Direct isomorphism between $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$ and direct sum.

Question

I "know" that $\mathbb{C} \otimes_\mathbb{R} \mathbb{C} \cong \mathbb{C} \oplus \mathbb{C}$ as rings, but I don't really know it, what I mean with this is that I don't know any explicit isomorphism $f: \mathbb{C} \otimes_\mathbb{R} \mathbb{C} \rightarrow \mathbb{C} \oplus \mathbb{C}$. I suspect that such an isomorphism should be easy to find, but I am really not finding anyone. Could anyone please help me?

Answer 1

You don't mean the direct sum of rings (or algebras), you mean the direct product of rings. See also here.

In general, if $K/k$ is a field extension and $f \in k[x]$ is a polynomial which splits over $K$ into $n$ distinct linear factors $x-\alpha_i$, then there is an isomorphism of $K$-algebras

$$k[x]/(f) \otimes_k K \cong K[x]/(f) \cong \prod_i K[x]/(x-\alpha_i) \cong \prod_i K = K^n.$$ It is given by mapping $x \otimes 1$ to $(\alpha_1,\dotsc,\alpha_n)$, the rest is given by the information that it is an $K$-algebra isomorphism. Explicitly, we have $p \otimes \lambda \mapsto (\lambda p(\alpha_1),\dotsc,\lambda p(\alpha_n))$ for $\lambda \in K$ and $p \in k[x]$.

The inverse can be found just by looking at the proof of the Chinese Remainder Theorem which we have used above: Since $\alpha_1,\dotsc,\alpha_n$ are pairwise distinct, there are $p_j \in K[x]$ such that $p_j(\alpha_i)=\delta_{ij}$ (e.g. Lagrange polynomial). Then $\overline{p_j} \in K[x]/(f)$ gets mapped to the unit vector $e_j \in K^n$. Hence, the inverse map $K^n \to K[x]/(f)$ is given by $(\lambda_1,\dotsc,\lambda_n) \mapsto \sum_j \lambda_j p_j$.

Example: If $f = x^2+1 \in k[x]$ is irreducible and $\mathrm{char}(k) \neq 2$, we get $k(i) \otimes_k k(i) \cong k(i) \times k(i)$ as $k(i)$-algebras (where $k(i)$ acts on the right tensor factor) given by $i \otimes 1 \mapsto (i,-i)$ (and hence $1 \otimes i \mapsto (i,i)$ and $i \otimes i \mapsto (-1,1)$). We compute $p_1=\dfrac{x+i}{2i}$ and $p_2 = \overline{p_1} = \dfrac{i-x}{2i}$ in $k(i)[x]$. The images in $k(i)[x]/(x^2+1) \cong k(i) \otimes_k k(i)$ are $p_1 = 1 \otimes \frac{i}{2i} + i \otimes \frac{1}{2i} = \frac{1}{2} (1 \otimes 1 - i \otimes i)$ and $p_2 = 1 \otimes \frac{i}{2i} -i \otimes \frac{1}{2i} = \frac{1}{2} (1 \otimes 1 + i \otimes i)$.

These are the two orthogonal idempotents mentioned in the other answers; as you can see you don't have to guess them etc., you can compute them following a general algorithm. It is useful that many theorems have constructive proofs, such as here the Chinese Remainder Theorem.

Let me just indicate what happens in characteristic $2$. Here we have $k(i) \otimes_k k(i) \cong k(i)[x]/(x+i)^2 \cong k[a,b]/(a^2,b^2)$ with $b=x+i$ and $a=i+1$.

Answer 2

This isomorphism would be of $\Bbb R$-algebras. Clearly both ${\Bbb C}\otimes_{\Bbb R}\Bbb C$ and $\Bbb C\oplus\Bbb C$ are four-dimensional as $\Bbb R$-vector spaces. If ${\Bbb C}\otimes_{\Bbb R}{\Bbb C}\cong{\Bbb C}\oplus{\Bbb C}$ then there must be central orthogonal idempotents in the algebra ${\Bbb C}\otimes_{\Bbb R}{\Bbb C}$ corresponding to $(1,0)$ and $(0,1)$. I boldface COI because these elements are critical to decomposing algebras in very general situations. Of course everything is commutative so you don't need to worry about central in this situation.

