Polynomial vanishing on $\mathbb{A}^2$ Suppose that $f \in k[x,y]$ is such that $f(x,xy)$ vanishes everywhere on $\mathbb{A}_k^2$ ($k$ algebraically closed).  Does this imply that $f$ is the zero polynomial?  
 A: You don't need the Nullstellensatz for this question.  
Write $f(x,y) = \sum f_n(x) y^n$, where the $f_n$ are polynomials in $x$.
Then $f(x,xy)  = \sum x^n f_n(x) y^n$.  This function is supposed to vanish
identically on $k^2$.  As long as $k$ is infinite, you should find it easy 
to deduce that each $f_n$ itself vanishes identically, and thus that
$f$ is the zero polynomial.
[A previous version of this answer was based on a misreading of the question.]

A geometric perspective [added]: I will now use the language of algebraic geometry, so it is best to imagine that $k$ is algebraically closed (or else to use an appropriately sophisticated set of foundations; but if we assume that $k$ is alg. closed, then the foundations of Hartshorne Ch. I, or any other similar introductory treatment, will do.)  
The map $(x,y) \mapsto (x,xy)$ is a morphism $f$ from $\mathbb A^2$ to itself.  The ideal of polynomials in $k[x,y]$ which vanish on its image corresponds to an algebraic subset
of $\mathbb A^2$, which is precisely the Zariski closure of this image of $f$.
Now this ideal consists exactly of those $f$ such that $f(x,xy)$ is identically zero, and so the original problem amounts to showing this ideal is the zero ideal. But
the zero ideal corresponds precisely to the algebraic subset which is all of $\mathbb A^2$, so what we have to show, in the end, is that $f$ has Zariski dense
image.  
(This is quite a lot of words I've just written, but it is good practice to work through them carefully enough to understand this translation; it is the kind
of reasoning/translating that should become intuitive if you want to understand the relationship between algebraic geometry and algebra.)
Now it's easy to compute the image of $f$, just by looking at the formula:
it consists of those points $(x,y)$ for which $x \neq 0$, together with the point $(0,0)$.  If we write this image as $X$, then in particular we see
that 
$$\mathbb A^2 = X \cup \text{ the line $x = 0$},$$
and so if we let $\overline{X}$ denote the Zar. closure of $X$,
then $$\mathbb A^2 = \overline{X} \cup \text{ the line } x = 0.$$
Now $\mathbb A^2$ is irreducible, and hence cannot be written as a union
of two proper Zar. closed subsets.  Since the line $x  = 0$ is a proper
closed subset, we see that $\overline{X} = \mathbb A^2$, as required.

Actually, this geometric argument is not completely unrelated to the 
algebraic argument.  For example, a suitable adaptation of the algebraic
argument proves that $\mathbb A^2$ is irreducible.  (See here for example.)
Also, if you work the algebraic argument carefully, you will find
first that for each $n$, the polynomial $x^n f_n(x)$ induces the zero function
on $k$.
Next, you have to use the fact that $k$ has infinitely many elements
that are non-zero to conclude that $f_n(x)$ is an identically zero polynomial.
This step, where you use the fact that there are plenty of points $(x,y)$ for which
$x \neq 0$, corresponds to the step in the geometric argument in which we observe
that the line $x = 0$ is a proper closed subset of $\mathbb A^2$.
A: Yes. By the Nullstellensatz, $f (x, x y)$ vanishes everywhere if and only if $f (x, x y)$ is the zero polynomial, so it suffices to verify that the $k$-algebra homomorphism $k [x, y] \to k [x, y]$ defined by $x \mapsto x$ and $y \mapsto x y$ is injective. (Hint: Choose a convenient $k$-linear basis for $k [x, y]$.)
