Let $(K,v)$ be a discrete valuation field ($v$ is surjective). Let $\mathcal O$ be the ring of integers of $v$ and moreover let $\mathfrak p$ be the unique maximal ideal of $\mathcal O$. Then we have the following theorem:

Let $R\subset\mathcal O$ be a set of representatives for the residue field $\mathcal O/\mathfrak p$ such that $0\in R$. Then the completion $\widehat K$ (of $K$ w.r.t. $v$) is the set: $$\mathscr L:=\left\{\sum_{i\ge -m} a_i\pi^i\,:\, a_i\in R,\;a_{-m}\neq 0,\ \right\}$$ where $\pi\in\mathcal O$ is an element such that $v(\pi)=1$. In other words every element of $\widehat K$ has a unique representation as convergent power series.

Fundamental remark: We can endow $\mathscr L$ with a field structure by transferring the field structure of $\widehat K$. Note that here the product in $\mathscr L$ is not the usual convolution of power series since $\widehat K$ is a ring of Cauchy sequences (modulo...). Here I must link this other question of mine.

Now consider the following example:

$k$ is a field and consider the ideal $\mathfrak p=(t)\subset k[t]$. Then $k(t)$ is a discrete valuation field w.r.t the $\mathfrak p$-adic valuation $v_{\mathfrak p}$; moreover $\mathcal O=k[t]$ and the residue field $\mathcal O/\mathfrak p$ is isomorphic to $k\subseteq k(t)$. It follows that $k$ is itself a system of representatives and $$\mathscr L=k((t))=\left\{\sum_{i\ge{-m}} a_it^i\,:\, a_i\in k\right\}$$

If we want to consider $k((t))$ as a field through the above theorem, we notice that this field structure isn't the same as the "usual" field structure on $k((t))$. Here, with the term usual I mean the component wise addition and the convolution product of power series.

In other words I don't agree with the following proposition:

"$k((t))$ is the completion of $k(t)$ w.r.t. $v_{\mathfrak p}$"

Why is it instead true?

Edit: The question has been edited because it contained many imprecisions.

  • $\begingroup$ I’m not sure I understand. Are you observing that the definitions seem very different, but the results are the same? Or are you fearful that the results are in fact different? $\endgroup$ – Lubin Jan 18 '16 at 13:48
  • $\begingroup$ I was trying to check that $k((t))$ is the completion of $k(t)$, but it seems that I concluded that this is not the case. $\endgroup$ – Dubious Jan 18 '16 at 13:59
  • $\begingroup$ Then I have to ask: have you found a pair $f$ and $g$ such that the sum or product is different in the two rings? Or a sequence that’s convergent in one but not the other? The admitted fact that the two definitions look very different is not sufficient for concluding that the two are different. I find your question reasonable, even though I disagree with you, because I know of a parallel situation where two things look alike but are different, it’s the case of power series over $\Bbb Q_p$ and look at the $p$-adic topology induced from $\Bbb Z[[x]]$, and the coefficientwise topology. $\endgroup$ – Lubin Jan 18 '16 at 18:51
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    $\begingroup$ @Dubious if the limit is the same, then they differ by a null sequence, so they will be equal in the completion. $\endgroup$ – Mathmo123 Jan 18 '16 at 20:14
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    $\begingroup$ Yes, the comment of @Mathmo123 may well be pointing out the source of your misunderstanding. $\endgroup$ – Lubin Jan 18 '16 at 23:40

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