Obviously $1\otimes1$ is idempotent. What about $i\otimes i$? Well $(i\otimes i)^2=-1\otimes-1=1\otimes 1$, so this is not an idempotent. However it is a nontrivial square root of one. This tells us that

$$1\otimes1=(i\otimes i)^2\iff (1\otimes1+i\otimes i)(1\otimes 1-i\otimes i)=0.$$

That is, $\alpha=1\otimes1+i\otimes i$ and $\beta=1\otimes1-i\otimes i$ are orthogonal, i.e. $\alpha\beta=0$. Are they idempotent? Not quite; check $\alpha^2=2\alpha\Leftrightarrow (\alpha/2)^2=\alpha/2$ and $\beta^2=2\beta\Leftrightarrow(\beta/2)^2=\beta/2$ so $\alpha/2$ and $\beta/2$ are central orthogonal idempotents in $\Bbb C\otimes_{\Bbb R}\Bbb C$.

In general if $R=Re\oplus R(1-e)$ where $e\in R$ is a central idempotent, then an isomorphism is given by $r\mapsto (re,r-re)$ (this works even if $R$ is non-unital!).

This gives us the algebra isomorphism $A=A\frac{\alpha}{2}\oplus A\frac{\beta}{2}$. Now we must find algebra isomorphisms between the ideals $(\alpha/2)$ and $(\beta/2)$ of $A=\Bbb C\otimes_{\Bbb R}\Bbb C$ and $\Bbb C$. The idempotents $\alpha/2,\beta/2$ correspond to $1\in\Bbb C$, so we only need to find square roots of the idempotents' negatives which correspond to the same in $\Bbb C$, i.e. $i\in\Bbb C$. Since $(\lambda\frac{\alpha}{2})^2=\lambda^2\frac{\alpha}{2}$ and $(\mu\frac{\beta}{2})^2=\mu^2\frac{\beta}{2}$, it suffices to find square roots $\lambda,\mu$ of $-1\otimes1$, which can easily be identified as $\lambda,\mu=1\otimes i,i\otimes1$ (it doesn't matter which of the four configurations we choose). For fun choose $\lambda,\mu=i\otimes1$. Compute

$$\lambda\frac{\alpha}{2}=i\otimes1\frac{1\otimes1+i\otimes i}{2}=\frac{i\otimes1-1\otimes i}{2} $$

$$\mu\frac{\beta}{2}=i\otimes1\frac{1\otimes1-i\otimes i}{2}=\frac{i\otimes1+1\otimes i}{2} $$

Therefore, after some (possibly mental) linear algebra,

$$\begin{array}{ll} a(1\otimes1)+b(i\otimes1)+c(1\otimes i)+d(i\otimes i) & = \color{Blue}{\frac{a+d}{2}}\left[\frac{1\otimes1+i\otimes i}{2}\right]+\color{Green}{\frac{a-d}{2}}\left[\frac{1\otimes1-i\otimes i}{2}\right] \\ & \,+\, \color{Magenta}{\frac{b-c}{2}}\left[\frac{i\otimes1-1\otimes i}{2}\right]+\color{Red}{\frac{b+c}{2}}\left[\frac{i\otimes1+1\otimes i}{2}\right]\end{array}$$

gets mapped to

$$\left(\color{Blue}{\frac{a+d}{2}}+\color{Magenta}{\frac{b-c}{2}}i,\color{Green}{\frac{a-d}{2}}+\color{Red}{\frac{b+c}{2}}i\right).$$

This completes our $\Bbb R$-algebra isomorphism $\Bbb C\otimes_{\Bbb R}\Bbb C\cong \Bbb C\oplus\Bbb C$.

Note the four configurations of $\lambda,\mu\in\{1\otimes i,i\otimes1\}$ correspond to the four distinct elements $(i,0),(-i,0),(0,i),(0,-i)$ (not in any particular order) in $\Bbb C\oplus\Bbb C$, analogous to how the conjugation map $i\leftrightarrow -i$ yields a $\Bbb R$-algebra automorphism of $\Bbb C$.

Answer 3

One example of an isomorphism $\varphi: \Bbb C \oplus \Bbb C \longrightarrow \Bbb C \otimes_{\Bbb R} \Bbb C$ is given on generators by $$\varphi(1, 0) = \tfrac{1}{2}(1 \otimes 1 + i \otimes i),$$ $$\varphi(0, 1) = \tfrac{1}{2}(1 \otimes 1 - i \otimes i).$$