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A monic relation makes the last generator integral over the earlier ones

Statement

Let $k$ be a field, let $A$ be a $k$ -algebra of finite type, and let $y_{1}, \dots, y_{n} \in A$ generate $A$ as a $k$ -algebra. Suppose there exists a monic polynomial $Y^{m} + b_{m - 1} Y^{m - 1} + \dots + b_{0} \in k [y_{1}, \dots, y_{n - 1}] [Y]$ that vanishes at $Y = y_{n}$ . Then $y_{n}$ is integral over the subalgebra $k [y_{1}, \dots, y_{n - 1}]$ , and hence $A$ is integral over that subalgebra.

Facts & Assumptions

Given: A field $k$ , a finite-type $k$ -algebra $A$ , generators $y_{1}, \dots, y_{n} \in A$ , and a monic polynomial relation for $y_{n}$ over $k [y_{1}, \dots, y_{n - 1}]$ .

[L1]

The notation $k [y_{1}, \dots, y_{n - 1}]$ denotes the $k$ -subalgebra generated by $y_{1}, \dots, y_{n - 1}$ , and finite type means generated by finitely many elements as an algebra (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

An element is integral over a base ring exactly when it satisfies a monic polynomial over that ring (Integrality and finite-module characterizations for one element).

Proof

technique · direct

1.1L1L2given

Let $B = k [y_{1}, \dots, y_{n - 1}]$ . By [L1], $B$ is a subalgebra of $A$ . The displayed relation is a monic polynomial in $B [Y]$ with value zero at $y_{n}$ , so [L2] shows that $y_{n}$ is integral over $B$ .

2.1step 1.1L1algebra

Since $A$ is generated by $y_{1}, \dots, y_{n}$ over $k$ , it is generated by $y_{n}$ as a $B$ -algebra: $A = B [y_{n}]$ . Every element of $B$ is integral over $B$ , and step 1.1 gives integrality of $y_{n}$ over $B$ , so every element of $A = B [y_{n}]$ is integral over $B$ .

3.1step 2.1∎

Therefore $A$ is integral over the subalgebra $k [y_{1}, \dots, y_{n - 1}]$ .

Depends on

Dependency tree · 2 levels

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, Lemma 8.2