$\text{Spec}~\mathbb C[x,y]/(xy)$ is connected in Zariski topology I want to prove that $\text{Spec}~\mathbb C[x,y]/(xy)$ is connected w.r.t. Zariski topology.(so that it is an example reducible space but connected under Zariski topology).
Intuitively, $\text{Spec}~\mathbb C[x,y]/(xy)$ is the "union of $x$-axis and $y$-axis" and clearly "connected", but I wish to prove this rigorously in the Zariski topology. To begin with, we identify $\text{Spec}~\mathbb C[x,y]/(xy)$ with the subspace of $\mathbb A^2_\mathbb C$ containing $(xy)$ and write it as a disjoint union of $V(I)$ and $V(J)$. But I get stuck in here and don't know how to continue. I wish the solution doesn't use theorems that are too advanced.
 A: Using this well known description of when the spectrum of a ring is connected, you only have to show that there is no non-trivial idempotent in $R=k[x,y]/(xy)$. Let $e^2=e$ in $R$, i.e. $e(e-1) \in (xy)$.
In particular $e(e-1) \in (x)$ and since $(x)$ is a prime-ideal, we have w.l.o.g $e \in (x)$. We want to prove $e \in (y)$, because then we have $e=0 \in R$ and we are done.
If not, we have $e-1 \in (y)$ and consequently $R=(e,e-1) \subset (x,y)$, contradiction! This finishes the proof.
A: $\def\Spec{\operatorname{Spec}}\def\C{\mathbb{C}}$Well, the map $\C[x,y]/(xy)\to \C[t]$ which maps $x$ to $t$ and $y$ to zero induces a map $\Spec \C[t]\to \Spec \C[x,y]/(xy)$ which is a homeomorphism from the line $\Spec \C[t]$ to a closed subset of $\Spec \C[x,y]/(xy)$. That image, call it $X$, is connected, because $\Spec \C[t]$ is connected (it is irreducible, in fact)
Similarly, you can construct a closed connected subset as the image of the map induced by the morphism $\C[x,y]/(xy)\to \C[t]$ which maps $x$ and $y$ to $0$ and to $t$, respectively.  
Now show that $X\cup Y$ is $\Spec \C[x,y]/(xy)$, and that the intersection $X\cap Y$ is not empty. A basic fact from topology then implies that $\Spec \C[x,y]/(xy)$ is connected.
