Ring homomorphism $\mathbb{Z}[[t]]\otimes_\mathbb{Z} \mathbb{Q}\rightarrow \mathbb{Q}[[t]]$ I want to prove, that we have a natural ring homomorphism $\mathbb{Z}[[t]]\otimes_\mathbb{Z} \mathbb{Q}\rightarrow \mathbb{Q}[[t]]$
I'm thinking, I somehow need to create a map, that induces such a ring homomorphism. Here my guess would be show that we have a ring homomorphism $\mathbb{Z}[[t]]\times\mathbb{Q}\rightarrow\mathbb{Q}[[t]]$, $(f(x),\frac{a}{b})\mapsto\frac{a}{b}f(x)$, where $f(x)$ is a power series. Would this be correct?
 A: By the universal property of the tensor product, all you need to do
is to construct ring homomorphisms $\phi:\Bbb Z[[t]]\to\Bbb Q[[t]]$
and $\psi:\Bbb Q\to\Bbb Q[[t]]$ with the property that $\phi(a)=\psi(a)$
for all $a\in\Bbb Z$. There are some obvious candidates for $\phi$
and $\psi$....
A: The map $\mathbb{Z}[[t]]\times\mathbb{Q}\to\mathbb{Q}[[t]]$ you use is not a ring homomorphism, but a $\mathbb{Z}$-bilinear map of abelian groups.
This map certainly defines a unique map $\mathbb{Z}[[t]]\otimes\mathbb{Q}\to\mathbb{Q}[[t]]$, which is a group homomorphism. It is also, by direct inspection, a ring homomorphism.
Another way to check the claim is to observe that $\mathbb{Z}[[t]]\otimes\mathbb{Q}$ is the coproduct ring of $\mathbb{Z}[[t]]$ and $\mathbb{Q}$ in the category of commutative rings (with identity), so all you need is to give ring homomorphisms $\alpha\colon\mathbb{Z}[[t]]\to \mathbb{Q}[[t]]$ and $\beta\colon\mathbb{Q}\to\mathbb{Q}[[t]]$.
Then, by the universal property of coproducts, these induce a unique ring homomorphism $\gamma\colon\mathbb{Z}[[t]]\otimes\mathbb{Q}\to\mathbb{Q}[[t]]$ such that $\gamma\circ i=\alpha$ and $\gamma\circ j=\beta$, where $i\colon\mathbb{Z}[[t]]\to\mathbb{Z}[[t]]\otimes\mathbb{Q}$ and $j\colon\mathbb{Q}\to\mathbb{Z}[[t]]\otimes\mathbb{Q}$ are the natural maps $i(f)=f\otimes 1$ and $j(q)=1\otimes q$.
