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My question is linked with normalization of reduced algebraic curves that are not necessarily irreducible.

Let $(A,\mathfrak{m})$ be a local reduced noetherian ring with Krull dimension $1$, let $\mathfrak{p}_1, \dots, \mathfrak{p}_n$ be the minimal primes of $A$ and let $S$ the multiplicative subset made up of regular elements of $A$, i.e. $S = A \setminus (\mathfrak{p}_1 \cup \cdots \cup \mathfrak{p}_n)$. ($S^{-1}A$ is called the total ring of fractions of $A$.) It is quite easy to prove that $A \subseteq S^{-1}A$ and that $S^{-1}A \simeq k(\mathfrak{p}_1) \times \cdots \times k(\mathfrak{p}_n)$ as rings. It is quite obvious that $A_{\mathfrak{p}_i} = B_{\mathfrak{p}_i}$, where $B$ is the integral closure of $A$ in $S^{-1}A$.

My question is: when $A$ is integrally closed in $S^{-1}A$? More precisely, I have the suspect that the following assertion holds: If $A$ is integrally closed in its total ring of fractions, then $A$ is a domain, i.e. $A$ has only one minimal prime.

Could one prove or disprove this assertion? Thanks to all!

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up vote 4 down vote accepted

Because $A$ is reduced, the inclusion $A\subseteq S^{-1}A$ can be decomposed canonically into $$ A \hookrightarrow \oplus_{1\le i \le n} A/\mathfrak p_i \hookrightarrow \oplus_{1\le i\le n} k(\mathfrak p_i).$$ As $A\to A/\mathfrak p_i$ is finite for each $i$, the first arrow is finite, hence integral.

If $A$ is integrally closed in $S^{-1}A$, then the first arrow must be an isomorphism by definition. As $A$ is local, it has only two idempotent elements $0$ and $1$, and the middle term has $2^n$ idempotent elements. So $n=1$. Note that the hypothesis on $\dim A$ is useless.

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Wonderfully crisp, dear QiL: congratulations! Wouldn't you consider writing $\Pi_{1\le i \le n} A/\mathfrak p_i$ instead of $\oplus_{1\le i \le n} A/\mathfrak p_i$ ? After all, you mean the product $A$-algebra here (the coproduct [=sum] for algebras being the tensor product over $A$). – Georges Elencwajg Jan 9 '12 at 21:49
@GeorgesElencwajg, you are right. I started writing with the product, but finally I found more natural to use the sum to show it is finite over $A$. Salutations ! – user18119 Jan 10 '12 at 20:38
Thank you very much! – user21962 Mar 15 '12 at 19:46

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