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Let $S$ be an integral extension ring of $R$. Then $S[x_1,\ldots, x_n]$ is an integral extension ring of $R[x_1,\ldots, x_n]$?

My idea is to prove when $S[x_1]$ is an integral extension ring of $R[x_1]$ and then extends to case when $S[x_1, \dots, x_n]$ and $R[x_1, \dots, x_n]$.

To prove for all $S[x_1]$ is an integral extension of $R[x_1]$, it is sufficient to check $x_1$ is a root of some monic polynomial $f(x)\in R[x_1][x]$. However since $x_1=1\cdot x$ there exists a monic function $g(x)\in R[x]$ such that $g(1)=0$. And I don't know how to extend $g$ to $f$.

user26857
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  • The monic polynomial you seek is $x-x_1$. – Angina Seng Apr 16 '20 at 05:58
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    Why do you need to check the $x_i$'s? If they're in the bottom ring they are integral "for free" (e.g. the linear polynomial from @AnginaSeng). Or consider, why are you not checking that each $r\in R$ is also integral viewed in $S$? –  Apr 20 '20 at 04:00

1 Answers1

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To show an extension ring is an integral extension, it suffices to prove that it is generated (as a ring) by integral elements.

Here $S[X_1,\ldots,X_n]$ is generated by the elements of $S$ and the $X_i$. Each element of $S$ is integral over $R$ and a fortiori over $R[X_1,\ldots,X_n]$. The $X_i$ are elements of $R[X_1,\ldots,X_n]$ and so are integral over $R[X_1,\ldots,X_n]$.

We conclude that $S[X_1,\ldots,X_n]$ is integral over $R[X_1,\ldots,X_n]$.

Angina Seng
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