TITLE: Normal Ring and Prime Ideal whose Square is Principal
QUESTION [5 upvotes]: Let $k$ be a field of characteristic $\neq2$ and consider $R=k[x,y]/(y^2-x^3+x)$. Then
(a) Show that $R$ is normal.
(b) Let $P=(x,y)$ be a prime ideal of $R$. Show that $P^2$ is a principal ideal.
(c) Show that $P$ is not a principal ideal. 
I came across this in some commutative algebra reading recently. Here are my thoughts so far (really only for part (c)). Assume that $P$ is generated by $\alpha=f(x)+g(x)y$. Then if $\sigma$ is the nontrivial element in the Galois group, $\sigma(\alpha)$ will generate $\sigma(P)=P$, so the norm of $\alpha$ will generate $P^2$. I thought to compare this supposed generator with the one found in part (b) and then arrive at a contradiction.   
Any help with this problem would be appreciated!

REPLY [3 votes]: This answer continues the OP's thoughts for part (c).
(c) Let $\sigma:k[X,Y]\to k[X,Y]$ be the $k$-automorphism given by $\sigma(X)=X$, $\sigma(Y)=-Y$ (which is probably the automorphism mentioned by the OP in his attempt). Since $\sigma(Y^2-X^3+X)=Y^2-X^3+X$, we can "extend" $\sigma$ to a $k$-automorphism of $R$. Obviously, $\sigma(P)=P$. 
Now note that the elements of $R$ can be uniquely written as $u(x)+v(x)y$ and such an element is in $P$ if and only if $ x\mid u(x)$. 
If $P=(xa(x)+b(x)y)$, then $P=\sigma(P)=(\sigma(xa(x)+b(x)y))=(xa(x)-b(x)y)$. It follows $(x)=P^2=(x^2a^2(x)-b^2(x)y^2)=(x^2a^2(x)-b^2(x)(x^3-x))$ and therefore there are $u,v\in k[X]$ and $w\in k[X,Y]$ such that $$X=[u(X)+v(X)Y][X^2a^2(X)-b^2(X)(X^3-X)]+w(X,Y)(Y^2-X^3+X).$$ By taking the degree in $Y$ of both sides we get $w=0$. In the end we have $$X^2a^2(X)-b^2(X)(X^3-X)=1 \text{ or } X,$$ and both cases are clearly impossible.