2

There is a similar problem here at:

Find the integral closure of an integral domain in its field of fractions

The problem from the link has a nice proof because it has a parametrization of $t=\frac{y}{x}$, and it's easy to show the integral closure is at least $k[t]$ which is a UFD, and therefore integrally closed.

Suppose given $\mathrm{char}(k)\not=2$. I want to find the integral closure of $k[x]$ in $k(x)[y]/(y^2-x^3+x)$.

The problem I have here is that I can only show $y$ is integral but not $\frac{y}{x}$. Therefore I know the integral closure is at least $k[x,y]/(y^2-x^3+x)$ and I think it is integrally closed, but I don't know how to prove it since I cannot find a nice parametrization to show it's isomorphic to a UFD.

user26857
  • 53,190
ZTransformer
  • 172
  • 3
    It's not isomorphic to a UFD; its the coordinate ring of (an affine patch of) an elliptic curve. An important point that shows the ring is integrally closed is that it is a regular curve, which can be checked by noting that $y^2 - x^3 + x = 0$ and $\mathrm{d}(y^2 - x^3 + x) = 0$ (that is, taking $\mathrm{d}y$ and $\mathrm{d}x$ as a basis for differentials, the pair of equations $2y = 0$ and $-3x^2 + 1 = 0$) don't have a simultaneous solution. I don't recall how the proof of everything goes off the top of my head. –  Apr 19 '17 at 03:50
  • @user26857. Thank you for the answer. My only problem was to show $k[x,y]/(y^2-x^3+x)$ is integrally closed. Hurkyl gave me a nice hint and I worked it out. – ZTransformer Apr 19 '17 at 08:47

0 Answers0