How to show Spec$\frac{\Bbb C [x, y]}{(xy)}$ is connected in the Zariski topology In exercise 3.6.E of Vakil's book, it is asked for the reader to find a ring for which its spec is connected and non-irreducible. Taking his hint, I  thought of $A = \frac{\Bbb C [x, y]}{(xy)}$. It is not irreducible, because $V((x)) \cup V((y)) = V((xy)) = \operatorname{Spec}A$. To prove connectedness I tried to use the fact that if $ \operatorname{Spec}A= V(I) \cup V(J) = V(IJ)$ where $V(I) \cap V(J) = \emptyset$ would imply $I + J = A$. Can anyone give me some hint?

The comments give you a completely algebraic way to view connectedness: a scheme $X$ is disconnected if and only if there exists $e\in\Gamma(X,\mathcal{O}_X)$ such that $e\neq 0,1$ and $e^2 = e.$ Proving that no such $e$ exists will do the trick.
Here's another method:
Hint: View $\operatorname{Spec}\Bbb{C}[x,y]/(xy)$ as a closed subscheme of $\operatorname{Spec}\Bbb{C}[x,y] = \Bbb{A}^2_\Bbb{C}$ (the particular subscheme structure doesn't matter since we're concerned about connectedness, which is a purely topological property). Recall that if $U_1, U_2\subseteq X$ are connected and $U_1\cap U_2\neq\emptyset,$ then $U_1\cup U_2$ is connected.
