I am trying to prove that a polynomial in two variables is constantly zero. In an infinite field $k$
Suppose $f\in k[x,y]$ has as roots the following set $k^2\setminus \{(x,0):x\in k\}$. Then for every $a\in k$ the polynomial $f(a,.)\in k[y]$ has infinite amount of roots $\implies f(a,.)\in k[y]$ is the zero polynomial. Let $b\in k\setminus \{0\}$, then $f(.,b)\in k[x]$ has infinite amount of roots $\implies f(.,b)\in k[x]$ is the zero polynomial.
My question is it possible to conclude that $f\in k[x,y]$ is the zero polynomial? If not can someone give me a tip to steer me into the right direction.
You are almost there.
Treat $f \in k[x,y]$ as a polynomial $f(x,y) = a_m(y)x^m + a_{m-1}(y)x^{m-1}...+a_0(y) \in k[y][x]$.
For each $b \in k - \{0\}$, $f(x,b) \in k[x]$ has infinitely many roots and hence $f(x,b) = 0$. Thus, each of the coefficients $a_m(b), a_{m-1}(b),..., a_0(b)$ are $0$, for every $b \in k - \{0\}$.
Hence, for each $0 \leq i \leq m$, $a_i(y) \in k[y]$ has infinitely many roots in $k$. Thus, each $a_i(y) = 0$.
Hence, $f(x,y) = 0$.