$f,g\in k[x,y]$ are relatively prime , $p\in k[x]$ and $p\in (f,g)\subseteq k[x,y]$ ; if $f(x_0,y_0)=g(x_0,y_0)=1$ , then $p(x_0)=0$?

Question

Let $f,g\in k[x,y]$ be non-zero non-unit relatively prime polynomials , where $k$ is a field . Then $f,g$ are relatively prime also in $k(x)[y]$ , so $\exists a,b \in k(x)[y]$ such that $af+bg=1$ . Clearing denominators , there is a non-zero polynomial $p\in k[x]$ such that $a'f+b'g=p$ , where $a',b' \in k[x,y]$ , so $p\in (f,g)\subseteq k[x,y]$ . Now suppose that $f(x_0,y_0)=g(x_0,y_0)=1$ , then is it true that $p(x_0)=0$ ?

Answer 1

No, and I don't know why you would expect this to be true. For instance, if $f=g=1$, then you can take $p=1$ and so $f(x_0,y_0)=g(x_0,y_0)=1$ for all $x_0$ and $y_0$ but $p(x_0)$ is never $0$.

If you want $f$ and $g$ to be non-units, just let $f_0$ and $g_0$ be any polynomials such that $1\in(f_0,g_0)$ (for instance, $f_0=y$ and $g_0=y+1$) and choose $x_0,y_0$ such that $f_0(x_0,y_0)$ and $g_0(x_0,y_0)$ are both nonzero. Then define $f=f_0/f_0(x_0,y_0)$ and $g=g_0/g_0(x_0,y_0)$ and you can again take $p=1$ as above.