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I'm reading page $59$ of Reid's "Undergraduate commutative algebra" book.

In example (ii) it says, $k[x^{2}] \subset k[x]$ is an integral extension.

How do we know this? I mean, in order to show this we must take a polynomial $f(x) \in k[x]$ and show there is a monic polynomial $g(x) \in k[x^{2}]$ such that $g(f(x))=0$, right? Why can we do this?

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No, we want to find a $g(T)\in k[x^2][T]$ such that $g(f(x))=0\in k[x]$. – Zev Chonoles May 16 '11 at 1:29
@Zev Chonoles: thanks, gonna post a question about local rings, can you please have a look? – user6495 May 16 '11 at 1:55
@user6495: Sure thing. – Zev Chonoles May 16 '11 at 1:59
up vote 4 down vote accepted

It's enough to prove that $x$ is integral over $k[x^2]$, which it clearly is, being a root of $T^2-x^2 \in k[x^2][T]$.

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@user6495: A reference for the fact that this suffices is Corollary 5.2 in Atiyah-Macdonald. – Zev Chonoles May 16 '11 at 1:34
@Zev, more precisely, I meant that you only need to know that if $u$ and $v$ are integral over $A$ then so are $u+v$ and $uv$, which is essentially Corollary 5.3. Thanks for the reference anyway. – lhf May 16 '11 at 1:42
@lhf: another doubt: why is this a finite extension? is it because $k[x^{2}]$ is finitely generated? – user6495 May 16 '11 at 1:54
@user6495: every polynomial in $k[x]$ can be written as $p(x^2)+xq(x^2)$. – lhf May 16 '11 at 2:00
@lhf: Thanks. I get confused sometimes with the definitions. To show that $k[x]$ is a finite extension of $k[x^{2}]$ we need to show that every polynomial in $k[x]$ is a finite linear combination of elements of $k[x]$ and $k[x^{2}]$ , i.e as a $k[x^{2}]$-module? – user6495 May 16 '11 at 2:40

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