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Let $K$ be a field. Let $p(x_1,x_2),q(x_1,x_2)\in K[x_1,x_2]$ such that $p(x_1,x_2),q(x_1,x_2)$ are linearly independent over $K[x_1,x_2]$ (considering $K[x_1,x_2]$ as a $K[x_1,x_2]$-module). Must the set $S:=\{(u_1,u_2)\in K\times K\mid p(u_1,u_2)=q(u_1,u_2)=0\}$ be finite ?

If the answer is yes, is it possible to bound the size of $S$ by $\deg(p),\deg(q)$ ?

Amr
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  • Aren't $p$ and $q$ always dependent over $K[x_1,x_2]$? After all, we have $$p\cdot q-q\cdot p=0.$$ It feels more likely that you want them to not have any common factors, when Bezout's theorem is your friend. – Jyrki Lahtonen Jul 29 '13 at 08:34
  • @JyrkiLahtonen Yes you are right. The condition of linear independence was a bad choice. I will edit the question. – Amr Jul 29 '13 at 12:56

1 Answers1

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Any two $p,q$ will be linearly dependent over $k[x_1,x_2]$ so what I think you want is if $p,q$ have no common factors. Now if $(p,q) = (1)$ then their vanishing locus will be empty so we may suppose that $p,q$ have no common factors and $(p,q) \neq (1)$. Now let $F = k(x_1)$. Then Gauss' Lemma says that $p,q$ have no common factors (relatively prime) in $k[x_1,x_2]$ iff they have no common factors in $F[x_2] = k(x_1)[x_2]$. This is a polynomial ring over a field, so having no common factors says that $(p,q)F[x_2] = 1$, i.e. there is $f_1$ and $f_2$ in $F[x_2]$ so that

$$f_1p + f_2q = 1.$$

Clearing denominators we get that $(p,q)$ contains a polynomial $r \in k[x_1]$. But now $V(p,q) \subseteq V(r)$ the latter being a finite set. This shows that your set is finite.

If you are looking for a bound on your set, you should look at Corollary 4 Section 1.7 of Fulton's Algebraic Curves which says that the number of points in $V(p,q)$ is at most the dimension of $k[x_1,x_2]/(p,q)$ (which is necessarily finite).

  • That $V(p,q)$ is finite is already proved on this site: http://math.stackexchange.com/questions/384740/irreducible-polynomials-and-affine-variety –  Jul 29 '13 at 10:12