If a polynomial $f(X,Y)$ vanishes on all $(x,y)$, can we conclude $f(X,Y)=0$?

Question

Let $k$ be an infinite field and $f(X,Y) \in k[X,Y]$ be a polynomial with two variables. If $f(X,Y)$ vanishes on all points $(x,y) \in k^2$, i.e., $$f(x,y) = 0, \forall (x,y) \in k^2$$ can we conclude that $f(X,Y) = 0$?

Put another way, let $k$ be an infinite field, and let $$k[X,Y]^* := \{f : k^2 \longrightarrow k, (x,y) \mapsto f(x,y) \mid f(X,Y) \in k[X,Y]\}$$ denote the ring of polynomial functions. Then do we have $k[X,Y] \cong k[X,Y]^*$?

Answer 1

To answer your first question, yes, if $ f (x,y) $ vanishes for all $(x,y) $ then indeed $f $ must be the identically zero polynomial.

The second question is also true, it's very easy to construct the isomorphism. Just take $f\in k [X,Y]^*$ and match it with $f (X,Y) \in k [X,Y] $ from where $f$ takes it's correspondence rule in $k [X,Y]^*$.