## Extension of Scalars(Pt. II)

**Point of Post: **This is a continuation of this post.

So, let’s look at some examples of extension of scalars, to get our heads around things.

Let’s extend the scalars on from to . What should we expect the answer to be? Well, we know that naturally sits inside as an -space, and so seems like a pretty good candidate. From the previous theorem we see that we have an -map such that , we claim that is an isomorphism. Indeed, the fact that is surjective follows from the fact that is -linear and so that hits all the generators of . To see that is injective we merely note that if is in then . Now, write and so we have that

and so we must have that is zero. But, we see then that

and so has trivial kernel, so is injective. Thus, is a -isomorphism and so we see that .

In fact, it’s easy to see that the above argument really had nothing to do with or and can really be stated:

**Theorem: ***Let be unital rings. Then, as left -modules.*

In a different direction of generalizing, we note that the above extensions of scalars from to is so common place, that it’s given a name. Namely, if is some -space, then the extension of scalars on to (e.g. regarded as a -space) is called the *complexification* of and is often denoted .

In a similar vein, let’s see what the complexification of looks like. I bet you already have a good guess. Namely, since we can include into in a -linear way we have a -map such that . Once again, it’s clear that this map is going to be surjective since it is -linear and hits the usual generators of (e.g. the matrices corresponding to the standard basis vectors in ) and similarly to the case of the polynomials, if is in the kernel of this map, then and so . Doing the same trick as the polynomial casewhere we break into a double sum one over and one over , interchanging the order of summation, and bringing the ‘s to the left shows that and so has trivial kernel and so is an isomorphism. Once again, there was nothing special in this argument about or so that we can definitely say that

**Theorem: ***Let be unital rings. Then, as left -modules.*

As a last general example let’s see what happens when we extend scalars on free modules. In particular, we merely note that if then by previous discussion we know that as left -modules. Or to be more precise:

**Theorem: ***Let be rings. Then, as left -modules. In particular, the extension of scalars on a free module results in a free module, and in the case that (and thus ) have the IBN property they have the same rank.*

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. *Module Theory.* Clarendon, 1990. Print.

[4] Lang, Serge. *Algebra*. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. *Abstract Algebra*. New York: Springer, 2007. Print.

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