I've been working the last section of Chapter 7 of Vakil's FOAG, namely the one on the Fundamental Theorem of Elimination Theory. I'm not entirely sure if I've understood the proof correctly, especially the following paragraph:
Here's what I make of this. First of all, since $Z$ is closed, it's something of the form $V(f_1,\ldots)$. Now we want to see if a particular $p\in \text{Spec} A$ is also in $\pi(Z)$. Now $p$ is in this image iff there's some $p'$ in each affine open $U_j = \text {Spec }A[x_0/x_j,\ldots,x_n/x_j]$, which as a prime ideal contracts to $p$ under the usual map of rings produced in the opposite direction. Since we're interested in only primes that contract to $p$, we may "throw out" everything else by working only with $k(p)[x_0/x_j,\ldots,x_n/x_j]$ in each of these cases, and as a result, we may just work with $Proj_{\bullet} k(p)[x_0,\ldots,x_n]$. If there is a point $q$ at which the images of the $f_i$ vanish then there's a corresponding prime ideal which contracts to $p$ via the corresponding map of rings, and thus $p\in \pi(Z)$.
Next, there is a proof of the fact that if we have homogeneous elements $g_i$ cutting out a closed subset of $\mathbb{P}^n_k$, for some field $k$, then the coefficients of these $g_i$ satisfy a Zariski closed condition. I think I understand this bit of the proof and I don't include it here.
Finally, there is the following exercise:
To this I would say that from the work done in the first paragraph, we work with $k = k(p)$ in the last paragraph above. This would give us that the images of the coefficients of $f_i$ in $k(p)$ satisfy a Zariski closed condition (namely a number of determinants being $0$, in some $k(p)$'s). This means that we're looking for precisely those prime ideals where the determinant is $0$, which is of course closed and we are done.
Is my understanding of the situation correct?