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I am trying to show that $k[X]$ is integral over $k[X^2]$, where $k$ is a field.

Taking an element $b=b_nx^n+b_{n-1}x^{n-1}+...b_1x+b_0 \in K[X]$ we want to find $a_i \in K[X^2]$ such that $a^nb^n+a_{n-1}b^{n-1}+...a_1b+a_0=0$. I am stuck because if a square b for example then I am still going to end up with a polynomial with odd powers and I have no idea of how to get rid of them using coeffecients in $k[X^2]$.

Any tips would be great!

Community
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user127214
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2 Answers2

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The easiest way is to note that $\,x\,$ is integral over $\,k[x^2],\,$ being a root of $\,y^2 - x^2.\,$ For a more direct proof, bisect $\,y\in k[x]\,$ into even+odd part $\, y = a + b\:\! x,\ a,b\in k[x^2],\,$ so $\, (y-a)^2 = b^2 x^2\,$ shows $\,y\,$ is a root of a monic quadratic over $\,k]x^2],\,$ so $\,y\,$ is integral over $\,k]x^2]$.

Bill Dubuque
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We can write $k[x] = (k[x^2])[x]$ and so $k[x]$ is an extension ring of $k[x^2]$. Now show that $x$ is integral over $k[x^2]$ and use the fact that a ring extension $A \subset A[b]$ is integral if and only if $b$ is integral over $A$.

Manos
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  • Thanks. I know that b is integral over A iff A[b] is a finitely generated A module, but how does A[b] being integral over A follow from this? – user127214 Mar 06 '14 at 17:08
  • In order to see this you need to know the Caylay-Hamilton theorem for finitely generated modules. Do you have Atiyah-MacDonald? You can find the equivalence $b$ integral over $A$ iff $A[b]$ f.g. iff $A \subset A[b]$ integral there as well as the theorem i mention. – Manos Mar 06 '14 at 18:08