Abstract Nonsense

Crushing one theorem at a time

Extension of Scalars (Pt. I)

Point of Post: In this post we discuss some of the fundamental ideas concerning extension of scalars.

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We finally can now get around to discussing a use of tensor products that I touted in my original introduction: extension of scalars. Namely, the idea is that we are handed some R-module M and some superring S\supseteq R and would like to see to what extent we can consider M to be an S-module. In other words, are are considering the problem which is dual to taking an S-module and considering it as an R-module by just “restricting” scalars. Unfortunately, this is often impossible to do. For example, \mathbb{Z} can’t be made into a \mathbb{Q}-space since, if it could, whatever \frac{1}{2}\cdot 1 would be (and note, it has to be an INTEGER since the multiplication is a map \mathbb{Q}\times\mathbb{Z}\to\mathbb{Z}) it would satisfy 2(\frac{1}{2}\cdot1)=1 which is impossible! So, the next best thing one could hope to do is perhaps extend the module M in some “minimal” way so that it can be naturally imbued with an S-module structure. In other words, we want to find some S-module N for which M embeds into N as an R-module, and doing this in some minimal sort of way.  For example, while \mathbb{Z} cannot be given the structure of an \mathbb{Q}-space it can surely be \mathbb{Z}-embedded into such a space, namely \mathbb{Q} itself. Once again, this may not always be possible, for example if A is a finite abelian group (e.g. \mathbb{Z}-module) then A can never be \mathbb{Z}-embedded into a \mathbb{Q}-space since (as can be easily proven!) every element of a \mathbb{Q}-space has infinite order. Thus, we can really only hope to ask for a “best case scenario”. What S-module maximizes both the ability to faithfully (to some degree) embed M and is minimal in some sense. In the case of our abelian group A it’s clear that we’re going to have to take 0 to be our \mathbb{Q}-space since this is the only such space in which A can be “embedded” (albeit very unfaithfully). This is what we mean by extension of scalars, such an S-module N.

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If the obvious intellectual curiosity isn’t enough to motivate this problem I can mention that it has many uses. For example, I have in the past discussed the notion of induced representations which can be seen as extension of scalars problem. Namely, we suppose that we have some group G and some subgroup H\leqslant G. Roughly then what we wish to do is pass from an H-representation to a G-representation, which can be thought of as extending an \mathcal{A}(H)-module (where \mathcal{A}(H) is the group algebra) to an \mathcal{A}(G)-module.

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So, why might we expect that the tensor product is the correct route for such an extension of scalars? There is actually a quite natural way one might realize this. The first is the naive attempt that one might actually try to make a given R-module M into an S-module in the most brutish way. Namely, let’s define a “formal multiplication” of S and M elements. Namely, given s\in S and m\in M let s\star m just be  formal symbol, our “multiplication”. We then see that if this “multiplication” is to create a valid S-module structure extending that of M‘s preexisting R-module structure, we’re going to need certain identities to hold. For example, by mere definition of a module we are going to need that \star is linear in each entry (this is because we should have that (s+s')\star m=s\star m+s'\star m, etc.). Moreover, since we want r\star m=rm (since we are extending the R-module structure) and (ss')\star m=s\star(s'\star m) we see that we are going to have (sr)\star m=s\star(r\star m)=s\star(rm) for all s\in S, r\in R, and m\in M. Thus, we see that \star is an R-biadditive map S\times M\to S\star M. Therefore if we’d like to consider a “universal” way to define an S-module structure on M it seems that we should be looking for a “universal” R-biadditive map S\times M\to S\star M and so really we want S\star M to just be S\otimes_R M and that this should be our extension of scalars.

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Extensions of Scalars

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Let R be some unital ring and S\supseteq R. If M is some left R-module, we define the extension of sclars from R to S on M to be the left S-module S\otimes_R M where we are thinking of S as an (S,R)-bimodule in the usual way.

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Note that we have an R-module map \iota:M\to S\otimes_R M given by m\mapsto 1\otimes m. This is, indeed, an R-map since

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rm\mapsto 1\otimes rm=r\otimes m=r(1\otimes m)

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The reason then we can really justify rigorously that S\otimes_R M is the solution to the “extension of scalars problem” as outlined above:

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Theorem: Let R be a unital subring of S, and M a left R-module. Then, if \iota:M\to S\otimes_R M is the above-described R-map then M/\ker\iota is the largest quotient of M R-embeddable into an S-module. In particular, M/\ker\iota is R-embeddable into and S-module and if M/\ker\phi is another such quotient then \ker\iota\subseteq\ker\phi.

Proof: The fact that M/\ker\iota can be R-embedded into an S-module is clear by the first isomorphism theorem. Suppose now that N is some left S-module and we have some R-map \phi:M\to N. We then consider the map f:S\times M\to N given by (s,m)\mapsto s\phi(m) and note that it is clearly linear in each entry and f(sr,m)=sr\phi(m)=s\phi(rm)=f(s,rm) for all s,r,m so that f is actually R-biadditive. By the universal property of tensor products we get an S-map j:S\otimes_R M\to N such that j(s\otimes m)=s\phi(m). Thus, we see that if m\in\ker\iota then \iota(m)=0 so that 0=j(\iota(m))=j(1\otimes m)=1\phi(m)=\phi(m) so that m\in\ker\phi. Thus, \ker\iota\subseteq\ker\phi as desired. \blacksquare

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In particular, we see that M is R-embeddable into an S-module if and only if \ker\iota is trivial.

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[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.


January 24, 2012 - Posted by | Uncategorized | , , , ,

1 Comment »

  1. […] Extension of Scalars(Pt. II) Point of Post: This is a continuation of this post. […]

    Pingback by Extension of Scalars(Pt. II) « Abstract Nonsense | January 25, 2012 | Reply

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