$k$ is algebraically closed $\Leftrightarrow$ $\forall f \in k[X,Y], V(f)\neq\{(0,0)\}$ Let $k$ be a field. I want to show that
$k$ is algebraically closed if and only if for all $f\in k[X,Y],V(f)=\{(a,b)\in k^2|f(a,b)=0\}\neq\{(0,0)\}$.
I proved that for all $f\in k[X,Y],V(f)=\{(a,b)\in k^2|f(a,b)=0\}\neq\{(0,0)\}$ if $k$ is algebraically closed, so I have to prove the other direction.
My idea is to find some polynomial $f\in k[X,Y]$ such that $V(f)=\{(0,0)\}$ by using non-constant polynomial $g\in k[t]$ which has no root in $k$. But I can't find it.
How to prove?

Let $p\in k[X]$ be an irreducible polynomial of degree $\ge 2$. Then, $f(X,Y)=Y^{\deg p}p\left(\frac XY\right)$ is a polynomial such as the ones you are looking for, because $(a,b)\in V(f)\setminus\{(0,0)\}$ if and only if $\frac ab$ is a root of $p$. On the other hand, $f$ is a homogeneous polynomial in two variables of degree $\deg p$, therefore $(0,0)\in V(f)$.
