# Tensor product algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$

I want to understand the tensor product $\mathbb C$-algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$. Of course it must be isomorphic to $\mathbb{C}\times\mathbb{C}.$ How can one construct an explicit isomorphism?

• Take obvious basis vectors in tensor product map them into basis vectors of direct sum Aug 12, 2012 at 18:46
• I mean an isomorphism of $\mathbf{C}$-algebras with respect to the left $\mathbf{C}$ in the tensor product... Aug 12, 2012 at 18:54
• You should write this in your question. Aug 12, 2012 at 18:57
• Why do you say "of course it must be isomorphic to..."? Because of dimension reasons or because of galois theory? Aug 12, 2012 at 21:29
• Nov 22, 2018 at 20:35

Write $\mathbb C=\mathbb R[x]/\langle x^2+1\rangle$ for one of the copies. Then, using a universal property of tensor products, $$\mathbb C\otimes_{\mathbb R} \mathbb C \;\approx\; \mathbb R[x]/\langle x^2+1\rangle \otimes_{\mathbb R}\mathbb C \;\approx\; \mathbb C[x]/\langle (x+i)(x-i)\rangle \;\approx\; \mathbb C[x]/\langle x+i\rangle \oplus \mathbb C[x]/\langle x-i\rangle$$ the last isomorphism via Sun-Ze's theorem (a.k.a. "Chinese Remainder Theorem"). That last isomorphism can be made explicit by choice of polynomials $A(x),B(x)$ such that $A(x)\cdot (x+i) + B(x)\cdot (x-i)=1$.

Edit: this treats "right" $\mathbb C$-algebra, but/and reversing the roles gives the same outcome as "left" algebra.

An explicit isomorphism of $\mathbb C$-algebras is given (on generators) by $\mathbb C\otimes _\mathbb R \mathbb C\stackrel {\cong }{\to} \mathbb C\times \mathbb C: z\otimes w \mapsto (z\cdot w,z\cdot\bar w)$.
Here $\mathbb C \otimes _\mathbb R \mathbb C$ is considered as a $\mathbb C$-algebra through its first factor: $z_1\cdot (z\otimes w)=z_1 z\otimes w$

• On the isomorphism $\mathbf{C}\otimes_\mathbf{R} \mathbf{C}\cong \mathbf{C}\times \mathbf{C}$: math.stackexchange.com/a/118275/3217 Aug 13, 2012 at 6:18
• Are these also isomorphic as $\mathbb{R}$-algebras? How can that be established?
– gen
Feb 9, 2022 at 21:20

As a minor complement to Georges's answer, I'll try to convince the reader that it's better to write $$\mathbb C\otimes_{\mathbb R}\mathbb C\simeq\mathbb C\times\overline{\mathbb C}$$ than $$\mathbb C\otimes_{\mathbb R}\mathbb C\simeq\mathbb C\times\mathbb C.$$ For each complex vector space $V$ define the conjugate vector space $\overline V$ as being the abelian group $V$ with the multiplication by (complex) scalars $*$ defined by $z*v:=\overline zv$. [Of course $\overline{\mathbb C}$ is canonically isomorphic to $\mathbb C$, but in general $\overline V$ is not canonically isomorphic to $V$.] Then the map $$\mathbb C\otimes_{\mathbb R}V\to V\times\overline V,\quad z\otimes v\mapsto(zv,z*v)$$ is a $\mathbb C$-linear isomorphism.

More generally, let $L/K$ be a finite Galois extension with Galois group $G$ and let $A$ be an $L$-algebra. [In this post, an $L$-algebra is just an $L$-vector space $A$ together with an $L$-bilinear map $A\times A\to A$.] For each $\sigma$ in $G$ let $*_\sigma$ be the multiplication by scalars in $L$ defined on $A$ by $z*_\sigma a:=\sigma(z)\,a$, and let $A_\sigma$ be the resulting $L$-algebra. Then the map $$\phi_A:L\otimes_KA\to\prod_{\sigma\in G}A_\sigma,\quad z\otimes a\mapsto(z*_\sigma a)_{\sigma\in G}$$ is an $L$-algebra isomorphism.

This is proved as Proposition 8 on page A.V.64 of the book Algebra, Chapters 4-7 by Bourbaki, book freely and legally available here.

Here is a simple proof:

It suffices to prove the statement obtained by replacing the $L$-algebra $A$ with an $L$-vector space $V$. If $(V_i)$ is a family of $L$-vector spaces such that each $\phi_{V_i}$ is an isomorphism, then $\phi_{\oplus V_i}$ is also an isomorphism. It suffices thus to prove that $\phi_L$ is an isomorphism.

Let $B$ be a $K$-basis of $L$. It suffices to show that the $\phi_L(1\otimes b)$, when $b$ runs over $B$, form an $L$-basis of $\prod L_\sigma$. It even suffices to check that the $\phi_L(1\otimes b)$ are linearly independent over $L$.

Suppose by contradiction that there are $x_b$ in $L$, not all zero, such that $\sum_bx_b\,\phi_L(1\otimes b)=0$, that is $\sum_b\sigma(x_b)\,b=0$ for all $\sigma$. The latter condition is equivalent to $\sum_bx_b\,\sigma(b)=0$ for all $\sigma$. If this condition holds, then the square matrix $(\sigma(b))_{\sigma,b}$ is singular, and there are $y_\sigma$ in $L$, not all zero, such that $\sum_\sigma y_\sigma\,\sigma(b)=0$ for all $b$, which contradicts Dedekind Independence Theorem.

