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Is there any idea to prove that $F + xk[[x]]$ is not integrally closed when the field $k$ is a proper extension of the field $F.$

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  • $\begingroup$ I find it strange to believe this when $k/F$ is purely transcendental... $\endgroup$ – darij grinberg Feb 29 '12 at 4:26
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I claim the integral closure of the ring $R:=F+xk[[x]]$ is $R':=F'+xk[[x]]$, where $F'$ is the algebraic closure of $F$ inside $k$. For one, the field of fractions of $R$ is $k((x))$, and $F'$ is integral over $R$, so the integral closure of $R$ is at least as large as $R'$. On the other hand, $R$ is contained in power series ring $k[[x]]$, which is integrally closed, so any element $f$ of the integral closure of $R$ had better be a power series in $x$. By evaluating its minimal polynomial at $x=0$, we see that the constant term of $f$ is integral over $F$, so $f\in R'$, as desired.

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  • $\begingroup$ It seems if $k$ is finite then the integral closure is just $k[[x]]$. Is that true or not? If that is true then how can we prove it? Thank you. $\endgroup$ – Rajnish Feb 29 '12 at 16:55
  • $\begingroup$ Yes, it is true, because a finite field (like $k$) is algebraic over any subfield (like $F$). $\endgroup$ – Tiankai Mar 9 '12 at 5:50

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