# Abstract Nonsense

## 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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Motivation

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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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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.