To fill in some details to paul garret's answer, to prove $\mathbb{R}[x]/⟨x^2+1⟩ \otimes_{\mathbb{R}} \mathbb{C} ≈ \mathbb{C}[x]/⟨x^2+1⟩$, prove the general fact $R[x_1, . . . , x_n]/I \otimes_R R[y_1, . . . , y_m]/J ≈ R[x_1, . . . , x_n, y_1, . . . , y_m]/⟨I, J⟩$.

Then observe that $$\mathbb{R}[x]/⟨x^2+1⟩ \otimes_{\mathbb{R}} \mathbb{C} ≈ \mathbb{R}[x]/⟨x^2+1⟩ \otimes_{\mathbb{R}} \mathbb{R}[y]/⟨y^2+1⟩ \\≈ \mathbb{R}[x,y]/⟨x^2+1,y^2+1⟩ ≈ \mathbb{R}[y,x]/⟨y^2+1,x^2+1⟩ ≈ \mathbb{C}[x]/ ⟨x^2+1⟩.$$

This isomorphism is conceptually important, so let's carefully write down all the consecutive isomorphisms involved together with all the formulas. $$\newcommand{\RR}{\mathbb{R}}$$ $$\newcommand{\CC}{\mathbb{C}}$$

We claim that $$\CC\otimes_\RR\CC$$ can be identified with the direct product $$\CC\times\CC$$. Let us first see why we have this isomorphism and then track down the exact formulas. We have $$\begin{equation*} \begin{split} \CC\otimes_\RR\CC &\cong \RR[x]/\langle x^2+1\rangle\otimes_\RR\CC\\ &\cong \CC[x]/\langle x^2+1\rangle\\ & \cong \CC[x]/\langle x-i\rangle\cdot\langle x+i\rangle\\ &\cong \CC[x]/\langle x-i\rangle\times\CC[x]/\langle x+i\rangle\\ &\cong \CC\times\CC \end{split} \end{equation*}$$ where the fourth isomorphism follows from the Chinese Remainder Theorem (the ideals $$\langle x-i \rangle$$ and $$\langle x+i\rangle$$ are coprime). Now let us write down the exact formulas on the level of simple tensors, i.e. on the generators of tensor product. Let $$z=a+bi$$ and $$w=c+di$$ for some $$a,b,c,d\in\RR$$.

(1) The isomorphism $$\CC\otimes_\RR\CC\to\RR[x]/\langle x^2+1\rangle\otimes_\RR\CC$$ sends simple tensor $$(a+bi)\otimes_\RR(c+di)$$ to $$(a+b\bar{x}\mod \langle x^2+1\rangle)\otimes_\RR(c+di).$$

(2) Next, the isomorphism $$\RR[x]/\langle x^2+1\rangle\otimes_\RR\CC\to\CC[x]/\langle x^2+1\rangle$$ sends $$(a+b\bar{x}\mod\langle x^2+1\rangle)\otimes_\RR(c+di)$$ to $$(ac+adi)+(bc+bdi)\cdot\bar{x}\mod\langle x^2+1\rangle.$$

(3) Now the isomorphism (coming from Chinese Remainder Theorem) $$\CC[x]/\langle x^2+1\rangle\to \CC[x]/\langle x-i\rangle\times\CC[x]/\langle x+i\rangle$$ sends $$(ac+adi)+(bc+bdi)\cdot\bar{x}\mod\langle x^2+1\rangle$$ to $$((ac+adi)+(bc+bdi)\cdot\bar{x}\mod\langle x-i\rangle, (ac+adi)+(bc+bdi)\cdot\bar{x}\mod\langle x+i\rangle).$$

(4) Finally, the isomorphism $$\CC[x]/\langle x-i\rangle\times\CC[x]/\langle x+i\rangle\to\CC\times\CC$$ sends the previous element, which is equal to

$$((ac+adi)+(bc+bdi)\cdot(\bar{x}-i)+(bc+bdi)\cdot i\mod\langle x-i\rangle, (ac+adi)+(bc+bdi)\cdot(\bar{x}+i)-(bc+bdi)\cdot i\mod\langle x+i\rangle)$$ to $$(ac+adi+(bc+bdi)\cdot i, ac+adi-(bc+bdi)\cdot i)=(ac-bd+(ad+bc)\cdot i, ac+bd+(ad-bc)\cdot i)=(zw,\bar{z}w).$$

Hence, the explicit formula for the isomorphism $$\CC\otimes_\RR\CC\to \CC\times\CC$$ is $$(z,w)\mapsto(zw,\bar{z}w)$$.

An explicit isomorphism is given by \newcommand\R{\mathbb{R}} \newcommand\C{\mathbb{C}} \begin{align} \varphi: \C \oplus \C & \to \C \otimes_{\R} \C, \\ (1,0) & \mapsto \frac{1}{2} (1 \otimes 1 + i \otimes i) \\ (i,0) & \mapsto \frac{1}{2} (1 \otimes i - i \otimes 1) \\ (0,1) & \mapsto \frac{1}{2} (1 \otimes 1 - i \otimes i) \\ (0,i) & \mapsto \frac{1}{2} (1 \otimes i + i \otimes 1) \\ \end{align}