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Let $A,B$ be commutative rings with identity. Let $f:A \rightarrow B$ be a ring homomorphism and let $M$ be an $A$-module. Since $B$ can be viewed as an $A$-module with the operation $A \times B \rightarrow B$ given by $(\alpha,b) \mapsto f(\alpha)b$, we can define the $A$-module $M_B = B \otimes_{A} M$. How can i show that $M_B$ is also a $B$-module with action such that $(b',b \otimes_{A} x) \mapsto (b'b \otimes_A x)$?

Added: In order to show that $M_B$ is a $B$-module, we need to construct a ring homomorphism $B \rightarrow End(M_B)$. How can we do that?

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there are several axioms defining a $B$-module (underlying abelian group, distributivity, actions of 0 and 1, etc.). Is checking one of these giving you difficulty, or are you asking if you really have to check them (I think you do)? If it's checking one of them, which one? –  user29743 Apr 23 '12 at 20:49
    
What i don't understand is why the map $(b',b \otimes x) \mapsto (b'b \otimes x)$ is well-defined and moreover why this induces an endomorphism. –  Manos Apr 23 '12 at 20:59

1 Answer 1

up vote 6 down vote accepted

For each $b \in B$, consider the map

$$[b]:B \times M \to B\otimes_A M$$

$$(b',m) \mapsto bb'\otimes m.$$

It is immediate that $[b]$ is bilinear and $A$-balanced, and thus factors through a homomorphism (which we also call $[b]$):

$$[b]: B\otimes_A M \to B\otimes_A M$$

given on simple tensors by $[b](b'\otimes m) =bb'\otimes m$.

It is immediate from this that $[b_1]\circ [b_2] = [b_1b_2]$, and that the collection of all these homomorphisms makes $B\otimes_A M$ into a $B$-module.

Note: to remember what module structures has a given tensor product, a good trick is this: if $A,B,C$ are any three rings (possibly non-commutative), and if $M$ is an $(A,B)$-bimodule (hence $M$ is a left $A$-module and a right $B$-module, and the two module structures are compatible), and $N$ is a $(B,C)$-bimodule, then $M\otimes_B N$ is an $(A,C)$-bimodule (just like the product of an $n\times m$ matrix with a $m \times p$ matrix has dimensions $n\times p$).

Also: (1) if $R$ is commutative, then every $R$-module is naturally an $(R,R)$-bimodule; (2) in general, a left $R$-module is the same thing as a $(R,\mathbf{Z})$-bimodule; (3) if $M$ is a right $R$-module and $N$ is a left $R$-module, $M\otimes_R N$ only has the natural structure of an abelian group (which is a $(\mathbf{Z},\mathbf{Z})$-bimodule).

In your problem $B$ is a $(B,A)$-bimodule, and $M$ an $(A,A)$-bimodule; hence $B\otimes_AM$ is a $(B,A)$-bimodule.

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Of course, this an interesting particular case of the very intriguing Eilenberg-Watt's theorem. –  Alex Youcis Apr 24 '12 at 1:17

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