Polynomial vanishing on $\mathbb{A}^2$

Question

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?

Answer 1

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.]

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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.

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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 $\mathbb A^2$.

Answer 2

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 $k [x, y]$.)