2
$\begingroup$

Let $k$ be a field. What are the $k$-rational points of the affine $k$-scheme $\mathrm{Spec}(k[[t]])$, where $k[[t]]$ is the power series ring over $k$ (equivalently, what are the $k$-algebra morphisms $k[[t]] \rightarrow k$?)

I'm only sure about one point, namely the map $t \mapsto 0$. Do I have to assume some sort of completeness of $k$ to get more points?

Is there a nice presentation of $k[[t]]$, i.e. a quotient of some polynomial ring that is isomorphic to $k[[t]]$?

Improve this question
$\endgroup$
8
  • $\begingroup$ Okay, a question into another direction: Why do people consider deformations parametrized by $k[[t]]$? Parametrizing over $k[t]$ makes perfect sense to me; I have a fiber over any $\alpha \in k$. But in case of $k[[t]]$? $\endgroup$
    – Georg S.
    Commented Jul 31, 2010 at 13:24
  • $\begingroup$ Perhaps you should post this comment as a separate question. I know there are times when you can embed a complete DVR with residue field $k$ into a field $K$ with nicer properties than $k$ (e.g., characteristic zero rather than characteristic $p$), but there are probably much better answers out there. $\endgroup$
    – Charles Staats
    Commented Jul 31, 2010 at 15:19
  • 2
    $\begingroup$ Well, it is easier to give deformations over $k[[t]]$ because its spectrum is small compared to $k[t]$. The ring $k[[t]]$ can be written as projective limit over $k[t]/(t^n)$ and those are local artin algebras, i.e. their spectrum are just points with some tangent vectors attached. Now, you can use deformation theory (in the sense of Schlessinger) to produce deformations those over artin algebras. If you have a system of deformation (say over each $k[t]/(t^n)$ then there are techniques (like Grothendieck's existence theorem) which sometimes allow you to pass to familiy over $k[[t]]$. $\endgroup$
    – Holger Partsch
    Commented Jul 31, 2010 at 15:20
  • 2
    $\begingroup$ Dear Georg, Regarding your question about deformations: the topic you are (implicitly) asking about is whether deformations can be algebraized. It would be easier to answer if you posted it as a separate question (and there are several people on MO who could give you good answers about it). Here I will just say that if C is any smooth curve over k, and you had a family over C, then looking at the formal n.h. of a point will give you something over k[[t]]. In other words, k[t] is not the unique way of algebraizing k[[t]]; any smooth curve will do. Thus you shouldn't prejudge the ... $\endgroup$
    – Emerton
    Commented Aug 1, 2010 at 4:24
  • 2
    $\begingroup$ ... situation and expect to have a family over k[t], just because there is one over k[[t]]. For example, if you look at families of elliptic curves with an 11-torsion point, there are is no interesting such family over an affine line, but there is an interesting such family over a (several times punctured) elliptic curve. In any event, people are very often interested in algebraic families of the type you are wondering about (this is the study of moduli problems), but computing formal deformations is typically much easier, and an important first step even if the moduli space is your goal. $\endgroup$
    – Emerton
    Commented Aug 1, 2010 at 4:28

2 Answers 2

Reset to default
5
$\begingroup$

$k[[t]]$ is a local ring with maximal ideal $(t)$ and the kernel of every $k$-homomorphism $k[[t]] \to k$ is a maximal ideal, thus the maximal ideal. Thus it factors as $k[[t]] \to k[[t]]/(t) = k \to k$ and $t \mapsto 0$ is the unique $k$-rational point.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Why is the kernel maximal? (This is probably obvious...) $\endgroup$
    – Georg S.
    Commented Jul 31, 2010 at 13:26
  • $\begingroup$ Every $k$-homomorphism to $k$ is surjective. This also shows: Every $k$-rational point of a $k$-scheme is closed. $\endgroup$
    – Martin Brandenburg
    Commented Jul 31, 2010 at 13:31
  • $\begingroup$ Ah, of course! Thanks. Can you also give me a hint concerning my deformation question above? $\endgroup$
    – Georg S.
    Commented Jul 31, 2010 at 13:34
1
$\begingroup$

Perhaps one answer to your question about deformations is something like the following. A deformation over a complete local ring A (such as k[[t]]) is just a family X $\to$ Spec(A). Suppose that the fibers belong to some sort of moduli space M, such as the moduli space of curves. In the functorial point of view of moduli spaces, the family X $\to$ Spec(A) corresponds to a morphism Spec(A) $\to$ M that assigns to a point of Spec(A) the moduli of the fiber over this point. So, one parameter formal deformations (by this I just mean that A = k[[t]]) correspond precisely to the morphisms Spec(k[[t]]) $\to$ M. The scheme Hom(k[[t]], M) is called the space of arcs in M. If we fix the central fiber of the deformation then we get the space of arcs in M at the point corresponding to the central fiber. The space of arcs is a subtle and important invariant of a singularity. One can think of an arc (that is, a morphism Spec(k[[t]]) $\to$ M) as follows: if we had a curve in M then the arc would be the collection of jets this curve determines, ie all the derivatives of all orders of the curve (think of the way morphisms Spec$(k[t]/(t^2)) \to M$ determine the tangent vectors at the image of the closed point). So these deformations are telling us something significant about the local structure of the moduli space.

The construction of these one parameter formal deformations works regardless of the existence of any moduli space. It tells us what the space of arcs on the moduli space of whatever it is you are deforming should be.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .