Proof. Part (A). Let $k \subset K$ be a finitely generated field extension. Let $R$ be a $k$-algebra which is a local complete intersection. We can find a global complete intersection $A = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ over $k$ such that $K$ is isomorphic to the fraction field of $A$, see Algebra, Lemma 10.156.11. Then $R \to R \otimes _ k A$ is a relative global complete intersection. It follows that $R \otimes _ k A$ is a local complete intersection by Divided Power Algebra, Lemma 23.8.9.
Proof of (B). This is clear because a ring is a local complete intersection if and only if all of its local rings are complete intersections.
Part (C). Let $A \to B \to C$ be flat maps of Noetherian rings. Assume the fibres of $A \to B$ are local complete intersections and $B \to C$ is regular. We have to show the fibres of $A \to C$ are local complete intersections. Clearly, we may assume $A = k$ is a field. Then we may assume that $B \to C$ is a regular local homomorphism of Noetherian local rings. Then $B$ is a complete intersection and $C/\mathfrak m_ B C$ is regular, in particular a complete intersection (by definition). Then $C$ is a complete intersection by Divided Power Algebra, Lemma 23.8.9.
Part (D). This follows by the same arguments as in (C) from the other implication in Divided Power Algebra, Lemma 23.8.9. Part (E) is immediate as the condition does not refer to the ground field. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like
$\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.