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The problem arises from reading this answer.

The Problem: Let $k$ be a field, let $R=k[X, Y]/(Y^2-X^3+X^2)$. Henceforth use lowercase letters to denote the elements in the quotient ring (e.g., $x=\overline{X}$ in $R$). Let $T=Y/X$. Define a ring homomorphism $\varphi: k[X, Y, T]\to k[x, y, y/x]$ by $X\mapsto x$, $Y\mapsto y$, $T\mapsto y/x$. Note that $(Y^2-X^3+X^2, XT-Y, T^2-X+1)\subseteq\ker\varphi$. Then $\mathbf{\ker\varphi}$ is a prime ideal with height $\mathbf{1}$.

My Question: I don't see it being too difficult to show that $k[X, Y, T]/\ker\varphi$ is an integral domain and then conclude that $\ker\varphi$ is prime; but why is it of height $1$? Any help would be greatly appreciated.

Dick Grayson
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    Do you know that in an affine domain (a domain that is finitely generated over a field as a ring), any maximal chain of primes has length equal to the dimension of the domain? Using this fact, you can evaluate the height of a prime by looking at the dimension of the quotient ring. – Acrobatic May 01 '24 at 04:50
  • @Acrobatic I find it equally daunting to show that $k[X, Y, T]/\ker\varphi$ has dimension $0$ (I believe you meant Krull dimension). There must be some crucial links that I'm missing. – Dick Grayson May 01 '24 at 20:53

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