Abstract Nonsense

Crushing one theorem at a time

Extension of Scalars(Pt. II)

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

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

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Let’s extend the scalars on \mathbb{R}[x] from \mathbb{R} to \mathbb{C}. What should we expect the answer to be? Well, we know that \mathbb{R}[x] naturally sits inside \mathbb{C}[x] as an \mathbb{R}-space, and so \mathbb{C}[x] seems like a pretty good candidate. From the previous theorem we see that we have an \mathbb{C}-map j:\mathbb{C}\otimes_\mathbb{R}\mathbb{R}[x]\to\mathbb{C}[x] such that j(z\otimes p(x))=zp(x), we claim that j is an isomorphism. Indeed, the fact that j is surjective follows from the fact that j is \mathbb{C}-linear and j(1\otimes x^n)=x^n so that j hits all the generators of \mathbb{C}[x]. To see that j is injective we merely note that if \displaystyle \sum z_i\otimes p_i(x) is in \ker j then \displaystyle \sum z_i p_i(x)=0. Now, write \displaystyle p_i(x)=\sum_{j=1}^n a_{i,j}x^j and so we have that

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\displaystyle 0=\sum_i z_i\otimes p_i(x)=\sum_i\sum_{j=1}^nz_ia_{i,j}x^j=\sum_{j=1}^{n}\sum_i z_ia_{i,j}x^j=0

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and so we must have that \displaystyle \sum_i z_ia_{i,j} is zero. But, we see then that

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\displaystyle \begin{aligned}\sum_i z_i\otimes p_i(x) &=\sum_i z_i\otimes\left(\sum_j a_{i,j}x^j\right)\\ &=\sum_i\sum_jz_i\otimes a_{i,j}x^j\\ &=\sum_j\sum_i (z_ia_{i,j})\otimes x^j\\ &=\sum_j\left(\sum_i z_ia_{i,j}\right)\otimes x^j\\ &=\sum_j\otimes x^j\\ &=0\end{aligned}

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and so j has trivial kernel, so is injective. Thus, j is a \mathbb{C}-isomorphism and so we see that \mathbb{C}\otimes_\mathbb{R}\mathbb{R}[x]\cong\mathbb{C}[x].

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

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Theorem: Let R\subseteq S be unital rings. Then, S\otimes_R R[x]\cong S[x] as left S-modules.

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

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In a similar vein, let’s see what the complexification of \text{Mat}_n(\mathbb{R}) looks like. I bet you already have a good guess. Namely, since we can include \text{Mat}_n(\mathbb{R}) into \text{Mat}_n(\mathbb{C}) in a \mathbb{R}-linear way we have a \mathbb{C}-map j:\mathbb{C}\otimes_\mathbb{R}\text{Mat}_n(\mathbb{R})\to\text{Mat}_n(\mathbb{C}) such that j(z\otimes (a_{i,j}))=(za_{i,j}). Once again, it’s clear that this map is going to be surjective since it is \mathbb{C}-linear and hits the usual generators of \text{Mat}_n(\mathbb{C}) (e.g. the n^2 matrices corresponding to the standard basis vectors e_i in \mathbb{C}^{n^2}) and similarly to the case of the polynomials, if \displaystyle \sum z_k\otimes (a_{i,j,k}) is in the kernel of this map, then \displaystyle \sum (z_k a_{i,j,k}) and so \displaystyle \sum z_k a_{i,j,k}=0. Doing the same trick as the polynomial casewhere we break \displaystyle \sum z_k\otimes (a_{i,j,k}) into a double sum one over k and one over i,j, interchanging the order of summation, and bringing the a_{i,j,k}‘s to the left shows that \displaystyle \sum z_k\otimes (a_{i,j,k})=0 and so j has trivial kernel and so is an isomorphism. Once again, there was nothing special in this argument about \mathbb{R} or \mathbb{C} so that we can definitely say that

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Theorem: Let R\subseteq S be unital rings. Then, S\otimes_R\text{Mat}_n(R)\cong\text{Mat}_n(S) as left S-modules.

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As a last general example let’s see what happens when we extend scalars on free modules. In particular, we merely note that if R\subseteq S then by previous discussion we know that S\otimes_R R^{\oplus\lambda}\cong (S\otimes_R R)^{\oplus\lambda}\cong S^{\oplus\lambda} as left S-modules. Or to be more precise:

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Theorem: Let R\subseteq S be rings. Then, S\otimes_R R^{\oplus\lambda}\cong S^{\oplus\lambda} as left S-modules. In particular, the extension of scalars on a free module results in a free module, and in the case that S (and thus R) have the IBN property they have the same rank.

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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 25, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , ,

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