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I know that there have been quite a few questions about showing $R$ is integrally closed over $\mathrm{Frac}(R)$ (the field of fraction of an integral domain $R$). Often $R$ is some polynomial ring with coefficients from a field quotiented out by some relation and it reduces to showing there is some isomorphism which makes it into a polynomial ring of a field and then remarking that it is an UFD.

But in general, how does one show that given domains $R\subset S$, $S$ is integrally closed over $R$?

(i) Is there a useful sufficient condition?

(ii) What if there are further restrictions on $R$ or $S$?

For example, in number fields, I know there is a characterization for the ring of integers (integral closure of $\mathbb{Z}$) of quadratic fields depending on whether $d\equiv 1 \bmod{4}$ or $d \equiv 2,3 \bmod{4}$ but this is somewhat number theoretic rather than algebraic.

Just to give an example,

Let $R=\mathbb{C}[X]$ and $S=\mathbb{C}[X,Y]/(X^2+Y^3-1)$. Show that $S$ is integrally closed over $R$ in $\mathrm{Frac}(S)$.

$S$ being integral over $R$ is straightforward since $X$ is integral over $R$. But I am not sure how one would show that it's integrally closed.

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    $\begingroup$ If you are in the geometric case (like the setting in your first paragraph), you can use the Jacobian criterion. This is applicable to the example you posed in the end. $\endgroup$ – Youngsu Jul 30 '16 at 23:30
  • $\begingroup$ @Youngsu I checked up on Jacobian criterion for integral closure. Is this the one you were thinking of? > Let $k$ be a field and $R$ and equidimensional finitely generated $k$- algebra. Let $J$ be the Jacobian ideal $J_{R/k}$. Let $P$ be a prime ideal in $R$. If $J$ is not contained in $P$ then, $R_P$ is a regular ring. $\endgroup$ – daruma Jul 31 '16 at 12:08
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    $\begingroup$ That's correct. If the codimension of such J is at least 2, then R is normal by Serre's characterization of normal domains. See math.stackexchange.com/questions/206099/… for other characterizations. $\endgroup$ – Youngsu Jul 31 '16 at 21:51
  • $\begingroup$ @Youngsu I am not quite sure if this is the sort of theorem that I am looking for considering that Jacobian ideals and regular rings haven't turned up yet. Would you be kind enough to write up an answer to how this all fits in? $\endgroup$ – daruma Aug 1 '16 at 1:45
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    $\begingroup$ Daruma, I'm fairly sure that the argument about $y^3-1$ not having square factors is that this is equivalent (look up elliptic curves in Weierstrass form, or just verify the absence of singular points) to the curve being smooth. For curves integral closure of the coordinate ring is equivalent to smoothness (IIRC, because integral closure here reduces to local conditions, i.e. smoothness). $\endgroup$ – Jyrki Lahtonen Aug 3 '16 at 10:29

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