# Normal Ring and Prime Ideal whose Square is Principal

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!

• Part (a) is because it is a degree 2 extension (formed by adding $\sqrt{x^3-x}$. Not sure about the others though. – Prahlad Vaidyanathan Oct 12 '13 at 3:57
• @PrahladVaidyanathan This is not asking about the extension of function fields, but of rings. Normality is slightly more subtle there. – Slade Oct 12 '13 at 3:59
• @user33433: if you wish to delete your comments, hover over your comment and several icons should appear. Click on the grey/red "x" and that will delete your comment. – robjohn Oct 16 '13 at 14:49

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.

a) Regular implies normal.

Okay, maybe that's too high-tech. If $f=g+hy$ lies in the integral closure, then it satisfies the quadratic polynomial $(z-g)^2 - h^2(x^3-x)$. If the coefficients are polynomials (which they must be, by Gauss's Lemma), then $g$ is a polynomial, and $h^2 (x^3-x)$ is a polynomial. But $x^3-x$ is square-free (here we use that the characteristic is not $2$), so $h$ is a polynomial and $f\in R$.

b) Since $y^2,x^3 \in P^2$, it is straightforward that $x\in P^2$ and $P^2 = (x)$.

c) (Note: it is sufficient to address this problem for $R\otimes_k \overline{k}$, so we assume that $k$ is algebraically closed to avoid some technicalities.)

First of all, $X=\operatorname{Spec}k[x,y]/(y^2-x^3+x)$ is an open subset of an elliptic curve $\overline{X} = Y\cup \{\infty\}$. This is a rather important observation, because it implies that $R=k[x,y]/(y^2-x^3+x)$ doesn't have any principal prime ideals whatsoever (except, of course, for $(0)$).

We will demonstate this, real mathematician-style, by looking at the global behavior of a hypothetical generator of $P$. If $P=(f)$ is principal, then $f$ extends to a rational function on $\overline{X}$ with a simple pole at $\infty$. Therefore $1/f$ is a non-constant rational function on $\overline{X}$ with only a simple pole at $P$. This implies that $l(P) \geq 2$.

Here $l(P)$ is shorthand for the dimension of the space of rational functions on $\overline{X}$ which have at worst a simple pole at $P$, and are regular elsewhere. In general, $l(D)$ is defined for $D$ a Weil divisor, i.e. a formal sum of points. For example, $l(P+2Q-3R)$ counts the number of linearly independent functions which must have a triple zero at $R$, and are allowed to have a pole of order $1$ at $P$, and $2$ at $Q$. The sum of the coefficients of $D$ is called its degree.

We will calculate $l(P)$ exactly using the Riemann-Roch theorem, which says that, for any Weil divisor $D$, $l(D)-l(K-D) = \operatorname{deg}{D} + 1 - g$, where $g$ is the genus of $\overline{X}$, and $K$ is the divisor associated to any differential form on $\overline{X}$. For an elliptic curve, we must have $K=0$ and $g=1$, so this equation takes the simple form $l(D)-l(-D) = \operatorname{deg}{D}$.

We set $D=P$. Since $\operatorname{deg}{P} = 1$, and $l(-P)=0$ (in fact, $l(-D)=0$ for any divisor of positive degree), we conclude that $l(P) = 1$. Since earlier we found that $l(P)\geq 2$ for any principal maximal ideal on $X$ (in fact, the proof works for any projective curve with a single point removed), we have our contradiction.

(The above does have a nice, concrete interpretation: when you strip away the machinery, the argument is that, if $P=(f)$, then the map $k[x]\to R$ sending $x$ to $f$ is an isomorphism. We use basic facts about divisors to avoid some difficult calculations, and genus to show that $R$ cannot be a polynomial ring.)