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Let $\mathbb R$ be the field of real numbers and $\mathbb C$ be the field of complex numbers. Consider the complexification of the real matrix algebra $M_n(\mathbb R)$ that is $\mathbb C\otimes_{\mathbb R}M_n(\mathbb R)$. It is known that $$\mathbb C\otimes_{\mathbb R}M_n(\mathbb R)\cong M_n(\mathbb C).$$

What is an example of an isomorphism from $\mathbb C\otimes_{\mathbb R}M_n(\mathbb R)$ to $M_n(\mathbb C)?$

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2 Answers 2

up vote 1 down vote accepted

Let $e_{jk}$ denote the "unit matrix" which is zero everywhere exept in the $j,k$ entry, where it has a 1.

Map each basis element of the form $x\otimes e_{jk}$ where $ x\in\{1,i\}$ and $1\leq j,k\leq n $ to $xe_{j,k}\in M_n(\mathbb{C})$. This produces an $\mathbb{R}$ algebra isomorphism.

If you want a $\mathbb{C}$ algebra isomorphism, you can check that the same map is a $\mathbb{C}$ algebra isomorphism, but the elements of $\mathbb{C}\otimes M_n(\mathbb{R})$ would no longer be linearly independent over $\mathbb{C}$. Long story short, the map would be:

$$ (\alpha i+\beta)\otimes M\rightarrow (\alpha i+\beta)M $$

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I think there is a problem: "Map each basis element of the form $x\otimes e_{jk}$ where $x\in${$1,i$}".This means that $dim(\mathbb C\otimes_{\mathbb R}M_n(\mathbb R))=2n^2$, as a $\mathbb C$-algebra. But $dim(M_n(\mathbb C))=n^2$ as a $\mathbb C$-algebra. –  zacarias Sep 10 '12 at 17:28
    
@zacaria No, the left is $2n^2$ dimensional as an $\mathbb{R}$ algebra (and so is the right). The set of things of the form $1\otimes e_{jk}$ is a $\mathbb{C}$ basis for the left hand side... that only has $n^2$ elements. –  rschwieb Sep 10 '12 at 18:28
    
Yes. Thank you rschwieb –  zacarias Sep 10 '12 at 19:57

A little bit more abstract, though without dealing with bases:

Let $A:=M_n(\mathbb{R})$, $A_\mathbb{C}=A\otimes_\mathbb{R}\mathbb{C}$ its complexification and $B:=M_n(\mathbb{C})$. We have a canonical embedding $\iota:A\hookrightarrow B$ of real algebras, which uniquely lifts via a morphism $\iota_\mathbb{C}:A_\mathbb{C}\to B$ of $\mathbb{C}$-algebras, by the universal property of $A_\mathbb{C}$, i.e. we have $\iota_\mathbb{C}\circ\alpha=\iota$, where $\alpha$ is the canonical map $A\ni a\mapsto a\otimes 1\in A_\mathbb{C}$. Now since $\iota_\mathbb{C}(A_\mathbb{C})$ is a $\mathbb{C}$-subalgebra of $B$ containing $\iota(A)$, $\iota_\mathbb{C}$ must be surjective and hence is the desired isomorphism, by counting dimensions.

Cheers, Robert

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