What are prime ideals in $k[[t]]$ and $k((t))$?

Question

Let $k$ be a field. What are prime ideals in $k[[t]]$ and $k((t))$? I think that $(t-a), a \in k$ and irreducible polynomials in $k[[t]]$ are prime ideals of $k[[t]]$. Are there other prime ideals? Thank you very much.

Answer 1

Are you so sure that $t-a$ with $a \in k^\times$ generates a prime ideal? Do you know that in $k[[t]]$ we have

$$\frac{1}{t-a} = \frac{-1}{a}\frac{t}{1-\frac{a}{t}} = \frac{-1}{a}(1 + (a/t) + (a/t)^2 + \ldots )?$$

There are actually not many prime ideals in $k[[t]]$: The ring of formal power series in one variable over a field is actually a discrete valuation ring! This basically comes two facts. First: Anything with non-zero constant coefficient is invertible. Second: A DVR is a Noetherian local domain of dimension $1$ in which every non-zero, non-unit can be written in the form

$$ut^n, \text{ $n\in\Bbb{Z}$, $u$ a unit}$$

and $t$ the uniformizing parameter. It follows the only prime ideals of $k[[t]]$ are the zero ideal and the ideal generated by $t$. I.e. $\operatorname{Spec} k[[t]]$ consists of two points, the generic point and closed point. Also, let me add why it is not a surprise that $\operatorname{Spec} k[[t]]$ consists of only two points. In the following discussion, I will assume $k = \Bbb{C}$.

We would like, theoretically speaking some algebraic object that approximates the ring of holomorphic functions on some analytic neighbourhood $U$ of zero. The polynomial ring in one variable $\Bbb{C}[z]$ is simply too small, as there are holomorphic functions that are not polynomials like say $\sin z$. So the next algebraic object to consider would be the power series ring. However the problem is now you have allowed arbitrary power series that don't converge, so this is simply too big! In other words we have inclusions

$$k[t] \subseteq \operatorname{Hol}(U,\Bbb{C}) \subseteq k[[t]]$$

and so taking spec we have $\operatorname{Spec} k[[t]] \subseteq \operatorname{Spec} \operatorname{Hol}(U,\Bbb{C}) \subseteq \operatorname{Spec} k[t].$ But now notice that $U$ is an arbitrary neighbourhood about $0$, hence the comment of Alex Youcis that $k[[t]]$ is the ring of functions on a "hyperzoomed" neighbourhood of zero.