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$K$ is an algebraically closed field.

The coordinate ring is isomorphic to $K[t^2,t^3]$, whose Krull dimension is of at most $2$(by an hint in the exercise without proof), but how to show it’s exactly $1$ ?

Mugenen
  • 1,091
  • Do you know the Going Up theorem (and related friends)? $K[X]$ is an integral extension of $K[X^{2}, X^{3}]$ and so must have the same Krull dimension. But $K[X]$ is a PID which is not a field, and so has Krull dimension $1$. (We do not need to use that $K$ is algebraically closed.) – Alex Wertheim Oct 19 '19 at 14:31
  • If you want an argument which avoids using the full strength of Going Up (but uses the same main engine), my answer here is pretty explicit: https://math.stackexchange.com/a/3084021/73817 – Alex Wertheim Oct 19 '19 at 14:35

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