Basic solution (linear programming)

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 a 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 separable algebraic field extension L of k is 0-étale over k. the formal power series ring $k[[t_1,\dots ,t_n]]$ is 0-smooth only when $\mathrm{char}k=p$ and $[k:k^p]<\mathrm{\infty }$ (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, $B=A[[t_1,\dots ,t_n]]$ and $I=(t_1,\dots ,t_n).$ Then B is I-smooth over A.

Let A be a noetherian local ring with maximal ideal $\mathfrak{m}$ that is a k-algebra. Then A is $\mathfrak{m}$-smooth over k if and only if $A{\otimes }_k{k}^{\prime }$ is a regular local ring for any finite field extension $k^{\prime }$ of k.

See also

• étale morphism
• formally smooth morphism

References

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• H. Matsumura Commutative ring theory. Translated from the Japanese by M. Reid. Second edition. Cambridge Studies in Advanced Mathematics, 8.