Background/Motivation:
I was playing around with a certain construction that I am trying to generalize and therefore needed to compute some examples to get a feel for the situation. I realized that I don't really feel comfortable with $\text{Hom}_k(k[[t]],-)$ for some field $k$, i.e. with $k$-algebra morphisms from a formal power series ring. One of the first examples I considered led to the following question:
The question:
Let $k$ be an arbitrary field and consider the set of $k$-algebra homomorphisms $\text{Hom}_k(k[[t]],k)$. How do the elements look like?
My thoughts:
At first it feels like such a morphism is determined by the value of $t$ which cannot be since
1) The $k$-algebra morphism doesnt allow us to commute with our infinite (formal) sums.
2) Then we would have $k[[t]] \cong k[t]$ as this is the universal property of the polynomial ring as free $k$-algebra.
So that is not what we are looking for. We certainly have the $t \mapsto 0$ morphism, but if $t$ is not sent to $0$, I am confused. Somehow it feels like the set of these morphisms is given (rather: can be identified) by all elements $a \in k$, such that all these formals sums "converge" if I plug in $a$. But now $k$ is not necessarily a topological field and hence talking about convergence does not seem to be the correct way of thinking. Thus "converge" should mean define an element in $k$ here, but I don't really know.
$k[[t]]$ is actually local, its only maximal ideal is $\mathfrak{m}=(t)$. To see that, you take a power series $f=\sum_{i}a_it^i$ with $a_0 \neq 0$ (so that $f \notin \mathfrak{m}$) and recursively find its formal inverse (So every element outside of $\mathfrak{m}$ is invertible, hence $\mathfrak{m}$ is the unique maximal ideal).
Since you are surjecting onto a field (this is forced by the assumption that we have a $k$-algebra homomorphism), kernel of any such homomorphism is maximal, i.e. $\mathfrak{m}$.
So there is only one such homomorphism: the one sending $t$ to $0$ and constants to constants.
Regarding the general situation, i.e. $k$-algebra maps from $k[[t]]$ to a $k$-algebra $A$: For complete topological $k$-algebras $A$ and continuous maps, the answer should be "topologically nilpotent elements". Unfortunately, we do not assume topology on $A$, and I have trouble pinpointing what this should mean in an abstract $k$-algebra (so that the description would not be more or less tautological).
Let me start by describing the easy cases, which are maps $\varphi: k[[t]] \rightarrow A$ that are not injective. Since $k[[t]]$ is a DVR with a uniformizer $t$, the kernel has to be of the form $(t^n)$ for some $n$, and so $\varphi$ factors through $k[[t]]/(t^n)=k[t]/(t^n)$. From the standard description of the maps $k[t]\rightarrow A,$ one sees that these correspond to (actually) nilpotent elements of $A$.
Now we have to consider "only" the case of injections $\varphi: k[[t]] \hookrightarrow A$. Here I have trouble in describing these in terms of elements or anything else sensible - I guess the best is "complete DVR subrings of $A$ with residue field $k$ together with a choice of uniformizer" (here the uniformizer should be the topologically nilpotent element in question).
It also does not relate particularly well to the maps $k[t] \rightarrow A$:
For example, obviously not every map $k[t] \rightarrow A$ extends to $k[[t]]\rightarrow A$ (just consider the situation when $k[t]\rightarrow A$ maps $t$ to some $a \in A$, alebraic over $k$ and not nilpotent).
However, also some maps $k[t] \rightarrow A$ may allow for multiple extensions: just consider the structure map of $k[t] \rightarrow k[[t]]\otimes_{k[t]}k[[t]]=:A$. There are two obvious extensions, namely sending $k[[t]]$ to the copy of itself on the left and on the right, resp. (note that in this example, just choosing the uniformizer - $t$ in both cases, is not enough, one actually needs to choose the copy of $k[[t]]$ inside $A$. In topological setting, this would be already determined by the topology (at least when $A$ is "complete and separated")).
Some good news at the end: Let me also point out that a necessary condition for an injection $k[[t]] \rightarrow A$ to exist is that $A$ has uncountable transcendence degree over $k$. So this means e.g. that for all $k$-algebras of finite type, maps $k[[t]] \rightarrow A$ will neatly correspond precisely to nilpotent elements of $A$.