Let $k$ be a field and $s$ and $t$ be variables. Is the ring $k[s][[t]]$ integrally closed in $k[s,s^{-1}][[t]]$?
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1$\begingroup$ Welcome new contributor. Every power series ring over a regular Noetherian ring is a regular Noetherian ring. Every regular Noetherian ring is normal. $\endgroup$– Jason StarrCommented Jan 6, 2019 at 12:55
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1$\begingroup$ @JasonStarr But normal means integrally closed in its field of fractions, which isn't being asked. $\endgroup$– Will SawinCommented Jan 6, 2019 at 13:04
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1$\begingroup$ @WillSawin. I read the question in the title of the post. I see that the OP asks a different question in his post than is in the title of his post. $\endgroup$– Jason StarrCommented Jan 6, 2019 at 13:16
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$\begingroup$ Thank you for the answer. I am sorry that the title was misleading. $\endgroup$– P. GrapeCommented Jan 6, 2019 at 13:24
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2$\begingroup$ @JasonStarr Fair enough! $\endgroup$– Will SawinCommented Jan 6, 2019 at 13:24
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No. Let $\ell$ be a prime invertible in $k$ and consider
$$x= s (1+ t/s)^{1/\ell} = s + \frac{t }{\ell} - \frac{(\ell-1) t^2}{ 2s \ell^2} + \frac{ (\ell-1) (2\ell-1) t^3}{ 6 s^2 \ell^3} + \dots \in k[s,s^{-1}][[t]] $$
Clearly it does not lie in $k[s][[t]]$. But we have $$x^\ell = s^{\ell} (1+t/s) = s^\ell + s^{\ell-1} t$$ so it satisfies a monic polynomial equation over $k[s][[t]]$ (and even $k[s,t]$).
answered Jan 6, 2019 at 13:08