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I was having a hard time with this exercise. I now think I have solved it, but my solution seems simpler than the one alluded to in the hint that is given, which makes me suspicious.

Some context for the problem: We have an integral $k$-variety $X$ and a closed irreducible subset $Z \subsetneq X$ which is maximal among all irreducible closed subsets strictly contained in $X$. We want to prove that $\dim Z = \dim X - 1$. By Noether normalization, we have a morphism $\pi : X \to \mathbb{A}^d_k$ where $d = \dim X$, which corresponds to a finite extension of rings. The exercise is:

Show that it suffices to show that $\pi(Z)$ is a hypersurface. (Hint: the dimension of any hypersurface is $d − 1$ by Theorem 11.2.1 on dimension and transcendence degree. Exercise 11.1.E implies that $\dim \pi^{-1}(π(Z)) = \dim \pi(Z)$. But be careful: $Z$ is not $\pi^{-1}(\pi(Z))$ in general.)

My solution was the following. Write $X = \text{Spec} B$ and let $\mathfrak{p}_0$ be the prime ideal in $B$ corresponding to $Z$ and let $\mathfrak{q}_0$ be the prime ideal in $k[x_1, \ldots, x_d]$ corresponding to $\pi(Z)$. Since $\dim \pi(Z) = d-1$ (assuming $\pi(Z)$ is a hypersurface), we have a chain of prime ideals $\mathfrak{q}_0 \subsetneq \mathfrak{q}_1 \subsetneq \cdots \subsetneq \mathfrak{q}_{d-1}$ in $k[x_1, \ldots, x_n]$. Since $k[x_1, \ldots, x_n] \subset B$ is a finite extension, and since $\mathfrak{p}_0$ lies over $\mathfrak{q}_0$, we can apply the going-up theorem to obtain a chain $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_{d-1}$ in $B$, from which it follows that $\dim Z \geq d-1$ and therefore $\dim Z = d-1$.

Is this solution correct? If not, I would appreciate if someone could give a solution along the lines of the hint.

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Your solution is correct (assuming you can also prove that $\pi(Z)$ really is a hypersurface). I'm not sure what's going on with the hint--Exercise 11.1.E applies just as well to directly show $\dim Z=\dim \pi(Z)$, and indeed the argument you have made is essentially just reproving Exercise 11.1.E in that particular case.

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