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Suppose $A$ is a local domain, with field of fractions $K$, let $A'$ be the integral closure of $A$ in $K$, is $A'$ a local ring?

1 Answers1

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Not necessarily. E.g. consider $A = \mathbb Z_{(5)} + 5 \mathbb Z_{(5)}[i]$. (Here $\mathbb Z_{(5)}$ denotes $\mathbb Z$ localized at the prime ideal $(5)$.) Then $A' = \mathbb Z_{(5)}[i],$ which has two maximal ideals (the ideals generated by $2 \pm i$).

For a geometric example, consider the local ring at the node of the nodal cubic $y^2 = x^2(x +1).$

Matt E
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  • Thanks! In the second example, the local ring of the node is $$k[T^2-1,T(T^2-1)]_{(T^2-1,T(T^2-1))}$$ the field of fractions be $k(T)$, but how to calculate the integral closure? –  May 31 '14 at 03:27
  • Dear mqx, The normalization is the desingularization of the nodal curve, which is just $\mathbb P^1$. The map to the node is given by identifying two points. (So the formal structure is very similar to the number theoretic example, in which I constructed $A$ by gluing together the two points above $5$ in Spec $\mathbb Z[i]$.) Regards, – Matt E May 31 '14 at 03:29
  • If we consider the whole curve, the inverse image of the node are two points, with corresponding local ring $k[T+1]{(T+1)}$ and $k[T-1]{(T-1)}$, but the normalization is not their direct sum, since the normalization is a domain. Sorry I just want to understand it more concretely..what is the normalized local ring? –  May 31 '14 at 03:31
  • Dear mqx, normalization commutes with localization, so the normalization is the ``localization'' of $\mathbb P^1$ at the intersection of the prime ideals corresponding to the two points lying over the node. (Just as in the number theory example, the normalization is $\mathbb Z[i]$ localized at $5$.) You could get formulas by writing down the normalization map from $\mathbb P^1$ to the nodal curve. Regards, – Matt E May 31 '14 at 03:32
  • Thanks! It is clear now. Another question is how did you “glue” the two points? Can we always glue any two points (must they have with isomorphic local rings) and explicitly write the glued local ring down? –  May 31 '14 at 03:40
  • Dear mqx, In the affine context, suppose we have a $X_1 = $ Spec $A_1$ and $X_2 = $ Spec $A_2$, and a scheme $Y = $ Spec $B$ which has a closed embedding into each of them. Ring-theoretically, this means we have surjections $A_1, A_2 \to B$. So we can form the fibre product $A = A_1 \times_B A_2$ , i.e. the preimage of the diagonal copy of $B$ under the product surjection $A_1 \times A_2 \to B \times B$. Then $X =$ Spec $A$ is the ``gluing'' of $X_1$ and $X_2$ along $Y$. If you do this with two points on $\mathbb P^1$, you get a nodal curve. If you do it with the two primes above $5$ in – Matt E May 31 '14 at 03:49
  • $\mathbb Z[i]$, you get my first example. Regards, – Matt E May 31 '14 at 03:49
  • P.S. I should add that the $\mathbb P^1$ example is not literally affine, but all the gluing is happening in the affine open $\mathbb A^1 \subset \mathbb P^1$ (if we choose the copy of $\mathbb A^1$ to contain the two points we are gluing). – Matt E May 31 '14 at 03:51
  • Thanks! But to define a tensor product, do we need to have $B\to A_i$ instead of $A_i\to B$？Or is the fiber product here is not in the usual sense? –  May 31 '14 at 03:58
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    Dear mqx, I mean fibre product of rings, not of Specs (which would give something quite different). If you don't like the notation, just use the definition I gave: the preimage of the diagonal copy of $B$ under the subjection $A_1\times A_2 \to B \times B$. As you can see, this has nothing to do with a tensor product. If I just took Spec $A_1 \times A_2$, I would get the disjoint union of $X_1$ and $X_2$. But I am taking the slightly smaller ring $A_1 \times_B A_2,$ in which the two copies of $Y$ $(=$Spec $B$) have been identified --- hence the gluing of $X_1$ and $X_2$ along $Y$. Cheers, – Matt E May 31 '14 at 04:01
  • OK, thanks a lot !! –  May 31 '14 at 04:04