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I'd like to show the following claim:

The radical of a primary ideal $\mathfrak q$, $r(\mathfrak q)$, is the smallest prime ideal containing $\mathfrak q$.

Can you tell me if my proof is correct:

I'll use the following two facts:

(i) There is a bijection between ideals $J$ of $R$ containing $I$ and ideals $\overline{J}$ of $R/I$.

(ii) The radical of an ideal $r(I)$ equals the nilradical $n(R/I)$. (To see this, let $x \in r(I)$. Then $x^n \in I$ and hence $\overline{x}^n = 0$ in $R/I$. On the other hand, let $\overline{x}^n = 0$ then $x^n \in I$ and hence $x \in r(I)$.)

Now for the claim: By (ii), we have $r(\mathfrak q) = n(R/\mathfrak q)$. We know that $n(R/\mathfrak q) = \bigcap_{I \in R/\mathfrak q; I \text{ prime}} I$. Going back to $R$ and using (i) we hence see that $r(\mathfrak q) = \bigcap_{I \subset R; \mathfrak q \subset I } I$. At first I thought that I could write "$\mathfrak q \subset I; I \text{ prime}$" here but I think that might be wrong. (Do you know an example where that's wrong?)

Now we have established that $r(\mathfrak q)$ is the smallest ideal containing $\mathfrak q$ so to finish the proof we want to show that $r(\mathfrak q)$ is prime:

Let $xy \in r(\mathfrak q)$. Then $x^n y^n \in \mathfrak q$ for some $n$. Since $\mathfrak q$ is primary we hence know that for some $m$ we have $x^m$ or $y^m$ in $ \mathfrak q$. Hence $x \in r(\mathfrak q)$ or $y \in r(\mathfrak q)$.

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$r(\mathfrak q)$ is not "the smallest ideal containing $\mathfrak q$", but it is the smallest semiprime ideal containing $\mathfrak q$. –  rschwieb Jun 16 '12 at 20:41

1 Answer 1

The last paragraph is great: you've shown that $r(\mathfrak q)$ is a prime ideal. The other part is actually easier, for if $\mathfrak p$ is a prime ideal then one shows by induction that if $a \in A$ and $n \geq 1$ are such that $a^n \in \mathfrak p$ then $a \in \mathfrak p$. Hence if $\mathfrak q \subset \mathfrak p$ then $r(\mathfrak q) \subset \mathfrak p$.

Regarding the earlier stuff: it might be safer to write something like $r(I)/I = n(R/I)$. The correspondence between ideals containing $I$ and ideals of $R/I$ preserves the property of being prime, and taking pre-images commutes with intersections, so you should indeed write $r(\mathfrak q) = \bigcap_{I \text{ prime }\supset\ \mathfrak q} I$; but I suspect you already knew this identity. Note that intersecting over all ideals containing $\mathfrak q$ would just give you $\mathfrak q$!

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