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Suppose $I$ is an ideal in a polynomial ring $R=k[x,y]$. Let $\overline{k}$ be the algebraic closure of $k$ and let $S=\overline{k} [x,y]$. Then is $IS\cap R=I$?

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    Isn't $R\subset S$ faithfully flat?2014-03-02
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    Dear Zuben, could you please accept one of Makoto's answer: he more than deserves it.2016-03-13
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    But his/her profile shows "Last seen Nov 23 '11 at 0:58"...2017-01-10

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