Solution: Kaplansky's Commutative Rings 1.6.14

Let $R$ be a commutative ring. We say that $R$ is a valuation ring if for any $a,b\in R$, either $a | b$ or $b | a$. For an integral domain $R$, this is equivalent to say that for any nonzero $r$ in the fraction field of $R$, either $r\in R$ or $r^{-1}\in R$.

Examples of valuation rings include discrete valuation rings such as the ring of formal power series $k[[t]]$ for a field $k$ and the ring $\mathbb{Z}_p$ of $p$-adic integers for a prime $p$. Valuation rings also have unique maximal ideals (exercise). Can you think of a valuation ring that is not a discrete valuation ring and not a field? Anyways, here's the main problem for today:

Problem. Show that every radical ideal in a valuation ring is a prime ideal.

(Recall that a radical ideal is an ideal $I$ such that $r\in I$ whenever $r^n\in I$ for some positive integer $n$.)

Solution. Let $I$ be a radical ideal in a valuation ring $R$, and suppose that $ab\in I$ for some $a,b\in R$. We need to show that $a\in I$ or $b\in I$. Since we are in a valuation ring, without loss of generality we may assume $ca = b$ for some $c\in R$. Then $cab\in I$ since $I$ is an ideal, and so $b^2\in I$. Since $I$ is radical, $b\in I$.

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