Let $A\subset B$ with $B$ an integral domain. If $B$ is integral over $A$ can we say that $Q(B)$ is algebraic over $Q(A)$ ? (Here $Q(\dots)$ denotes the quotient field of $(\dots))$.)
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Yes this true. Localize at $S = A \setminus \{0\}$ to get an integral extension $S^{-1}A \subset S^{-1}B$. Since $S^{-1}A$ is a field (the quotient field of $A$) and the extension is integral, we obtain that $S^{-1}B$ is also a field, hence equal to the quotient field of $B$.
MooS
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yea I understand..thank you. – Via Mar 01 '15 at 13:11
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Are you sure $S^{-1}B$ is a field when $B$ is not finiteley generated over $A$? – Bernard Mar 01 '15 at 13:27
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If $R \subset S$ is an integral extension of integral domains, then $R$ is a field iff $S$ a field. A finite extension is not required in this theorem. – MooS Mar 01 '15 at 13:53