$\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}$ is isomorphic to the complex numbers I am new to tensor poducts (of modules over a commutative ring with identity) and need to understand the following example to continue with the actual exercises in my material.
Namely, 

I need to know why/prove that $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}\simeq \mathbb{C}$ as $\mathbb{Z}$-modules, by using the universal property of the tensor product $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}$.

I have no idea how to do this, but really need to know the general procedure. Any help is appreciated.
 A: The fact that $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}\cong\mathbb{C}$ only as $\mathbb{Z}$-modules follows as $\mathbb{Z}[i]\cong\mathbb{Z}^{\oplus 2}$ and $\mathbb{Z}^{\oplus 2}\otimes_{\mathbb{Z}}\mathbb{R}\cong\mathbb{R}^{\oplus 2}\cong\mathbb{C}$.
If you need to use the universal property of tensor products somewhere, it can be used in showing that $R^{\oplus n}\otimes_R M\cong M^{\oplus n}$ for a ring $R$ and $R$-module $M$.
A: Here is another way to do it: you can just write down a module homomorphism $\mathbb{Z}[i]\otimes \mathbb{R}\to \mathbb{C}$ by $a+bi\otimes r \mapsto ra+rbi$, and show that it is an isomorphism. To see injectivity for example, you can say that if $ra + rbi = 0$ then $ra = rb = 0$ and hence $r = 0$ or $a=b=0$ in which case the tensor $a+bi\otimes r = 0$, so the map is injective. Surjectivity is similarly not too difficult. 
You should also probably ensure that this is in fact a module homomorphism, and I would tend to go about doing this in a similar way as @neth, and this method does not use the universal property. 
You could also show that $\mathbb{C}$ satisfies the universal property of the tensor product in this case, which would imply that $\mathbb{C} = \mathbb{Z}[i]\otimes\mathbb{R}$
A: We have that $\mathbb Z[X]\otimes_\mathbb Z\mathbb R\cong \mathbb R[X]$ because both are free commutative $\mathbb R$-algebras over one element. It immediately follows that $\mathbb Z[i]\otimes_\mathbb Z\mathbb R\cong \mathbb R[i]\cong\mathbb C$
