Irreducible component of a reduced scheme

Question

Let $A$ be a reduced ring, commutative with unity. Let $C$ be an irreducible component of $Spec \ A$. Does it then follow that $C$ is isomorphic as a scheme to $Spec \ B$ of some ring $B$ that is an integral domain? Thank you!

Answer 1

As Richard points out in the comments, my old answer was bogus. Of course irreducible closed subschemes of reduced affine schemes can fail to be reduced (and hence fails to be integral). E.g., Richard gave the example of the subscheme of $k[x,y]/(xy)$ cut out by $x^2$.

Though, one should also note that "irreducible component" typically only applies to topological spaces. As Richard's example indicates, one can often place a number of different scheme structures on an irreducible component of even a reduced scheme.

The following old answer is rubbish: "Yes closed subschemes of affine schemes are affine. Closed subschemes of Spec $A$ correspond to ideals $I$ of $A$, and are themselves of the form Spec $A/I$. Irreducible components of affine schemes are closed, hence they also have such a form. Under the assumption that $A$ is reduced, the irreducible component only has one possible closed-subscheme structure (their closed subscheme structure is determined by the topology), and that is Spec $A/P$ where $P$ is a prime ideal, so the answer to your question is yes. (If $P$ weren't prime then $A/P$ wouldn't be irreducible)"