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 $r$?

$\endgroup$
2
  • $\begingroup$ The correct formula is $\wedge^n(M\oplus N) = \oplus_{p+q=n} \wedge^p(M) \otimes \wedge^q(N)$. $\endgroup$ Commented Jan 5, 2013 at 1:04
  • $\begingroup$ @MartinBrandenburg, Thanks, fixed. $\endgroup$
    – ougao
    Commented Jan 5, 2013 at 1:12

2 Answers 2

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 $2\lt 2$.

$\endgroup$
3
  • $\begingroup$ 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
    Commented Jan 5, 2013 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. $\endgroup$ Commented Jan 5, 2013 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
    Commented Jan 5, 2013 at 10:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .