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Let $\mathfrak a$ be an ideal of a Noetherian ring $A$ and $x$ an element in $A$. Define $(\mathfrak a:x)=\{r\in A:rx\in \mathfrak a\}$.

I try to understand one thing about the ideals in the set $(\mathfrak a:x)$ $(x\in A)$. The following is from Atiyah MacDonald (I'm not wondering about this proposition but the first statement in the proof):

Proposition 7.17. Let $\mathfrak a\neq (1)$ be an ideal in a Noetherian ring $A$. Then the prime ideals which belong to $\mathfrak a$ are precisely the prime ideals which occur in the set of ideals $(\mathfrak a:x)$ $(x\in A)$.

In the first line of the proof it is written: By passing to $A/\mathfrak a$ we may assume $\mathfrak a=0$. It is this sentence I'm wondering about.

By passing to $A/\mathfrak a$ I assume that they are talking about a correspondence $(\mathfrak a:x)\rightarrow (0:\overline{x})$ were $\overline{x}=x+\mathfrak a$? Since every ideal in the set $(\mathfrak a:x)$ $(x\in A)$ contains $\mathfrak a$ there is a one to one correspondence between prime ideals $\mathfrak p$ in this set and some of the prime ideals in $A/\mathfrak a$ given by $\mathfrak p\subset (\mathfrak a:x) (x\in A) \leftrightarrow \mathfrak p/\mathfrak a$. How do I see that this $\mathfrak p/\mathfrak a$ is an ideal on the form $(0:\overline{x})$ and every prime ideal on this form is the image of an ideal in the set $(\mathfrak a:x) (x\in A)$? If this even is the way to go about this?

And I am sorry if this is a completly retarded question but I don't understand this, so please be as detailed as possible in your answer. To specify, I wonder why, when considering the set of prime ideals in the set of ideals $(\mathfrak a:x)$ $(x\in A)$ we may assume $\mathfrak a=0$ and what is really ment by this statement. Thank you.

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up vote 0 down vote accepted

The obvious correspondence matches

  • ideals in $A$ containing $\mathfrak a$ with ideals in $A/\mathfrak a$
  • and prime ideals in $A$ containing $\mathfrak a$ with prime ideals in $A/\mathfrak a$
  • an ideal of the form $(\mathfrak b:x)$ with $(\mathfrak b/\mathfrak a:x+\mathfrak a)$

For the latter note that $rx\in \mathfrak b$ implies $(r+\mathfrak a)(x+\mathfrak a)\in \mathfrak b+\mathfrak a$ and vice versa.

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I don't understand point three, when considering the ideals occuring in the set $(a:x)$, $(x\in A)$ why are these on the form $(b:x)$ for an ideal $b$ containing $a$? – harajm Jan 11 '13 at 16:32
You may replace b by a+b. – Martin Brandenburg Jan 12 '13 at 19:19

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