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Let $F$ be a field and $\langle a_1,...,a_n \rangle \subset F$.

Then given a non-zero polynomial $f \in F[X_1,...,X_n]$ is it true that if $f(a_1,...,a_n)=0$ then $(X_i - a_i)$ divides $f$ for some $i\leq n$?

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4  
This is one of those "almost any example is a counter-example" situations.... –  Hurkyl Oct 30 '12 at 19:32
1  
Consider $f(x,y)=(x-a)+(y-b), a,b\in F.$ –  Andrew Oct 30 '12 at 19:33

1 Answer 1

Recall how the argument goes in the classical setting. Let $k$ be a field and $p(x) \in k[x]$ such that for some $a \in k$ we have $p(a)=0$. Then we can divide $p(x)$ by $(x-a)$ to leave us with an equation of the form $p(x)=q(x)(x-a)+c$ for some $c \in k$. Of course $0=p(a)=q(a)(x-a)+c=c$, so $(x-a) \mid p(x)$. The key point here is that we have a division algorithm, in particular that $k[x]$ is a Euclidean domain. Now in the case of even two variables this all begins to break down. We may think of $k[x,y]$ as $k[x][y]$ then we may still divide by $(y-a)$ by cancelling leading terms, but we are left with $p(x,y)=(y-a)q(x,y)+f(x)$. So instead of a constant term we have a polynomial in $x$, hence the last step fails.

Concretely, as mentioned in the comments, nearly any two-variable polynomial suffices. For instance $p(x,y)=1-xy$ and the point $(1,1)$.

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