# split linear factor from multivariate polynomial

This is a bit embarrassing (and hopefully hasn't been asked before). Let $$F$$ be a field and $$p(x,y)$$ be a polynomial in $$F[x,y]$$. Suppose that $$p(x,x)=0$$. For most authors it is obvious that this implies that $$x-y$$ is a factor of $$p$$. For me, however, it doesn't seem so easy. The only argument I found is the following: Consider $$p$$ as a univariate polynomial in $$y$$ over the function field $$F(x)$$. Then $$p$$ has the root $$x$$, so we can factor the linear term $$y-x$$ (using that $$F(x)[y]$$ is an Euclidean domain). Since $$y-x$$ is monic, the polynomial division does not introduce denominators. Hence, the complementary factor lies in $$F[x,y]$$.

Question: Is there a more direct argument?

It's an excellent question. One which I've thought about before. I believe your proof, of looking at $$p(x, y)$$ as a polynomial over $$F[x]$$ or $$F(x)$$ is the most direct argument.
The other argument is using the Nullstellensatz which—among other things—says that $$I(V(x - y)) = (x - y)$$. In words: a polynomial vanishes on the line $$y = x$$ if and only if it belongs to the ideal $$(x - y)$$. If you look at the (constructive) proofs of the Nullstellensatz, you get an idea of what sorts of arguments will apply to your question:
• Elimination theory/resultants, which for a single polynomial is the same thing as viewing it as a polynomial over $$F(x)$$
• Gröbner bases, which is a version of the univariate division theorem. Applied here, the argument is to write $$p(x,y) = (x-y)q(x,y) + r(x)$$ and set $$y = x$$ to get $$r(x) = 0$$; not too dissimilar from the "polynomial over $$F(x)$$" argument.