. Now, $M'/K \cong A/p$ thus there is an isomorphism $f:M'/K->A/p$ that generates a morphism $g:M'->A/p$ such that $g(m')=f(m'+K)$ and since f is one to one it holds that: $K={m' : m'+K=K}={m' : f(m'+K)=p}=ker(g)$ Now, $m' \in Ann_M'(s)$ implies $sm'=0$ and thus since g is a morphism it holds that $sg(m')=g(sm')=g(0)=p$. Since by definition of $g: g(m') \in A/p$, I can denote $g(m')=a+p$ and it holds that $sa+p=sg(m')=p$ and thus $sa\in p$. But $s\in A-p$ and $p$ is prime thus $a\in p$ and thus $g(m')=a+p=p$ which proves that $m' \in ker(g)=K$ as desired.
Thus it holds by the third isomorphism theorem (which required $K \supset Ann_M'(s)$ which i just proved) that $(M'/Ann_M'(s))/(K/Ann_M'(s)) \cong M'/K \cong A/p$. But $m'->sm'$ defines an isomorphism $M'/Ann_M'(s) \cong sM'$. $K/Ann_M'(s)$ is a submodule of $M'/Ann_M'(s)$ and thus there exists $K' \subset sM'$ such that $K' \cong K/Ann_M'(s)$. Thus we get $(M'/Ann_M'(s))/(K/Ann_M'(s)) \cong sM'/K' \cong A/p$. Which means that $sM'$ has a quotient isomorphic to $A/p$ as desired.

7)

I will prove that if $M$ is finitely generated and there is some $s\in A-p$, such that $sM$ has a quotient isomorphic to $A/p$, then it holds that $M$ has a quotient isomorphic to $A/p$. Proof: By assumption there is some $K \subset sM$ such that $sM/K \cong A/p$. We know as already stated that $sM \cong M/Ann(s)$ and thus $K \subset sM$ implies that there is some $K'\subset M/Ann(s)$ such that $K' \cong K$. But any submodule of $M/Ann(s)$ is of the form $T/Ann(s)$ with $Ann(s) \subset T \subset M$. Thus there is some $K''$ such that $Ann(s) \subset K'' \subset M$ and $K''/Ann(s)=K' \cong K$. Thus by the third isomorphism theorem: $M/K'' \cong (M/Ann(s))/(K''/Ann(s)) \cong sM/K \cong A/p$. I found a quotient of $M$ isomorphic of $A/p$ as desired.

8) LAST STEP!!

By definition of $M'$ and $W$, $M'$ has a quotient isomorphic to $A/p$. $M$ is finitely generated and $M' \subset M$ and $A$ is noetherian, thus $M'$ is finitely generated and thus by the 6) step it follows that $s^{k}M'$ has a quotient isomorphic to $A/p$. But according to (*) in step 5) it holds that $s^{k}M=s^{k}M'$ and thus $s^{k}M$ has a quotient isomorphic to $A/p$ ($s^k$ is in $A-p$ because $s\in A-p$ and $p$ is prime). Thus by 7) step we conclude that $M$ has a quotient isomorphic to $A/p$ and this is a contradiction to the maximality of $M'$ in $W$ (I assumed $M\neq M'$ and $M' \subset M$). I came to a contradiction in all cases which contradicts the assumption that $M' \neq M$ and thus $M'=M$ and thus $M$ has a quotient isomorphic to $A/p$.

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    $\begingroup$ You should take a counterexample and check by yourself at which step something goes awry. $\endgroup$ – YCor Aug 4 '14 at 18:06
  • $\begingroup$ @YCor My interest in this is if maybe my technique of proof can be used to prove some other more interesting question. So I posted my proof so that maybe someone will read it and tell me that it can be used for something. $\endgroup$ – Guy L. Aug 4 '14 at 19:37
  • $\begingroup$ @Student: Here is what I see as the main flaw in the argument (it is perhaps best not to begin an endless comment discussion, so take this as you will): the result (*) in step 5 is wrong. The stabilization of $0 :_M s^k$ (which you denote by $Ann(s^k)$) does not imply that $s^kM$ stabilizes - consider $M = A = k[x]$, $s = x$, so that $\text{ann}(s^k) = 0$ for all $k$, but $(x^k) \supsetneq (x^{k+1})$ (although they are isomorphic). $\endgroup$ – zcn Aug 5 '14 at 0:19
  • $\begingroup$ For the rest - I must admit that I was not patient enough to read past step 5 (or to even check the details of step 3). I will simply say that much of the argument could be dramatically shortened - e.g. step 1 can be proven in under 2 lines, and in step 4 one could simply take just the first and last two sentences. Unfortunately, I don't think the techniques in the rest of the argument can be used to make interesting deductions $\endgroup$ – zcn Aug 5 '14 at 0:22
  • $\begingroup$ @zcn : I see, well was worth an effort. Thanks for your time :) $\endgroup$ – Guy L. Aug 5 '14 at 19:52

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