TITLE: Smoothness over a field and regularity
QUESTION [6 upvotes]: Hartshorne, Algebraic Geometry
In example III.10.0.3, Hartshorne remarks that with k algebraically closed, X smooth of dimension n over Spec k is equivalent to X regular of dimension n.  He references II.8.8.
However II.8.8 requires that when one looks at a local ring B, the relative sheaf of differentials for this local ring must be a free B-module.  This is certainly true if X is irreducible.  But what if X is not irreducible?
So my overall question: is the statement from III.10.0.3 true for X not irreducible?

REPLY [3 votes]: After wrestling with definitions, I have the answer to my question.  If two irreducible components meet at a point, then $\dim_{k}(\Omega_{X/k} \otimes k)$ will jump, contradicting the definition of smooth.
Thus, while $X$ may have multiple irreducible components, each connected component is irreducible.  Since II.8.8 is a local property, we may use it on each irreducible component of $X$.