# Exercise 11.2.H in Vakil's FOAG

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.

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.