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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?

Henry Reed
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2 Answers2

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The high-school proofs of the Remainder (& Factor) theorem work over any commutative ring.

Consider $\,p(x,y)\,$ as polynomial as a polynomial in $\,x,\,$ whose coefficients are polynomials in $\,y,\,$ i.e. use $\,R[y,x]\cong (R[y])[x] = S[x].\,$ Factor Theorem in $\,S[x]\Rightarrow \,p(x) = (x-x_0)q(x)\,$ since $\, p(x_0) = 0,\,$ being $\,0\,$ for $\rm\color{#c00}{all}$ $\,y\,$ (here we need that $R\,$ is an $\rm\color{#c00}{infinite}$ $\rm\color{#0a0}{domain}$; recall that a nonzero polynomial over a $\rm\color{#0a0}{domain}$ has no more roots than its degree).

Beware not to confuse polynomial functions vs. formal polynomials.. For example, in the ring of bivariate polynomial functions over $\, R = \Bbb Z/p,\,$ notice $\, f(x,y) = x\,g(x,y) + y^p\!-y\ $ we have $\,f(0,y) = y^p\!-y\,$ is the $\,0\,$ function on $\, R = \Bbb Z/p,\ $ so $\ f(x,y) = x\,g(x,y)\,$ as functions on $ R^2.$ However, $\,x\,$ does not divide $\,f\,$ in the ring of formal polynomials $R[x,y],\,$ since $\, x\nmid y^p\!-y.\,$

Thus over a finite ring $\,R,\,$ rings of polynomial functions differ from rings of formal polynomials, so we need to be careful about which ring we intend. 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.

Bill Dubuque
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$p(x,y)=\sum_{i=0}^Np_i(y)x^i\in R[x]$ where $p_i(y)\in R:=A[y]$ for some commutative ring $A$. If (as an element of $R$, not as a function) $p(a,y)$ is zero for some $a\in A$ or more generally $a\in R$, instead of using the remainder theorem like in the "12 years old accepted answer", write directly the division without remainder of $x^i-a^i$ by $x-a$: it results in $p(x,y)=(x-a)q(x,y)$ where $$q(x,y)=\sum_{i=1}^Np_i(y)\sum_{k=0}^{i-1}x^ka^{i-1-k}\in R[x].$$

Anne Bauval
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  • But OP does not have $,p(a,y) = 0,$ as a formal polynomial. Also, why repeat the common proof of the ubiquitous Factor Theorem (vs. invoke it by name)? When you fix both issues you end up with the same argument as the $12$ year old accepted answer. – Bill Dubuque Jul 29 '24 at 17:39
  • I originally used the Remainder (vs. Factor theorem) theorem (in $2012$) because that's what the OP referenced (in title). They are trivially equivalent (cf. link in my answer). I updated my answer to mention both for better dupe handling (sparked by a recent dupe). Your answer is exactly the common "linearity" based proof of the Factor Theorem linked through my answer (in the special case when the coef ring is a polynomial ring $A[y]$, which plays no role since the proof works for any commutative coef ring). – Bill Dubuque Jul 30 '24 at 21:41
  • Btw, as I hinted above, your answer doesn't apply to the OP unless you justify why $,p(x_0,y)=0.,$ for all $,y,$ implies it is the zero polynomial (cf. my answer). – Bill Dubuque Jul 30 '24 at 21:42