$(f_1,\dots,f_r)$ coprime $\implies V(f_1,\dots , f_r)\subset \mathbb{A}^2_{\mathbb{K}}$ is finite

Question

I'm trying to solve the following exercise.

Let $(f_1,\dots,f_r)\in \mathbb{K}[x,y]$ and suppose that $\gcd\{f_1,\dots,f_r\}=1$. Show that $V(f_1,\dots , f_r)\subset \mathbb{A}^2_{\mathbb{K}}$ is finite.

I honestly don't even know how to start: intuitively I understand why it is true, because $V(f_1,\dots , f_r)$ is the intersection of two or more curves which aren't contained one in another, but I can't find a way to formalize this.

I also have a general doubt: I tried using Bezout's identity, which says that if $\gcd\{f_1,\dots,f_r\}=1$ then there exists $\alpha_1,\dots, \alpha_r \in \mathbb{K}[x,y]$ such that $\alpha_1f_1+\dots+ \alpha_r f_r=1$, but plugging in $x\in V(f_1,\dots , f_r)$ would imply $0=1$, a contradiction. So Bezout doesn't hold here?

Answer 1

Bézout's Theorem in algebraic geometry tells you that if $f(x,y)$ and $g(x,y)$ have no common factors in ${K}[x,y]$, then the projective varieties corresponding to $f(x,y)=0$ and $g(x,y)=0$ have $\deg(f)\times \deg(g)$ common solutions in $\mathbb{P}^2(\overline{K})$ counting multiplicities, where $\overline{K}$ is the algebraic closure of $K$.

See, for instance, this link.

Now, $\mathbb{A}^2_{K}\subset\mathbb{P}^2(K)\subset\mathbb{P}^2(\overline{K})$, so on $\mathbb{A}^2_{K}$ you will get at most $\deg(f)\times \deg(g)$ solutions.

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I recognise that your question does not suppose that $f_1,f_2,f_3,\ldots,f_r$ are pairwise co-prime. It supposes that $f_1,\ldots,f_r$ have no simultaneous non-trivial common factor. But adapting the theorem to your question, perhaps through induction, is not that difficult. Let $\text{gcd}(f_1,f_2)=g$. Then $f_1=h_1 g$ and $f_2=h_2 g$. Since $h_1$ and $h_2$ are co-prime, $h_1=h_2=0$ has at most $\deg(h_1)\times\deg(h_2)$ solutions. So add those potential solutions to $V(g,f_3,f_4,\ldots,f_r)$, which involves $(r-1)$ polynomials with no simultaneous non-trivial common factor.

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As for the last paragraph in your question,

Bézout's Identity works for polynomials of one variable, but not in general for two variables.

Can you find polynomials $p(x,y),q(x,y)$ such that $x\,p(x,y)+y\,q(x,y)=1\in K[x,y]$?

No, because the constant term of $x\,p(x,y)+y\,q(x,y)$ is $0$.

Answer 2

It suffices to prove the result for two coprime polynomials $f$ and $g$. Since $f$ and $g$ have no common factor in $\mathbb{K}[x,y],$ we see (by Gauss' lemma) that they have no common factor in $\left(\mathbb{K}(x)\right)[y].$ This is a PID, so there exist rational functions $a(x),b(x)$ such that $a(x) f(x,y) + b(x) g(x,y) =1.$ Clearing denominators, we have $$ \tilde{a}(x) f(x,y) + \tilde{b}(x) g(x,y) = d(x)$$ for some polynomials $\tilde{a}, \tilde{b}, d.$ From this we see that any common root $(x',y')$ of $f$ and $g$ forces $d(x)$ to have $x'$ as a root. Since $d$ has only finitely many roots, the set of common roots of $f$ and $g$ has finitely many $x$ values appearing in it. The same reasoning applies for the $y$ variable, so $V(f,g)$ is finite.