Computation of fraction field of formal series over the integers

Question

What is the fraction field $K$ of the domain $\mathbb Z[[X]]$?
It is strictly smaller than the field of Laurent series $L=\operatorname {Frac}\mathbb Q[[X]]$, since $\sum_{i\geq 0}\frac {X^i}{i!}\in L\setminus K$.
Indeed a Laurent series $f(X)=\sum_{i\in \mathbb Z} q_iX^i \in L \; (q_i=0 \operatorname {for} i\lt\lt 0)$ must have as its coefficients rational numbers whose denominators involve only finitely many primes (which depend only on $f(0)$) in order to belong to $K$.
But is that necessary condition sufficient?

Edit
No, that necessary condition is not sufficient.
For example if $p$ is a prime the series $\sum_{i\geq 0}\frac {X^i}{p^{i^2}}$ does not belong to $K=\operatorname {Frac}\mathbb Z[[X]]$, even though the denominators in its coefficients only involve the single prime $p$.
This follows from Elad Paran's Example 2.3(c) quoted by Arno in his very pertinent answer below.

Answer 1

For an integral domain $A$ with fraction field $\operatorname{Frac}(A)=K$ we can define $H=\bigcup_{0\neq a\in A}\frac{1}{a} A[[\frac{x}{a}]]$. We have the series of inclusions $$\operatorname{Frac}(A[[x]])\subseteq H[\frac{1}{x}]\subseteq K((x))$$ where $H[\frac{1}{x}]=\operatorname{Frac}(H)$ is the ring mentioned in the other answer. Moreover, we can describe explicitly when equality is achieved.

For instance a theorem of Gilmer [1] shows that $\operatorname{Frac}(A[[x]])=K((x))$ iff for any sequence $s_n$ of nonzero elements of $A$ we have $\bigcap s_nA\neq (0)$. A theorem of Benhissi [2] shows that $\operatorname{Frac}(A[[x]])=H[\frac{1}{x}]$ iff for any nonzero $a\in A$ we have $\bigcap a^nA\neq(0)$.

Everything is downhill from here. Obviously $A=\mathbb Z$ doesn't satisfy such a condition so unfortunately the inclusion $\operatorname{Frac}(\mathbb Z[[x]])\subset H[\frac{1}{x}]$ is strict. To make things worse, Rivet has a theorem [3] that shows that it is impossible to characterize which power series belong to $\operatorname{Frac}(A[[x]])$ when $A$ is a discrete valuation ring, using criteria that involve only knowing valuations of individual coefficients. So you can put conditions on which primes appear in denominators and to which multiplicities, yet this will not allow you to conclude whether your series belongs to the fraction field or not.

A textbook reference is the monograph [4], particularly chapter 2 on pathologies of $R[[x]]$. It goes over an exposition of constructions that are used in the literature to prove these sort of negative results. The main tool is the ability to explicitly construct power series in $H[1/x]$ that are so sparse that they can't be in $\operatorname{Frac}(A[[x]])$.

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[1] Gilmer, R. "A note on quotient fields of $D[[X]]$" Proc. Amer. Math. Soc. 18, 1138-1140 (1967)

[2] Benhissi, A. "Corps des fractions de $A[[X]]$" Communications in algebra. 25 (9), 2861-2879 (1997)

[3] Rivet, R. "Sur le corps des fractions d’un anneau de séries formelles à coefficients dans un anneau de valuation discrete" C. R. Acad. Sci., Paris, Sér. A 264, 1047-1049 (1967)

[4] Brewer, J.W. "Power Series Over Commutative Rings", CRC press (1981)

Answer 2

There exists a relatively simple necessary and sufficient condition. A power series $ f\in \mathbb Q((X))$ lies in the field of fractions of $\mathbb Z[[X]]$ if and only if it satisfies the following two conditions

• There exist only finitely many primes $p$ appearing in the denominators of coefficients of $f$.
• For each $p$ appearing in the denominators, there exists a monic polynomial $Q \in \mathbb Z_p[X]$ such that $Q f \in \mathbb Z_p[[X]] [\frac{1}{X}, \frac{1}{p} ]$, in other words the powers of $p$ appearing in the denominators of coefficients of $Q f$ are bounded.

