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I'm reading some notes that has the following denotation:

  • the set of formal power-series with coefficients in $\mathbb{F}_p$ is denoted by $\mathbb{F}_p[[t]]$.
  • the fraction field, $\operatorname{Frac}\mathbb{F}_p[[t]]$, is denoted by $\mathbb{F}_p((t))$.
  • the fraction field, $\operatorname{Frac}\mathbb{F}_p[t]$, is denoted by $\mathbb{F}_p(t)$.

I'm not sure what the difference in double bracket vs single bracket, and double parenthesis and single parenthesis refers to exactly. I looked for other answers, and found this post, but that hasn't quite elucidated the problem for me. In particular, can somebody help clarify for me what the relationship between $\mathbb{F}_p[[t]]$ and $\mathbb{F}_p[t]$ is?

Michael Hardy
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ctesta01
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  • An element in $\Bbb F_p[t]$ has only a finite number of (non-negative) powers of $t$ (or what amounts to the same thing, a finite number of non-zero coefficients). Elements of $\Bbb F_p[[t]]$ may have an infinite, or finite number of "terms". So $\Bbb F_p[t]$ is a sub-ring of $\Bbb F_p[[t]]$, – David Wheeler Jul 10 '16 at 22:58

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The ring $\mathbb{F}_p[t]$ consists of polynomials with coefficients in the finite field of order $p$: these are finite expressions of the form $\sum_{i=0}^n a_it^i$, with $a_i \in \mathbb{F}_p$.

In contrast, the ring of formal power series allows for infinite expressions: a typical element looks like $\sum_{i=0}^\infty a_i t^i$. So this ring strictly contains the former one.

The fraction field $\mathbb{F}_p(t)$, which is often called the field of rational functions, consists of all quotients $P(t)/Q(t)$ of polynomials, except when $Q(t) = 0$ is the null polynomial.

Finally, the fraction field $\mathbb{F}_p((t))$ may be called the field of formal Laurent series. A generic element in that field takes the form $\sum_{i \in \mathbb{Z}} a_i t^i$, where all but finitely many terms with negative exponent vanish.

Alex Provost
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    The last sum should have only finitely many terms with negative exponent. – Bill Dubuque Jul 10 '16 at 23:28
  • @BillDubuque You're right, of course! Thank you. – Alex Provost Jul 10 '16 at 23:31
  • @BillDubuque is it still a field or is it only a ring? I am currently reading a paper where they use this notation without defining what it is and they are reducing modulo some element... it would be weird if it were in a field where reduction is pretty trivial since there are only two ideals... – Evariste May 15 '19 at 12:25