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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 $a_i \in A$ .

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    $\begingroup$ 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..." $\endgroup$ Sep 22, 2022 at 6:47
  • $\begingroup$ A side remark if don't consider the finiteness, the arguement can be simpler as :math.stackexchange.com/a/844880/360262 $\endgroup$
    – yi li
    Nov 5, 2022 at 13:58

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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.

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  • $\begingroup$ very clear , thanks $\endgroup$
    – yi li
    Sep 22, 2022 at 6:57

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