1
$\begingroup$

I have two problems related to exterior powers, and I have little background on the exterior algebra of an $R-$mod.

Problem 1. For any ideal $J_i\subset R$, which is a commutative ring, consider $M=R/J_1\oplus\cdots\oplus R/J_k$ as an $R-$mod.

Does the exterior power $\wedge^nM=0$ implies $k<n?$ If Yes, Why?


I ask it because I want to solve the following problem:

Prove if $M$ is a finitely generated module over a PID, and $\wedge^{r+1}M=0$, then $M$ can be generated by $r$ elements.

To prove this, I used $\wedge^{r+1}(N_1\oplus N_2)=\bigoplus_{i+j=r+1}\wedge^{i}(N_1)\otimes \wedge^{j}(N_2)$(I think this is right) to reduce to the above problem.


Problem 2. Let $I_1\subset I_2\subset\cdots I_p$ and $J_1\subset J_2\subset\cdots J_p$ be chains of proper ideals in a commutative ring $A$. Prove that if $$A/I_1\oplus \cdots \oplus A/I_p\cong A/J_1\oplus \cdots \oplus A/J_q,$$ then $p=q$ and $I_r=J_r$ for all $r$.


Of course, when Problem 1 has a positive answer, then $p=q$, but why $I_r=J_r$ for all $r$?

$\endgroup$
  • $\begingroup$ The correct formula is $\wedge^n(M\oplus N) = \oplus_{p+q=n} \wedge^p(M) \otimes \wedge^q(N)$. $\endgroup$ – Martin Brandenburg Jan 5 '13 at 1:04
  • $\begingroup$ @MartinBrandenburg, Thanks, fixed. $\endgroup$ – ougao Jan 5 '13 at 1:12
1
$\begingroup$

P1. Let $M$ be generated by $k$ elements, say $M = R/J_1 \oplus \dotsc \oplus R/J_k$ with $J_1 \subseteq \dotsc \subseteq J_k \subseteq R$, and $\bigwedge^n(M)=0$. We claim that $M$ can be generated by $<n$ elements. So assume $k \geq n$. For $p>1$ we have $\bigwedge^p(R/J_i)=0$, since this is a quotient of $\bigwedge^p(R)=0$. Thus, we have $$0 = \bigwedge^{n}(M) = \bigoplus_{p_1+\dotsc+p_k=n,~p_i \leq 1} \bigwedge^{p_1}(R/J_1) \otimes \dotsc \otimes \bigwedge^{p_k}(R/J_k),$$ and all summands vanish. For $p_1=\dotsc=p_n=1, p_{n+1}=\dotsc=p_k=0$ we get $0 = R/J_1 \otimes \dotsc \otimes R/J_n=R/(J_1+\dotsc+J_n)=R/J_n$, i.e. $J_n=R$. It follows that $M=R/J_1 \oplus \dotsc \oplus R/J_{n-1}$ is generated by $<n$ elements.

P2. Hint: Think of annihilators and induction.

$\endgroup$
1
$\begingroup$

The answer to Problem 1 is "no": if $M=R/J_1\oplus\cdots\oplus R/J_k$, then $\wedge^nM=0$ does not imply $k<n$.

For example take $R=\mathbb Z$ and $M=\mathbb Z/(2) \oplus \mathbb Z/(3)$.
Then $\bigwedge_\mathbb Z^2 M=\mathbb Z/(2) \otimes_\mathbb Z \mathbb Z/(3)=0$ but we cannot conclude that $2\lt 2$.

$\endgroup$
  • $\begingroup$ Isn't $M \cong \mathbb{Z}/(6)$ generated by a single element? So maybe a minimality assumption on $k$ in the representation of $M$ as $M = R/J_1 \oplus \cdots \oplus R/J_k$ is missing. $\endgroup$ – Martin Jan 5 '13 at 9:51
  • $\begingroup$ I'm answering Problem 1 as it is stated, not the motivation below the horizontal line starting with "I ask it because I want..." I have edited my answer in order to emphasize this. $\endgroup$ – Georges Elencwajg Jan 5 '13 at 9:55
  • $\begingroup$ I see. You are of course perfectly right. I should have checked the question again before commenting. Sorry for the bother. $\endgroup$ – Martin Jan 5 '13 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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