# Proof that the image of $(x,y) \mapsto (x, xy)$ is neither open, closed nor locally closed.

Let $$k$$ be an algebraically closed field and $$f: \mathbb{A}^2_k \rightarrow \mathbb{A}^2_k$$, $$(x,y) \mapsto (x,xy)$$. I understand what the image looks like, i.e. $$\mathbb{A^2_k} - (\mathbb{A}^1-0)$$. By observing that: $$\mathbb{A}^2_k = \textrm{Im} \cup \mathbb{A}^1_k$$ and since $$\mathbb{A}^2_k$$ is irreducible, by taking the closure we get that $$\overline{\textrm{Im}}$$ is the whole plane and therefore that the image is not closed. If it was open, the intersection $$\textrm{Im} \cap \{X \neq 0\} = 0$$ would also be open, which is impossible because $$\mathbb{A}^2_k$$ is irreducible. Finally, since the image is dense and is not open in the whole set, it can't be locally closed either.

I find this reasoning to be a bit arbitrary. In particular, for the "locally closed" part, I initially wanted to look that at every neighbourhood of $$\{(0,0)\}$$, which would be the whole plane without some curves, but I feel like that is too tedious.

Is there a more intuitive (understand obvious) way to think about these things ? In general, is there an intuitive way to think about sets being closed or open in the Zariski topology?

• Thank you for the quick answer. Perhaps I should have specified that this is exactly what I don't completely get: justifying that something CANNOT be the zero set of any polynomial. You say that in this a case, the polynomial is "clearly" zero but it is not so clear for me why exactly. Commented Oct 12, 2023 at 20:10
• What I mean is that for the complement, though it is very natural to think that the polynomial should also vanish at the origin, I can't come up with a formal and direct argument that it should beyond the above proof. Commented Oct 12, 2023 at 20:15
• We can write such a polynomial by separating its coefficients $\sum k + \sum X + \sum Y + \sum XY$. By taking x=0, we're reducing the polynomial to the Y and constant parts, and hence to a polynomial on $k[Y]$, were it should only have finitely many roots. Hence, all coefficients of constants and the Y part are 0 and the polynomial can be factored by X, so it is zero everywhere. Is this it ? For the limit part, I'm unaware of what it refers to. Commented Oct 12, 2023 at 20:41
• By zero everywhere, I meant zero at the origin as well :) Commented Oct 12, 2023 at 20:51