Projection map sends closed sets to closed sets I am reading Algebraic Geometry, a first course by Joe Harris. In the section on projections, he talks about an application of elimination theory to prove that image of a projection map $\pi: Y \times \mathbb{P}^1 \rightarrow Y$ is a closed set. I do not have strong background in Algebraic Geometry, so I do not understand elimination theory and resultants. Can someone explain what is the author trying to say in the highlighted part and why is it true? Basically, I am interested in understanding its proof.
Also, does the projection map send an open set $Z \subseteq Y \times \mathbb{P}^1$ to an open set $\pi(Z)$? If yes, why?
Thank you very much in advance!

 A: Suppose $k$ is a field and $f,g\in k[x,y]$ are homogeneous polynomials. Claim: the homogeneous resultant of $f,g$ is zero iff $V(f,g)\subset\Bbb P^1_k$ is nonempty.
Proof: The resultant is the determinant of the map $(A,B)\mapsto Af+Bg$ from $k[x,y]_{\deg g}\times k[x,y]_{\deg f} \to k[x,y]_{\deg f+\deg g}$. But the image of this map is exactly $(f,g)\cap k[x,y]_{\deg f+\deg g}$, so $(A,B)\mapsto Af+Bg$ not being surjective is equivalent to $(f,g)$ being contained in homogeneous ideal who's radical does not contain the irrelevant ideal of $k[x,y]$. Such ideals define nonempty subsets of $\Bbb P^1_k$, so by the nullstellensatz we see that $V(f,g)=V(f)\cap V(g)$ is nonempty as a subset of $\Bbb P^1_k$. $\blacksquare$
Now we note that the homogeneous resultant plays nicely with ring homomorphisms, since it's the determinant of a particular matrix: if $\varphi:R\to S$ is a ring homomorphism and $f,g\in R[x,y]$ are homogeneous, then $\varphi(Res(f,g))=Res(\varphi(f),\varphi(g))$. Taking $R=k[Y]$, $S$ to be the residue field of a point $y$, and $\varphi$ the evaluation map, we see that $Res(f,g)$ vanishes at $y$ iff $f,g$ have a common zero in the fiber over $y$. The highlighted result follows.
As for the question about projections of open sets, this has been addressed on MSE before here. The upshot is that for any scheme over a field, the map $X\to\operatorname{Spec} k$ is open, so any projection $X\times Y\to Y$ is open because such a projection is the base change of $X\to\operatorname{Spec} k$ along $Y\to\operatorname{Spec} k$. See Stacks 01TZ for a full proof. A slightly more elementary approach can be found in the comments.
