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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!!!
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Here's an explanation of that error and what you can do to avoid getting it. –  joriki Sep 17 '11 at 12:15
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Perhaps because the question repeatedly includes the word "ass" ? –  Daniel McLaury Sep 17 '11 at 15:42

2 Answers 2

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.

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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?

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