Irreducibility of schemes under etale morphism Let $X$ and $Y$ be two schemes and let $Y$ be irreducible. 
If $X \hookrightarrow Y$ is an open immersion or in other words if $X$ is an open subscheme of $Y$ then we know that $X$ is also irreducible. My question is what happens when we have an etale (or maybe etale surjective) map instead of open immersion? Do we still have $X$ irreducible?
 A: Suppose that $f:X\to Y$ is finite etale with $X$ and $Y$ connected and $Y$ normal. Then, $X$ is connected. The key is the oldest trick in the book:

Theorem: Suppose that $Z$ is a connected scheme (uh..maybe locally Noetherian..probably not) with $\mathcal{O}_{Z,z}$ integral for all $z\in Z$. Then, $Z$ is irreducible. 


Remark: This trick has a funny place in algebraic geometry, it seems to me. Namely, it was a conjecture for a long time that $M_g$ (the coarse moduli space of genus $g$ curves) was irreducible. It was known to be connected, but one can't apply the above trick because it's singular. That said, it's only singular since coarse spaces can be poor representations of the actual geometry of the objects they parameterize. Namely, the stack $\mathscr{M}_g$ is connected and smooth and thus irreducible by the trick. Since we have a surjection $\mathscr{M}_g\to M_g$ this shows the latter is irreducible. It was actually this that motivated Deligne and Mumford to create stacks in their seminal paper on this subject.

The proof of this theorem is just that at the intersection of two components (which must intersect by connectedness) the local ring will have two minimal primes which contradicts that the local ring is integral.
So, why this makes the case of $f:X\to Y$ finite etale with $X$ and $Y$ connected, $Y$ normal, good is the following result from Matsumura:

Lemma: Let $f:X\to Y$ be smooth and let $y=f(x)$. Then, $\mathcal{O}_{X,x}$ is normal if and only if $\mathcal{O}_{Y,y}$ is.

In particular, if $f:X\to Y$ is finite etale with $Y$ normal, then $X$ is normal. So, if $X$ is normal and connected then by the theorem above, it's irreducible.
As an example when $Y$ is not normal, consider the projective nodal cubic $Y=\text{Proj}(k[x,y,z]/(y^2z-x^2(x+z))$--for simplicitly, let's assume that $k=\overline{k}$ and characteristic $0$. This has fundamental group $\widehat{\mathbb{Z}}$ with the finite etale cover of degree $n$ being given by a  Neron $n$-gon-- a Neron $n$-gonwhich is essentially a chain of $\mathbb{P}^1$'s glued together in a 'pearl necklace' formation. These are certainly not irreducible.
This question is actually important because it basically allows one to deduce when $\pi_1^{\mathrm{\acute{e}t}}(U,\overline{x})\to\pi_1^{\mathrm{\acute{e}t}}(X,\overline{x})$ is surjective for $U\subseteq X$ (both connected) open. The above shows that this is true when $X$ is normal. For example, you can use the above Neron $n$-gon example to show that $U=X-\{[0:0:1],[0:1:0]\}$ is isomorphic to $\mathbf{G}_m$ and so has fundamental group $\widehat{\mathbb{Z}}$ as well. But, the map $\pi_1^{\mathrm{\acute{e}t}}(U,\overline{x})\to\pi_1^{\mathrm{\acute{e}t}}(X,\overline{x})$ is the zero map!
