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.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
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 $F$.
-
$\begingroup$ How do you prove the linear independence of $\{1\otimes e_i\}$? $\endgroup$– BogdanJul 31 at 15:11