# Zeros of multivariate polynomials

Consider the ring of polynomials $k[x,y]$ where $k$ is an infinite field.

(1) If $f$ and $g$ are two non-constant polynomials with no common irreducible factors then $V(f,g)$ is finite.

(2) If $V$ is a proper irreducible algebraic set in $k^2$ then either $V$ contains one point or $V=V(f)$ for some irreducible polynomial $f$ in $k[x,y]$.

For (1), it is the Bezout's Theorem, but is there any elementary proof?

For (2), since $V$ is irreducible, $I(V)$ is prime in $k[x,y]$. Moreover $I(V)=(f_1,f_2,\cdots,f_n)$ for some $f_i\in k[x,y]$ since $k[x,y]$ is Noetherian. Now $V=V(f_1,\cdots,f_n)$. How can I continue?

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Bezout's theorem is stronger than (1) because it tells you exactly how many points there are in $V(f,g)$.

In fact, one can show (1) directly. First note that it's enough to prove the claim for $k$ algebraically closed. This is obvious if your definition of $V(f)$ is $\{(a,b)\in k^2\colon f(a,b)=0\}$. If you don't know scheme theory, just skip the next paragraph.

If for you $V(f)=\{\mathfrak p\in \text{Spec } k[x,y]\colon (f)\in \mathfrak p\}$ then maybe you have to use the fact that the inclusion $k[x,y]\to \overline{k}[x,y]$ induces a surjective map on the spectrums because the extension is integral. So from now on, $k=\overline{k}$. Note that $\text{Spec }k[x,y]$ has 3 type of elements: $(0)$, the maximal ideals $(x-a,y-b)$ for $a,b\in k$ and the principal prime ideals $(g)$ generated by an irreducible element $g$. Now $V(f)$ can contain only a finite number of this last type of ideals, because $k[x,y]$ is a UFD. So since $V(f,g)=V(f)\cap V(g)$, then $V(f,g)$ can contain only a finite number of nonmaximal ideals. Now a maximal ideal $(x-a,y-b)$ lies in $V(f,g)$ precisely when $f(a,b)=g(a,b)=0$.

We'll argue by contradiction. Suppose there are infinitely many elements $(a,b)\in k^2$ s.t. $f(a,b)=g(a,b)=0$. Now let's use the following trick: we can think of $f,g$ as elements of $k(y)[x]$, namely the ring of polynomials in the variable $x$ with coefficients in $k(y)$. I claim that $f,g$ are coprime in $k(y)[x]$. In fact, suppose $d(x)\in k(y)[x]$ is a common divisor. Then there exist $e(x),e'(x)\in k(y)[x]$ such that $f(x,y)=d(x)e(x)$, $g(x,y)=d(x)e'(x)$. Now clearing denominators we end up with polynomials $h,h'\in k[y]$, $d',l,l'\in k[x,y]$ such that $h(y)f(x,y)=d'(x,y)l(x,y)$ and $h'(y)g(x,y)=d'(x,y)l'(x,y)$. This equality holds in $k[x,y]$ which is a UFD. Therefore, if $r(x,y)\in k[x,y]$ is an irreducible component of $d'(x,y)$, it can't divide both $f,g$ because by hypothesis they don't have irreducible components in common. Therefore we can assume $r(x,y)\mid h(y)$ in $k[x,y]$ and this proves that $r(x,y)$ and so also $d'(x,y)$ is a polynomial in $k[y]$. But now $d'(x,y)=m(y)d(x)$ in $k(y)[x]$ for some $m(y)\in k[y]$. This proves that $d(x)$ has degree $0$ in $x$, and is therefore a constant in $k(y)[x]$. Thus $f,g$ are coprime in $k(y)[x]$.

Now since this ring is an Euclidean domain there are $r(x),s(x)\in k(y)[x]$ with $r(x)f(x,y)+s(x)g(x,y)=1$. Clearing denominators again we get an expression of the type $r(x,y)f(x,y)+s(x,y)g(x,y)=n(y)$ which holds in $k[x,y]$. By hypothesis, there are infinitely many $(a,b)\in k^2$ with $f(a,b)=g(a,b)=0$, so there are infinitely many $(a,b)$ with $n(b)=0$. Since $n(y)\neq 0$, this means that for some fixed $\bar{b}\in k$ there are infinitely many $a\in k$ such that $f(a,\bar{b})=0$. But then $f(x,\bar{b})=0$ for all $x$. This proves that $f(x,y)$ has degree $0$ in $x$, so it must lie in $k[y]$. To conclude, note that the whole construction is symmetric in $x,y$, so doing everything again replacing $x$ and $y$ tells us that $f(x,y)=0$, and we're done.

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Thank Ferra a lot. It is clear and detailed enough. Have you had some idea for proving (2)? – 9999 Feb 7 '14 at 21:09
The idea for (2) is the following: if $V(f_1,\dots,f_n)$ is finite then it must have exactly one point, otherwise it would be reducible. Assume it has more than 1 point. Then it is infinite, so by point (1) $f_1,\dots,f_n$ must all have one irreducible component in common. Now show that they have exactly one irreducible component in common, say $f$, and prove that $V(f_1,\dots,f_n)=V(f)$. I gthink that there is some subtlety in the scheme-theoretic case in proving that if $V(f_1,\dots,f_n)$ is finite than it's reducible unless it has only one point, but I guess one can work that out. – Ferra Feb 8 '14 at 17:48