Artin–Tate lemma

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

Let A be a Noetherian ring and $B \sub C$ 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 $y_iy_j = \sum_{k}b_{ijk}y_k$ 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

1. Eisenbud, Exercise 4.32
2. E Artin, J.T Tate, "A note on finite ring extensions," J. Math. Soc Japan, Volume 3, 1951, pp. 74–77
3. 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