Abstract Nonsense

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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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.