0
$\begingroup$

For an $A$-module $M$ and a morphism of rings $F:A\to B$, if I wanted to give $B\otimes_A M$ the structure of a $B$-module and verify it induces a morphism $Mod_A\to Mod_B$ , I note that a general element of $B\otimes_A M$ is of the form $\sum_k a_k(b_k\otimes m_k)$ and I identify this with $\sum_k (F(a_k)b_k)\otimes m_k$ where $F(a_k)b_k$ is just usual multiplication in $B$, right?

But it is still not clear how this will describe a functor $Mod_A\to Mod_B$, because in general, I won't know that $F(a_k)x_k$ will be well defined when the $x_k$ aren't in $B$.

$\endgroup$

1 Answer 1

1
$\begingroup$

The functor looks like this:

$B\otimes _A \_ : Mod_A \to Mod_B $

$(M \in Mod_A) \mapsto B\otimes _A M$

$(f : M \to N ) \mapsto ((id \otimes f): B \otimes _ A M \to B \otimes _A N)$

Writing it out this is the map that sends $a\otimes m \mapsto a \otimes f(m)$.

And $B\otimes _A M$ is $B$-Module by $ b (a\otimes m) := ba \otimes m$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .