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How do you know which rings $R$ and fields $F$ satisfy

Given a free $R$-module $M$, $\textrm{rank}(M) = \dim_F(F\otimes_{R}M)$?

E.g., $(\mathbb{Z},\mathbb{Q})$ seems to work.

1 Answers 1

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To give a sense to the right hand side, you need a ring homomorphism $R\to F$. Then the equality holds. Actually, if $\{ e_i\}_i$ is a basis of $M$ over $R$, then $\{ 1\otimes e_i\}_i$ is a basis of $F\otimes_R M$ over $F$.