Artin–Tate lemma

In algebra, the Artin–Tate lemma, named after Emil Artin and John Tate, states:^[1]

B\subset C

algebras over A. If C is of finite type over A and if C is finite over B, then B is of finite type over A.

(Here, "of finite type" means "finitely generated algebra" and "finite" means "finitely generated module".) The lemma was introduced by E. Artin and J. Tate in 1951^[2] to give a proof of Hilbert's Nullstellensatz.

Proof

The following proof can be found in Atiyah–MacDonald.^[3] Let $x_{1},...,x_{m}$ generate $C$ as an $A$ -algebra and let $y_{1},...,y_{n}$ generate $C$ as a $B$ -module. Then we can write $x_{i}=\sum _{j}b_{{ij}}y_{j}$ and $y_{i}y_{j}=\sum _{{k}}b_{{ijk}}y_{k}$ with $b_{{ij}},b_{{ijk}}\in B$ . Then $C$ is finite over the $A$ -algebra $B_{0}$ generated by the $b_{{ij}},b_{{ijk}}$ . Using that $A$ and hence $B_{0}$ is Noetherian, also $B$ is finite over $B_{0}$ . Since $B_{0}$ is a finitely generated $A$ -algebra, also $B$ is a finitely generated $A$ -algebra.

Noetherian necessary

Without the assumption that A is Noetherian, the statement of the Artin-Tate lemma is no longer true. Indeed, for any non-Noetherian ring A we can define an A-algebra structure on $C=A\oplus A$ by declaring $(a,x)(b,y)=(ab,bx+ay)$ . Then for any ideal $I\subset A$ which is not finitely generated, $B=A\oplus I\subset C$ is not of finite type over A, but all conditions as in the lemma are satisfied.

References

↑ Eisenbud, Exercise 4.32
↑ E Artin, J.T Tate, "A note on finite ring extensions," J. Math. Soc Japan, Volume 3, 1951, pp. 74–77
↑ Atiyah–MacDonald 1969, Proposition 7.9

Eisenbud, David, Commutative Algebra with a View Toward Algebraic Geometry, Graduate Texts in Mathematics, 150, Springer-Verlag, 1995, ISBN 0-387-94268-8.
M. Atiyah, I.G. Macdonald, Introduction to Commutative Algebra, Addison–Wesley, 1994. ISBN 0-201-40751-5

External links

http://commalg.subwiki.org/wiki/Artin-Tate_lemma

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