What is the integral closure of the ring $k[t^3-t, t^2-1]$ in $k(t)$, where $k(t)$ is the field of fractions for $k[t]$?
I have read a post on this site saying that the integral closure is just $k[t]$, but I don't quite understand why. Thus if this is a trivial result, how do I show/prove it ?
Any help or insights is deeply appreciated.
Cheers
$k[t]$ is integrally closed and contains $k[t^3 - t, t^2 - 1]$ so it is an obvious candidate for the integral closure. Then the only question is: is $t$ integral over $k[t^3 - t, t^2 - 1]$? Which is indeed the case because $t$ is a root of $x^2 - (t^2 - 1) - 1$.
Geometrically, you are looking at the graph $\{(t^3 - t, t^2 - 1) : t \in k\}$, which is singular at $(0,0)$. Then you draw a line from $(0,0)$ in some direction. This will hit the graph in exactly one other point (except for the horizontal line). Thus the graph is isomorphic to the set of lines through the origin except for one. I.e. to $\mathbf{P}^1 - \{\text{point}\} = \mathbf{A}^1$. $\mathbf{A}^1$ corresponds to $k[t]$ via the standard algebraic geometry correspondence.