# Does finite integral extension of a ring will induced finite algebraic extension over the quotient field?

Let $$A,B$$ be commutative unital ring. Let $$A \subset B$$ be a finite integral extension , and both of them are integral domain, hence we can consider the quotient field of $$A$$ and $$B$$.

Does the quotient field $$Q(B)$$ be the finite algebraic extension of $$Q(A)$$

By definition finite integral extension implies $$B$$ is a finite generated $$A$$ module. i.e. exist a finite set of $$\{e_1,...,e_n\}\subset B$$ such that any elements in $$B$$ can be written as $$b = \sum a_i e_i$$ for some $$a_i \in A$$ .

• Yes. Your definition of an integral extension is incorrect; the quantifiers are "there exists a finite set $\{ e_1, \dots e_k \} \subset B$ such that..." Sep 22, 2022 at 6:47
• A side remark if don't consider the finiteness, the arguement can be simpler as :math.stackexchange.com/a/844880/360262 Nov 5, 2022 at 13:58

Tensoring $$\operatorname{Frac}(A)$$ with the inclusion $$A\to B$$ over $$A$$ gives rise to a morphism $$\operatorname{Frac}(A)\to \operatorname{Frac}(A) \otimes _A B$$. The latter is localization of B (so, an integral domain) and is generated over $$\operatorname{Frac}(A)$$ by a finite number of integral elements, namely $$1\otimes b_i$$, where $$b_i$$ are a set of generators of $$B$$ over $$A$$. Thus it is a field and moreover finite over $$\operatorname{Frac}(A)$$. Finally, the canonical morphism $$\operatorname{Frac}(A) \otimes _A B \to \operatorname{Frac}(B)$$ is an isomorphism since it has the canonical inverse.