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Let $R$ be a commutative ring, with 1.

Prove that if $R$ is reduced, then $R$ is integrally closed in $R[X]$, i.e. $R \subset R[X]$ is an integral extension of rings.

I found this problem in many introductory courses, but I simply can't solve it.

Thanks in advance.

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    Oh :) The truth is that I am off track in ComAlg, as a whole, unfortunately. That's why I posted some questions tonight, which are not difficult, I'm sure, but I can't get them solved :)2012-06-03

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The task is to show that the only elements of $R[x]$ integral over $R$ are those already in $R$.

Here's a start:

Suppose $p(x)\in R[x]\setminus{R}$ is integral over $R$, and say the degree of $p(x)$ is $n>0$.

Key observation: Since $R$ is reduced, the $i$'th power of $p(x)$ has degree $n*i$.

We have that $p(x)$ satisfies some monic polynomial over $R$, say of degree $k$. Rewrite that equation as $(p(x))^k=$ (sum of lower powers of $p(x)$ with coefficients from $R$).

Can you see why this results in a contradiction?

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    Got it :) Your help was extremely useful, thanks a lot and, if you please, let's just stop here with this discussion :)2012-06-03