Why are $xy=0$ and $x^2-x=0$ not isomorphic?

Question

I'm not sure how to rigourously back up my intuition that the curves given by $x^2-x=0$ and $xy=0$ in $\mathbb{A}^2_k$ are not isomorphic.

If I denote the varieties as $V_1=V(xy)$ and $V_2=V(x^2-x)$, then $V_1$ is the coordinate axes, and $V_2$ is the disjoint of the vertical lines $x=0$ and $x=1$. I suspect these curves are not isomorphic since the axes cross each other at the origin.

I know two algebraic varieties $V_1$ and $V_2$ are isomorphic if we have inverse regular maps $\varphi:V_1\to V_2$ and $\psi:V_2\to V_1$. I think there should be a problem defining inverse regular maps near $(0,0)$ so that no such regular maps could exist. Why are these curves not isomorphic?

Answer 1

a) The curve $V_2$ is disconnected because it is the disjoint union of its two non-empty disjont Zariski open subsets $x\neq 0$ and $x\neq 1$.
You can also see it algebraically: an affine algebraic variety is connected iff its ring of regular functions $A$ has no idempotent element ($a^2=a$) apart from $0$ and $1$.
This is not the case for $A_2=\frac{k[X,Y]}{(X^2-X)}$, since $x$= class of $X$ is such an idempotent $\neq0,1$.

b) A topological space is connected as soon as it is the union of two connected subspaces with non-empty intersection.
You can apply this to the two subspaces $x=0,y=0$ of $V_1$, both isomorphic to the connected space $\mathbb A^1_k$, and intersecting in $(0,0)$.
Here the algebraic criterion can also be applied, noticing that every element of $A_1=\frac{k[X,Y]}{(XY)}$ can uniquely be written as $c+xa(x)+yb(y)$, but the (easy) calculation is not particularly illuminating.

Answer 2

You could make your idea rigorous, but just looking at the pictures (which can be misleading -- you might think that $y^2 = x^3 - x$ was disconnected looking at the picture over $\mathbb R$, for example) there seems to be a basic topological difference between these two varieties.

The more difficult part will be rigorously showing that $V(xy)$ is connected. I think you can use the fact that each axis is isomorphic to $\mathbb{A}^1$ in order to do this without too much pain.

Answer 3

An isomorphism between two affine algebraic curves over a field yields an isomorphism of their corresponding rings of regular functions $k[x,y]/(x(x-1))$ and $k[x,y]/(xy)$. Can you show that these rings are not isomorphic? If you know the concept of localization, localize the second ring at the ideal $m=(x,y)$, i.e. invert all elements outside this ideal. Then linearize, i.e. for the ideal $m$ of the local ring consider the $k$-vector space $m/m^2$. Now pull this vector space back by the hypothetical isomorphism to get the vector space of the preimage. Can you distinguish these two vector spaces? What invariant best distinguishes two vector spaces?