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If $R\subseteq S$ is an integral extension, then $R[x]\subseteq S[x]$ is an integral extension as well. Why?

$R,S$ are both commutative rings with $1$.

Koto
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    Duplicate of http://math.stackexchange.com/questions/1201554/polynomial-ring-with-integral-coefficients-is-integral – user26857 Dec 14 '16 at 16:38

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Sums and products of integral elements are integral, and elements of $ S[x] $ are $ S $-linear combinations of powers of $ x $. Since $ S $ is integral over $ R $, and thus $ R[x] $, by assumption; and $ x $ (by virtue of being an element of $ R[x] $) is integral over $ R[x] $, it follows that any $ S $-linear combination of powers of $ x $ is also integral, and thus $ S[x] $ is an integral extension of $ R[x] $.

Ege Erdil
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  • It was really simple. Thanks, Starfall. I was on the wrong way, trying to find a nice polynomial in $R[x][y]$, you know... – Koto Oct 07 '16 at 14:59