I have solved (maybe) this exercise, but I'm not sure about my proof. The exercise text is:
Let $K$ be a field and let $f,g\in K[x_1,\dots ,x_n]$ with $x_n$ actually appearing in $f$ and $g$. Show that $f$ and $g$ have a common zero $(\alpha_1,\dots,\alpha_{n-1},\alpha_n)$ if and only if $(\alpha_1,\dots,\alpha_{n-1})$ is a zero of $R_{x_n}(f,g)$.
My solution is the following:
Claim. $$R_{x_n}(f,g)(\alpha_1,\dots,\alpha_{n-1})=R_{x_n}(f(\alpha_1,\dots,\alpha_{n-1},x_n),g(\alpha_1,\dots,\alpha_{n-1},x_n))$$
If we have the claim, then the thesis follows easily:
- $(\alpha_1,\dots,\alpha_{n-1})$ is a zero of $R_{x_n}(f,g)$ if and only if
- $R_{x_n}(f(\alpha_1,\dots,\alpha_{n-1},x_n),g(\alpha_1,\dots,\alpha_{n-1},x_n))=0$ if and only if
- $\exists \alpha_n$ common zero of $f(\alpha_1,\dots,\alpha_{n-1},x_n),g(\alpha_1,\dots,\alpha_{n-1},x_n)$ as polynomials in $K(\alpha_1,\dots,\alpha_{n-1})[x_n ]$, if and only if
- $\exists \alpha_n$ such that $f(\alpha_1,\dots,\alpha_n)=g(\alpha_1,\dots,\alpha_n)=0$.
I'm not sure about the Claim and about the sequence of equivalences (the all proof). It looks too easy, so I think I'm doing it wrong, maybe at formal level.