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Smooth algebra : ウィキペディア英語版
Smooth algebra

In algebra, a commutative ''k''-algebra ''A'' is said to be 0-smooth if it satisfies the following lifting property: given a ''k''-algebra ''C'', an ideal ''N'' of ''C'' whose square is zero and a ''k''-algebra map u: A \to C/N, there exists a ''k''-algebra map v: A \to C such that ''u'' is ''v'' followed by the canonical map. If there exists at most one such lifting ''v'', then ''A'' is said to be 0-unramified (or 0-neat). ''A'' is said to be 0-étale if it is 0-smooth and 0-unramified.
A finitely generated ''k''-algebra ''A'' is 0-smooth over ''k'' if and only if Spec ''A'' is a smooth scheme over ''k''.
A separable algebraic field extension ''L'' of ''k'' is 0-étale over ''k''. The formal power series ring k is 0-smooth only when \operatornamek = p > 0 and (k^p ) < \infty (i.e., ''k'' has a finite ''p''-basis.)
== ''I''-smooth ==
Let ''B'' be an ''A''-algebra and suppose ''B'' is given the ''I''-adic topology, ''I'' an ideal of ''B''. We say ''B'' is ''I''-smooth over ''A'' if it satisfies the lifting property: given an ''A''-algebra ''C'', an ideal ''N'' of ''C'' whose square is zero and an ''A''-algebra map u: B \to C/N that is continuous when C/N is given the discrete topology, there exists an ''A''-algebra map v: B \to C such that ''u'' is ''v'' followed by the canonical map. As before, if there exists at most one such lift ''v'', then ''B'' is said to be ''I''-unramified over ''A'' (or ''I''-neat). ''B'' is said to be ''I''-étale if it is ''I''-smooth and ''I''-unramified. If ''I'' is the zero ideal and ''A'' is a field, these notions coincide with 0-smooth etc. as defined above.
A standard example is this: let ''A'' be a ring, B = A and I = (t_1, \ldots, t_n). Then ''B'' is ''I''-smooth over ''A''.
Let ''A'' be a noetherian local ''k''-algebra with maximal ideal \mathfrak. Then ''A'' is \mathfrak-smooth over ''k'' if and only if A \otimes_k k' is a regular ring for any finite extension field k' of ''k''.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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