$k[X]$ is integral over $k[X^{2}]$

Question

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!

Answer 1

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 $\,k]x^2]$.

Answer 2

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 $A$.