Does there exist an integral domain $R$ of characteristic $p > 0$ that is perfect (i.e., $x \mapsto x^p$ is bijective on $R$) but not integrally closed in its field of fractions?
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1$\begingroup$ Tons of them. Let $R$ be a domain with integral closure $R'$ in its fraction field $K$ of characteristic $p$ such that ${\rm{Spec}}(R') \rightarrow {\rm{Spec}}(R)$ has a fiber with at least 2 points or a residue field extension not purely inseparable (i.e., $R'$ is not radiciel over $R$); easy to make such $R$. For the perfect closure $K_p$ of $K$, the ring $R_p$ of elements with a $p$-power in $R$ does the job (since $R_p$ is radiciel over $R$ and $R'_p$ is radiciel over $R_p$ with $R'_p$ normal and integral over $R_p$: if $R_p$ is normal then $R'_p=R_p$ and hence $R'$ is radiciel over $R$. $\endgroup$– user74230Commented Dec 15, 2014 at 17:49
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1$\begingroup$ I should have mentioned that taking $R$ to be the coordinate ring of a nodal cubic gives something slightly simpler than the variant in the example below. $\endgroup$– user74230Commented Dec 15, 2014 at 22:13
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$\begingroup$ Very nice! I like this family of examples most so far. $\endgroup$– Lisa S.Commented Dec 15, 2014 at 23:46
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1 Answer
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Let $S$ be the local ring of nodal curve, $R$ = inverse limit $Frob: S \to S$. For example:
- $k$ a perfect field,
- $f(x,y)=y^{p+1}-x^{p+1}(1+x)$,
- $R=k[x^{1/p^{\infty}},y^{1/p^{\infty}}]/(f^{1/p^{\infty}})$.
Here's a complete local example:
- $k$ a perfect field,
- $f(x,y)=y^{p}-x^{p}y-x^{p+1}$,
- $R=k[[x^{1/p^{\infty}},y^{1/p^{\infty}}]]/(f^{1/p^{\infty}})$.
In each example $(y/x)$ is integral over $R$.
answered Dec 15, 2014 at 12:25
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$\begingroup$ Though the formatting is actually pretty good, I think you should really TeX/MathJax this answer. $\endgroup$– jmcCommented Dec 15, 2014 at 12:28
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$\begingroup$ It appears to me that your ring is seminormal. Is every perfect integral domain a seminormal ring? $\endgroup$ Commented Dec 15, 2014 at 12:31
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1$\begingroup$ @JasonStarr Let k=perfect field char p>3, R=k[[x^(1/2p^n),x^(1/3p^n):n>=0]]. Then a=x^(1/2), b=x^(1/3), a^2=b^3, but x^(1/6) isn't in R so R isn't seminormal. R is another negative answer to the original question. (Sorry for the primitive formatting.) $\endgroup$ Commented Dec 15, 2014 at 13:35
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$\begingroup$ @DavidLampert — I did some reformatting for you. If you just add
$-signs around your maths, everything suddenly looks a lot nicer (-; $\endgroup$– jmcCommented Dec 15, 2014 at 13:50 -
$\begingroup$ @jmc Thank you. I'm a novice with formatting. $\endgroup$ Commented Dec 15, 2014 at 13:54