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Consider an algebraically closed field $k$, a finite field extension $K$ of $k(T)$, the integral closure $A$ of $k[T]$ in $K$, and the integral closure $A'$ of $k[T^{-1}]$ in $K$. Does it follow that $A \cap A' = k$?

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The level of questions on MO tends to be lower and lower. This highly upvoted question has been posted (by a vanishing user) and answered on M.SE few days ago. Moreover, the OP on M.SE has shown his own tries, and was very close to solve it. (As a matter of fact, on M.SE the question received 3 upvotes and the answer one.) Link: math.stackexchange.com/questions/1706915/…. – user26857 Mar 27 at 21:09
up vote 10 down vote accepted

Yes. Each element of $K$ satisfies a unique irreducible monic polynomial over $k(T)$. It is integral over $k[T]$ if and only if the coefficients lie in $k[T]$ and integral over $k[T^{-1}]$ if and only if the coefficients lie in $k[T^{-1}]$. If it is integral over both, the coefficients lie in $k[T] \cap k[T^{-1}]=k$. Because $k$ is algebraically closed, an irreducible polynomial with coefficients in $k$ is linear.

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