Past exam of the algebra course of the University of Oxford / 2011 / 2 / a / ii

Table of contents
- Solution ii

Solution

For

a \in A

,

m \in M

,

(f \cdot x) (am) = f (am) x = a f (m) x = a (f \cdot x) (m) .

Additivity and complex linearity follow from those of

f

and multiplication in

A

, so

f \cdot x

is an module homomorphism. In particular the action stays inside

V

. Associativity gives

((f \cdot x) \cdot y) (m) = f (m) x y = (f \cdot (x y)) (m), (f \cdot 1) (m) = f (m) .

Both distributive laws hold pointwise. These are exactly the axioms for a right module, proving $Hom_{A} (M, A)$ is a right $A$ -module.

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