Polynomial remainder theorem for bivariate polynomials.

Suppose $p(x,y)$ is a bivariate polynomial of degree at most $N$ in $x$ and $y$.

If $p(x_0,y) = 0$, for all y, then is it true that

$$p(x,y) = (x-x_0)q(x,y)?$$

I know this holds in the univariate case, but does this work?

• In the reals at least, yes. – André Nicolas Nov 29 '12 at 19:01

Yes, the high-school proof of the remainder theorem still works: simply consider the polynomial as a polynomial in $\rm\,x,\,$ whose coefficients are polynomials in $\rm\,y,\,$ i.e. use $\rm\:R[y,x]\cong (R[y])[x].\:$
The remainder theorem works for monic polynomials over any ring, since division with remainder is always possible for polynomials that are monic (i.e. leading coefficient $= 1$ or a unit).
Beware not to confuse polynomial functions vs. formal polynomials. For example, in the ring of bivariate polynomial functions over $\rm\, R = \Bbb Z/p,\:$ notice $\rm\,\ f(x,y) = x\,g(x,y) + y^p\!-y\$ has $\rm\:f(0,y) = y^p\!-y = 0\:$ for all $\rm\:y \in R = \Bbb Z/p,\:$ hence $\rm\:f(x,y) = x\,g(x,y)\:$ as functions on $\rm R^2.$ However, $\rm\,x\,$ does not divide $\rm\,f\,$ in the ring of formal polynomials $\rm\:R[x,y],\:$ since there $\rm\: x\nmid y^p\!-y.\:$
Over a finite ring $\rm R,$ rings of polynomial functions differ from rings of formal polynomials, so one needs to be careful to stay in one ring or the other; in particular, be careful not to mix equalities of functions with formal equalities (equal coefficients). The proof of the remainder theorem works fine in either ring, but, as we saw, not if we confuse the two.