Let be a field and let be a vector space over of some finite dimension . Show that the mapping given by is an -vector space isomorphism but not a ring isomorphism for . Exhibit an -algebra isomorphism .

We begin with a lemma.

Lemma: Let be a unital ring and let be left unital -modules. If is -bilinear, then the induced map given by is a well-defined -module homomorphism. Proof: To see well definedness, we need to verify that is a module homomorphism. To that end note that . Similarly, to show that is a module homomorphism, note that , so that .

[Note to self: In a similar way, if is a unital ring and unital modules, and is trilinear, then is bilinear. (So that the induced map is a module homomorphism, or unilinear- if you will.) That is to say, in a concrete fashion we can think of multilinear maps as the uncurried versions of higher order functions on modules. (!!!) (I just had a minor epiphany and it made me happy. Okay, so the usual isomorphism is just this lemma applied to the dot product … that’s cool.) Moreover, if and if and are -algebras, then the induced map is an algebra homomorphism if and only if and .]

Define by . This map is certainly bilinear, and so by the lemma induces the linear transformation . Since has finite dimension, and since its dual space has the same dimension, to see that is an isomorphism of vector spaces it suffices to show that the kernel is trivial. To that end, suppose . Then we have for all . In particular, we have for all . If there exists a nonzero element , then by the Building-up lemma there is a basis of containing . In particular, there is a linear transformation such that . That is, we have , so that . Hence is injective, and so an isomorphism of vector spaces.

Note that , while . If has dimension greater than 1, then is not a commutative ring. Thus these expressions need not be equal in general. In fact, if we choose , , and such that , , and , then clearly and . In particular, is not a ring isomorphism if . On the other hand, if , then is commutative, and is a ring isomorphism.

On the other hand, these rings are clearly isomorphic since and are vector spaces of the same dimension.

Note that and are both -algebras via the usual scalar multiplication by . Fix a basis of , and identify the linear transformation with its matrix with respect to this basis. (Likewise for .) Now define by . It is clear that is well defined, and moreover is an -vector space homomorphism. Note also that , so that . Thus is a ring homomorphism; since , we have , and indeed is an -algebra homomorphism. It remains to be seen that is an isomorphism; it suffices to show injectivity. To that end, suppose for all . Then for all , and so . Thus is an -algebra isomorphism . Note that depends essentially on our choice of a basis , and so is not “natural”.