i have some problems with these exercises, can you give me a hint?
Let $f:\mathbb A^n_k\rightarrow\mathbb A^m_k$ be a regular function. If $X\subset\mathbb A^n_k$ is an algebraic set, show that the graph $B=\{(p,f(p))|p\in X\}$ is an algebraic subset of $\mathbb A^{n+m}_k$.
Let $k$ be a field, and $f, g\in k[x, y]$ irreducible elements, none multiple of the other. Show that $V(f,g)=\{p\in \mathbb{A}^2_k|f(p)=g(p)=0\}$ is a finite set in $\mathbb A^2_k$.
I can not use Bezout. I have done this: For every $c\in k[x,y]$, there are $a,b\in k[x,y]$ such that $af+bg=c$, and ther i get stuck.
Thanks for your help.
Hints:
1) It's almost as if $B$ is $\{(x,y)\in \mathbb{A}_k^n\times\mathbb{A}_k^m:y-f(p)=0\}$ :)
2) Let $f,g\in k[x,y]$ be irreducible. We want to show that if $f\nmid g$ and $g\nmid f$ then $V(f)\cap V(g)$ is finite. Note that if $V(f)\cap V(g)=V(f)$ then this says that $V(f)\subseteq V(g)$ which says that $(f)\supseteq (g)$ or $f\mid g$. Similarly, we can't have $V(f)\cap V(g)=V(g)$. Now, since $V(f)\cap V(g)\subseteq V(f)$ is closed, we may decompose it as
$$V(f)\cap V(g)=C_1\cup\cdots\cup C_n$$
where each $C_i$ is an irreducible closed subset of $V(f)$. Now, if any of the $C_i$ had dimension $1=\dim V(f)$ then $V(f)=C_i$ (why?) which is bad. So, each $C_i$ must be dimension $0$. The only irreducible dimension $0$ spaces are points. So...