Prove that $\mathbb{Z}[i]\otimes_\mathbb{Z}\mathbb{R}\cong\mathbb{C}$ as rings. Exercise from "Abstract Algebra (pg. 377)"  of Dummit and Foote.

Prove that the extenstion of the scalars from $\mathbb{Z}$ to the Gaussian integers $\mathbb{Z}[i]$ of the ring $\mathbb{R}$ is isomorphic to $\mathbb{C}$ as a ring: $\mathbb{Z}[i]\otimes_\mathbb{Z}\mathbb{R}\cong\mathbb{C}$ as rings.

Here the definition of multiplication in $\mathbb{Z}[i]\otimes_\mathbb{Z}\mathbb{R}$ follows from a general definition for tensor product of $R$-algebras: $(a\otimes b)(a'\otimes b')=(aa'\otimes bb')$.
I'm looking for a solution, or at least a sketch. Thanks.
 A: I will give a direct solution but it's worth mentioning some general tensor product facts that I used to reach this solution. Getting used to these facts is very useful as it allows you to do tensor product calculations like these very quickly and without much computation.
I'll give the direct argument first. We define a map $f: \mathbb Z[i] \otimes_{\mathbb Z} \mathbb R \longrightarrow \mathbb C$ by universal property. We need a $\mathbb Z$ bilinear map $\mathbb Z[i] \times \mathbb R \longrightarrow \mathbb C$. I take this to be $(a + bi, t) \mapsto (a + bi)t$. This is bilinear by distributivity of multiplication so it extends to a map on the tensor product. This yields a well defined $\mathbb Z$-linear map $f: \mathbb Z[i] \otimes_\mathbb Z \mathbb R \longrightarrow \mathbb C$. It's a straight-forward check to show that this is a ring homomorphism using the definition of multiplication you cited.
We now claim that it's an isomorphism. Proving surjectivity isn't hard - the map turns out to be $\mathbb R$-linear and the image contains $1$ and $i$. However, computing kernels directly when dealing with tensor products is often hard, since it's difficult to deduce when something like $\sum (a_n + i b_n) \otimes t_n = 0$. As such, we instead find an inverse. Let $g: \mathbb C \longrightarrow \mathbb Z[i] \otimes_\mathbb Z \mathbb R$ via $t + si \mapsto 1 \otimes t + i \otimes s$. Once we compute both compositions, we'll be done. Indeed, $f(g(t + si)) = f(1 \otimes t + i \otimes s) = f(1 \otimes t) + f(i \otimes s) = t + si$. For $g \circ f$ we only compute this on simple tensors. This is sufficient as they generate the tensor product.
\begin{align*}
g(f((a + bi) \otimes t)) &= g(at + bti)\\
&= 1 \otimes at + i \otimes bt\\
&= a \otimes t + bi \otimes t\\
&= (a + bi) \otimes t$.
\end{align*}
Hence, this map is an isomorphism.
Here are the general facts I mentioned. All rings discussed from now on are assumed to be commutative. I will also use the fact that $\mathbb Z[i] \cong \mathbb Z[x]/(x^2 + 1)$ and $\mathbb C \cong \mathbb R[x]/(x^2 + 1) \cong \mathbb C$.

*

*Let $A$ be an $R$-algebra and $B$ be an $A$-algebra. Then $A[x] \otimes_R B \cong B[x]$ as $B$-algebras.


*Let $A, B$ be $R$ algebras and let $I$ be an ideal of $A$. Then $A/I \otimes_R B/J \cong (A \otimes_R B) / (I \otimes_R 1 + 1 \otimes_R J)$ as $R$-algebras. There's a good discussion of this fact here with a variety of proofs.
There are also explicit maps for each isomorphism.
These imply that $\mathbb Z[x]/(x^2 + 1) \otimes_{\mathbb Z} \mathbb R \cong (\mathbb Z[x] \otimes_{\mathbb Z} \mathbb R) / ((x^2 + 1) \otimes \mathbb R)$. The isomorphism of 1. is multiplication of simple tensors, so this expression is isomorphic to $\mathbb R[x] / (x^2 + 1)$. I skipped some details here of course, but the map this yields is exactly the same as in the direct solution.
A: Every element of $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}$ can be written as $1\otimes a+i\otimes b$. Moreover
$$
(1\otimes a+i\otimes b)(1\otimes c+i\otimes d)=1\otimes(ac-bd)+i\otimes(ad+bc)
$$
by direct computation.
Suppose $a\ne0$ or $b\ne0$; then
$$
(1\otimes a+i\otimes b)\Bigl(1\otimes\frac{a}{a^2+b^2}+i\otimes\frac{-b}{a^2+b^2}\Bigr)=1\otimes1
$$
by direct computation. Hence the ring is a field and it's straightforward to see that it's an algebraic extension of $\mathbb{R}$, where $-1$ is a square.
