Is there a polynomial which vanishes at only one point Assume $(a,b) \in \mathbb C^2$. Is there a polynomial $f(x,y)$ which vanishes only at $(a,b)$ ?
Thanks.
You could use the algebraic sledgehammer. If $V(f)$ is a point, $\sqrt{(f)}$ is a maximal ideal, which is a contradiction, since $(f)$ is always contained in a non-maximal prime ideal $(g)$, where $g$ is an irreducible factor of $f$.

Or you do it the more elementary way:

Without loss of generality assume $(a,b)=(0,0)$. For any $t \in \mathbb C \setminus \{0\}$ the polynomial $g(x) := f(x,t)$ has no root, hence is constant. The same holds for $f(t,y)$ and you can deduce that $f$ itself is constant from that.
No,

You are working on a plane, and one single equation $f(x,y)=0$ left you a 1-dimensional solutions subspace.