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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?

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1 Answer 1

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 $X$.

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