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?
 A: 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 $$\mathbb{R}[x]/⟨x^2+1⟩ \otimes_{\mathbb{R}} \mathbb{C} ≈ 
\mathbb{R}[x]/⟨x^2+1⟩ \otimes_{\mathbb{R}} \mathbb{R}[y]/⟨y^2+1⟩ \\≈ \mathbb{R}[x,y]/⟨x^2+1,y^2+1⟩ ≈ \mathbb{R}[y,x]/⟨y^2+1,x^2+1⟩ ≈ \mathbb{C}[x]/ ⟨x^2+1⟩. $$
A: 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.
A: 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 $
A: 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.
