I was trying to follow the strategy in this answer to compute the integral closure of the domain $$R := k[X, Y, Z]/(Y^3 + Y^2X^2 + YX^2 + X^3Z)$$ where $k$ is a field. During the computation, I realized I could tweak one of the arguments to reach an absurd conclusion, while the end conclusion is still true. So I'm wondering what is wrong in the proof below.
Let $f = Y^3 + Y^2X^2 + YX^2 + X^3Z$. Let $x, y, z$ denote the images of the free variables $X, Y, Z$ in the quotient. Note that $t := \frac{y}{x}$ is integral over $R$ as we have $t^3 + t^2x + t + z = 0$. My claim is that $S := R[t]$ is the integral closure of $R$.
Indeed, consider the surjective homomorphism $\phi: A = k[X, Y, Z, T] \to S$ given by $X \mapsto x$, $Y \mapsto y$, $Z \mapsto z$ and $T \mapsto t$, which exists by the universal property of polynomial rings. Since $S$ is an integral domain, $\ker \phi$ is a prime ideal of $A$. Moreover, we have $\dim A = 4$ and $\dim S \ge 3$, since $$(f) \subsetneq (X, Y + X^2) \subsetneq (X, Y) \subsetneq (X, Y, Z) $$ is a chain of prime ideals in $k[X, Y, Z]$ containing $f$ of length 3, so by the correspondence theorem it corresponds to a chain of primes in $R$, which by the going-up theorem can be extended to a chain in $S$. The height of $\ker \phi$ satisfies: $$\dim(A) - h(\ker \phi) \ge \dim (A/\ker \phi) = \dim S \ge 3$$ So we have that $h(\ker \phi) \le 4 - 3 = 1$. But $\ker \phi$ is nonzero, as it contains e.g. $TX - Y$, and $A$ is a domain, so its height is exactly 1.
We clearly have $(TX - Y) \subseteq \ker \phi$, and since the latter is a prime ideal of height one and the former is a nonzero ideal in a domain, it suffices to prove that $(TX - Y)$ is prime to conclude $(TX - Y) = \ker \phi$. And indeed: $$A/(TX - Y) = k[X, Y, Z, T]/(TX - Y) \cong k[X, Z, T]$$ which is a domain.
Now, this is the absurd conclusion, since $\ker \phi$ contains elements outside of $(XT - Y)$, such as $T^3 + T^2X + T + Z$. What is wrong in the above argument?
However, the final conclusion still holds, I've checked that $S$ is indeed the integral closure of $R$.
This last equation also tells us that $S \cong k[X, Z, T]$ which is a UFD, hence integrally closed. Since we have $R \subseteq S \subseteq \mathrm{Frac}(R)$ and $S$ is integral over $R$, we conclude that $S$ is the integral closure of $R$.
It is a theorem in algebraic geometry that an affine variety in $\mathbb{A}^n$ is of dimension $n - 1$ if and only if it is cut out by a single polynomial. Since $R$ satisfies this criterion, we have $\dim R = 2$. Moreover, integral extensions don't change the dimension, so $\dim S = \dim R = 2$.
My mistake was in writing the chain of prime ideals:
This inclusion is not strict, as they're the same ideal. Therefore the chain of prime ideals has length 2, so if I continued my argument from there I would get $h(\ker \phi) \le 2$.
Then, we could argue that the following is a chain of length 2 of prime ideals in $A$ contained in $\ker \phi$: $$ (0) \subsetneq (XT - Y) \subsetneq (XT - Y, T^3 + XT^2 + T + Z)$$ Since $\ker \phi$ has length at most 2, we conclude that $\ker\phi = (XT - Y, T^3 + XT^2 + T + Z)$. If we compute the quotient $A/\ker\phi$, we get $S \cong A/\ker\phi \cong k[X, T]$, and now we can use the same argument I used at the end above to finish the proof.