Let $R$ be a ring and $M$ a free $R$-module with basis $B$. If $P$ is a prime ideal of $R$, is $B_P$ a basis for the $R_P$-module $M_P$?
2026-04-06 03:19:39.1775445579
Basis of the localization of a free module is the localization of its basis?
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Let $f : R \to S$ be an arbitrary homomorphism of rings and let $M$ be a free $R$-module. Then $M \otimes_R S$ is a free $S$-module, and any basis of $M$ is a basis of $M \otimes_R S$. This follows more or less by the universal property of the tensor product together with the universal property of free modules.