Let $A$ be a commutative ring. I am thinking about the condition of being integrally closed in relation to being normal. How do the following conditions relate to each other?
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$A$ is normal
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$A/\mathfrak p$ is integrally closed in $\text{Frac}(A / \mathfrak{p})$ for each prime $\mathfrak p$
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$A_{\mathfrak{p}}/\mathfrak{q}$ is a valuation ring for each prime $\mathfrak{p}$ in $A$ and each prime $\mathfrak{q}$ in $A_{\mathfrak{p}}$.
I have found in my own thinking that (2) and (3) are equivalent, and that they imply (1), but I cannot get the other direction. Inn particular, is there an integrally closed domain $A$ such that $A/p$ is not integrally closed for some $p$?
Best Answer
Yes, there are tons of integrally closed domains $A$ with a prime ideal $\mathfrak{p}$ so that $A/\mathfrak{p}$ isn't integrally closed. Take any non-normal affine variety, like $V(x^2-y^3)\subset \Bbb A^2_k$: $k[x,y]$ is a UFD, hence normal, but $k[x,y]/(x^2-y^3)\cong k[t^2,t^3]$ which is not integrally closed in its field of fractions $k(t)$. So (1) does not imply (2).
Let me also point out that the condition in (3) of $A_\mathfrak{p}/\mathfrak{q}$ being a valuation ring for all $\mathfrak{p,q}$ is incredibly uncommon: even when $A$ is the coordinate ring of a smooth variety over a field, the local ring of any point $\mathfrak{p}$ of codimension greater than one with $\mathfrak{q}=0$ provides a counterexample. Condition 3 should be equivalent to $A$ being regular of dimension one.