Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found this problem and I don't understand the solution. I will appreciate your help. Let $A = \mathbb{Q}[X_1,...,X_n,...], a = (X_1^2,...,X_n^2,...)$ and $ M = A/a$. Show that $Ass_A (M) = \emptyset $. Why isn't $Ass_A(M) = (X_1,...,X_n,...)$ ?

I'm writing this because the site is giving me this error : "Oops! Your question couldn't be submitted because:

* It does not meet our quality standards.".  This feels verry strange!!!
share|cite|improve this question
Here's an explanation of that error and what you can do to avoid getting it. – joriki Sep 17 '11 at 12:15
Perhaps because the question repeatedly includes the word "ass" ? – Daniel McLaury Sep 17 '11 at 15:42
up vote 3 down vote accepted
  1. $Ass_A(M)$ denotes the set of associated primes of $M$, i.e. the set of prime ideals $\mathfrak p$ of $A$ for which there is an embedding $A/\mathfrak p \hookrightarrow M$. (Note that in particular that $Ass_A(M)$ is not an ideal of $A$ --- unlike the annihilator of $M$, which is an ideal of $A$. This is why the assertion $Ass_A(M) = \emptyset$ even makes sense.)

  2. A contextual remark: if $M$ is an module over a Noetherian ring, then $Ass_A(M)$ is always non-empty. The point of problem you are asking about is to show that this can be false for non-Noetherian $A$ (such as the $A$ in your question).

  3. If $\mathfrak p$ is an element of $Ass_A(M)$, then (as you implicitly observe in your post) it contains the annihilator $Ann_A(M)$ --- which in your case is $(X_1^2,X_2^2,\ldots)$, and hence, being prime, it contains the ideal $(X_1, X_2, \ldots)$. Thus, to solve the problem, you need to show that there is no embedding $\mathbb Q[X_1,\ldots]/(X_1,\ldots) \hookrightarrow M$, that is, that there is no non-zero element of $M$ annihilated by $(X_1,X_2, \ldots).$

  4. If you don't see how to do this straight away, try thinking about the case when $A$ has only finitely many indeterminates, i.e. when $A = \mathbb Q[X_1, \ldots,X_n],$ and when $M = A/(X_1^2,\ldots,X_n^2)$. In this case, by remark 2 above, it must be possible to find a non-zero element of $M$ which is annihilated by $(X_1,X_2,\ldots, X_n)$. Find this element explicitly. Once you have found it, look at it, and see why you can't construct an analogous element in the case of infinitely many variables.

share|cite|improve this answer

The annihilator of a module $M$ over a ring $A$ is the set of elements $a$ in $A$ such that $aM=0$. This is an ideal of $A$ usually denoted by $\textrm{Ann}_A(M)$ and I think this is what you mean by assasin. In this case, the annihilator of $M$ over $A$ is the zero ideal because if $X_i X_{i+1} \neq 0$ in $M$ for every $i \geq 1$.

If you actually mean the set of associated primes of $M$ over $A$ I think it works as follows. The ring $A/a$ has precisely one (prime) ideal: the image of $(X_1,X_2,\ldots)$ in $A/a$. Now, the annihilator of this submodule is just zero. This is a prime ideal of $A$ and thus, the set of associated primes should be a singleton.

Maybe I'm wrong?

share|cite|improve this answer

Your Answer


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.