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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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    Hmm, something that may be hampering you is that you might be confusing two things. Let $R\subseteq S$ be a ring extension and write the integral closure of $R$ as $\overline{R}$. Then $R$ is called integrally closed if $R=\overline{R}$. The extension is called integral if $S=\overline{R}$.2012-06-03
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    I should have qualified "write the integral closure of $R$ in $S$ as $\overline{R}$".2012-06-03
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    I translated the problem as faithful as I could (I'm Romanian, it's a problem our professor gave us). What I think I have to prove is that there exists a monic poly $g \in R[X]$ s.t. $\forall 0 \neq f \in R[X], \ g \circ f=0$, i.e. $f^n+\displaystyle\sum_{i=0}^{n-1} a_i \cdot f^i=0$, where $a_i \in R$ are the coeffs of $g$. Right? If so, what next?2012-06-03
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    I'm 99% certain that the goal is to show that $\overline{R}=R$. That means that *no* polynomial with degree>0 is integral over $R$.2012-06-03
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    The strategy you gave in the last comment is way off track. It looks like you are trying to show that there is a monic polynomial that has everything in $R[x]$ as a root. That should already be setting off some alarms :)2012-06-03
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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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    ...because of equality of degrees? $deg(p^k)=n\cdot k$, and deg of RHS cannot exceed $n \cdot (k-1)$..?2012-06-03
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    Right :) Good luck with your studies!2012-06-03
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    Thanks a million! I really need a lot of luck, since I'm getting ready for a really tough exam in ComAlg. But my personal interest is NONComAlg, which seems (to me) something from another world.2012-06-03
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    @AdrianM I'm a noncommutative algebraist too, and I wish my commutative algebra were better. The commutative algebraists are able to get really deep results because their rings are so nice.2012-06-03
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    I have an organic repulsion to commutative algebra, since my university course is very computational (even) and I like a lot more working with categories or homological algebra. You are right, though, that commutative results are deeper, but let's say it's just not my cup of tea.2012-06-03
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    @AdrianM I've come to the point were I feel like I can no longer avoid plunging into what the commutative algebraists have to say :)2012-06-03
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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