Reducibility of polynomial in two variables

Question

Given a polynomial $ P(x, y) $, and also I know that this polynomial is zero on all points of the unit circle. How can I prove that P can be represented as the multiplication of the equation of the unit circle and another polynomial ?

Answer 1

A bivariate real polynomial $P(x,y) = \sum_{i,j} a_{ij}x^iy^j$ can be written as a univariate polynomial in $x$ with coefficients polynomials in $y$, so $P(x,y)=P_y(x)=\sum_i p_i(y)\cdot x^i$ where $p_i(y)=\sum_j a_{ij}\,y^j$.

Since the polynomial $T_y(x)=x^2+y^2-1$ is monic in $x$, it is possible to do the euclidean division $P_y(x) = T_y(x) \, Q_y(x) + R_y(x)$ where $R_y(x)=r_1(y)\,x + r_0(y)$.

Combined with the premise, $T_y(x)=0 \implies P_y(x)=0 \implies R_y(x)=0$, so $R_y(x)$ has infinitely many roots, therefore must be the zero polynomial, which leaves $P_y(x) = T_y(x) \, Q_y(x)$.

Answer 2

Welcome to MSE!

Hint:

Since you mention reducibility in your title, do you see why it's enough to show $x^2 + y^2 - 1$ is irreducible? The idea is analogous to the fact that if a prime number $p$ divides $n$, then actually $p$ must be a factor of $n$.

To show irreducibility, we only need to consisder degree $1$ polynomials (do you see why?), so it suffices to ask

$$x^2 + y^2 - 1 \overset{?}{=} (ax + by + c)(dx + ey + f)$$

By expanding the right hand side and comparing coefficients, you can set up a linear system. Can you show that it is not solvable? That is, no such factorization exists?

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I hope this helps ^_^