DVR, power series expansion. Let $A$ be a discrete valuation ring with quotient field $K$, maximal ideal $\mathfrak{m}$, uniformizing parameter $t$. Let $k = A/\mathfrak{m}$, so $k$ is a field. How do I show that there is a homomorphism $\varphi$ of rings which maps $A$ to $k[[x]]$ with $\varphi(t) = x$?
 A: If the uniformizing parameter is a prime number multiple of the multiplicative identity, then there is no ring homomorphism $A\to k[[x]]$. In particular, if we let $A=\Bbb Z_p$ (the $p$-adic integers, which is complete) or $A=\Bbb Z_{(p)}=\{\frac{x}{y}\in\Bbb Q:p\nmid y\}$ (whose completion is $\Bbb Z_p$) then the uniformizing parameter will be $t=p$ and the residue field is $\Bbb F_p$, which has positive characteristic. There will be no homomorphisms $\phi:A\to\Bbb F_p[[x]]$ sending $p\mapsto x$, because then $x=\phi(p)=p\phi(1)=0$, which contradicts the fact that $x$ is not zero in $\Bbb F_p[[x]]$.
There is of course a superficial similarity: every element of $k[[x]]$ is a power series in $x$ whose coefficients are from $k$, and, choosing a set of representatives $T\subset A$ for the residues in $k$, every element of $A$ is uniquely a power series in $t$ whose coefficients are from $T$. But even addition of elements in these rings may not be compatible, as our above counterexamples show.
There is a way to put an algebraic structure on the set (!) $k^{\Bbb N}$ which makes it $A$, using a construction known as Witt vectors. This is very complicated though.
