9
$\begingroup$

I've found a few articles that write the ring of formal Laurent series in $t$ as $R((1/t))$, but what's the underlying meaning of $\cdot ((\cdot))$?

A mathematician of my acquaintance swears that $R((t))$, not $R((1/t))$, should be used to denote the ring of formal Laurent series in $t$. We can't decide who's right without knowing what $\cdot((\cdot))$ means. (We both agree that $R[[t]]$ denotes the ring of formal power series in $t$ with coefficients in $R$.)

$\endgroup$
5
  • 10
    $\begingroup$ $(())$ should be the fraction field of $[[]]$, just as $()$ is of $[]$. $\endgroup$ Jul 24, 2012 at 22:45
  • 12
    $\begingroup$ ... so $R((1/t))$ denotes the field of formal power series of the form $\sum_{k=-\infty}^ r c_k t^k$ (possibly infinitely many terms of negative degree, but only finitely many terms of positive degree) $\endgroup$ Jul 24, 2012 at 22:54
  • $\begingroup$ I've taken a closer look at the articles I mentioned in the original post, and it appears that they all are looking at formal sums in which the exponent of t is bounded from above, rather than from below. So it seems that my acquaintance was quite right about $R((1/t))$. Is it fair to say, then, that the definition of $R((x_1,x_2,…))$ is the quotient ring of $R[[x_1,x_2,…]]$? $\endgroup$ Jul 25, 2012 at 0:25
  • 5
    $\begingroup$ @James: What you write in your last comment is true if $R$ is a field. Note though that taking fraction fields of formal power series rings gets tricky: e.g. $K[[t_1,t_2]] = K[[t_1]][[t_2]]$, but $K((t_1,t_2)) \subsetneq K((t_1))((t_2))$. $\endgroup$ Jul 25, 2012 at 0:57
  • $\begingroup$ Just a (related) question: for $k$ a field, $k[[t]]$ is $\hat{\mathcal{O}}_{\mathbb{A}^1_k,0}$. $k((t))$ is the quotient field of $\hat{\mathcal{O}}_{\mathbb{A}^1_k,0}$. Is $k((t))$ also the completion of the stalk at the generic point of $\mathbb{A}^1_k$? Which of the definitions of "formal Laurent series" $k((x_1,\ldots,x_n))$ in the answer by @eithil has an analogous geometric interpretation in terms of $\mathbb{A}^n_k$? $\endgroup$
    – Qfwfq
    Sep 8, 2018 at 17:32

3 Answers 3

9
$\begingroup$

I agree with the mathematician of your acquaintance -- well, okay, I am the mathematician of your acquaintance.

Here are some references for the notation $K((x))$ for the field of formal Laurent series $\sum_{n \geq n_0} a_n x^n$ over $K$:

The wikipedia article on formal power series.

$\textbf{}$

Jacobson's Basic Algebra II, $\S 9.12$.

$\textbf{}$

Lam's Introduction to Quadratic Forms Over Fields, $\S VI.1$

$\textbf{}$

Neukirch's Algebraic Number Theory, $\S II.4$.

$\textbf{}$

Serre's Corps Locaux, $\S 1.1$.

$\endgroup$
7
$\begingroup$

I've seen $k((x,y))$ used to mean two different things: the field of fractions of $k[[x,y]]$ and also

$k((x,y)) = \{\sum_{i \geq n} \sum_{j \geq m} f_{i,j}x^iy^j : f_{i,j} \in k; n,m \in \mathbb{Z}\}$

These rings are distinct, e.g. the second one does not contain $\sum_{i \geq 0}y^{-(i+1)}x^i = (y-x)^{-1}$.

In one variable the distinction disappears, and then the only difference between $R((t))$ and $R((1/t))$ is whether the powers of $t$ go to $\infty$ or $-\infty$.

$\endgroup$
2
$\begingroup$

The ring of formal Laurent series is a particular case, for the group order $$ \ldots 1/x^{n+1} < 1/x^{n} < \ldots 1/x^2 <1/x < 1 < x < x^2 < \ldots x^n < x^{n+1} < \ldots $$ of Malcev-Neumann series (series with well-ordered support on an ordered group $\Gamma$). These series are denoted $k((\Gamma))$. So I agree with the mathematician of your acquaintance because this is in accordance with the ``double bracketing'' and also with the the order. Indeed $k((x))$ is well-suited for the classical Laurent series whereas $k((1/x))$ tends to indicate $1/x>1$ and then the series with finite number of positive powers and (possibly) infinite number of negative ones (see Pietro Majer's comment).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.