Another way of saying the second condition is that $f$ is the sum of an element in $Z_p[[X]] [ \frac{1}{p} ]$ plus an element in $\mathbb Q_p(X)$.

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Proof of "only if"

Given a ratio $\frac{A}{B}$ with $A, B \in \mathbb Z[[X]]$, the only primes appearing in the denominators of coefficients of $\frac{A}{B}$ are the primes dividing the leading coefficient of $B$, of which there are finitely many.

Now fix one of these primes. By the Weierstrass preparation theorem in $\mathbb Z_p[[X]]$, we can write $B$ as an invertible element $U$ of $\mathbb Z_p[[X]]$, times a power of $p$, times a monic polynomial $Q \in \mathbb Z_p[X]$. (Theorem 1.3 of the paper Jesse Elliott linked, after dividing by a suitable power of $p$ to make one of the coefficients zero. ) Writing $f =\frac{A}{B} = frac{A}{U Q p^n}$, we see that $Q f = \frac{A}{U p^n} \in \mathbb Z_p[[X]] [\frac{1}{p} ]$ because $U$ is invertible. This verifies the second condition.

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Proof of "if"

Let $p_1,\dots, p_m$ be the primes appearing in the denominator of $m$ and let $Q_{p_1}, \dots, Q_{p_m}$ be the $p$-adic power series guaranteed by the condition. We will show the existence for each $p$ of a power series $U_{p} \in \mathbb Z_p[[X]]^\times $ such that $U_{p} Q_{p} \in \mathbb Z[[X]]$.

To do this, we can assume by removing a factor of $X^k$ from $Q_{p}$ that the leading coefficient of $Q_{p}$ is nonzero. Because it is a nonzero $p$-adic number, it has the form $p^{n_i}$ times a $p$-adic unit for some natural number $n_i$. Let the first coefficient of $U_{p_i}$ be the inverse $p$-adic unit, and then choose inductively for each $n$ the coefficient of $X^n$ in $U_{p}$ so that the coefficient of $X^n$ in $Q_{p} U_{p}$ is an integer in the range $\{0,\dots, p^{n_i}-1\}$. This is possible since that range hits every element of $\mathbb Z_p / p^{n_i} \mathbb Z_p$.

Now $\prod_{i=1}^m (Q_{p_i} U_{p_i} ) \in \mathbb Z[[X]]$, and $$f \prod_{i=1}^m (Q_{p_i} U_{p_i} ) = Q_{p_i} f U_{p_i} \prod_{j=1}^{i-1}(Q_{p_j} U_{p_j} )\prod_{j=i+1}^{m}(Q_{p_j} U_{p_j} ) $$ is the product of $Q_{p_i} f \in \mathbb Z_p[[X]] [\frac{1}{X}, \frac{1}{p} ] $ with $U_{p_i} \in \mathbb Z_p[[X]]$ and several terms $Q_{p_j} U_{p_j}\in \mathbb Z[[X]] \subset \mathbb Z_p[[X]] $. Thus $f \prod_{i=1}^m (Q_{p_i} U_{p_i} )$ lies in $\mathbb Z_p[[X]] [\frac{1}{X}, \frac{1}{p} ] $, i.e. the powers of $p$ dividing the denominators of its coefficients are bounded. Since these primes $p$ are the only primes that divide the denominators of coefficients of $f$, and no primes divide the denominators of the coefficients of $(Q_{p_j} U_{p_j} )$, it follows that the coefficients of $f \prod_{i=1}^m (Q_{p_i} U_{p_i} )$ have bounded denominators, and thus we may clear denominators by multiplying by a natural number, exhibiting $f$ as a ratio in $\mathbb Z[[X]]$.

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Proof that the alternative second condition is equivalent:

That such $f$ satisfy the second condition is clear. For the converse, given $A \in \mathbb Z_p[[X]]$ and $Q$ monic in $\mathbb Z_p[X]$ of degree $n$, to show that $\frac{A}{Q}$ has this form, we may assume by dividing $Q$ by any factors that are a unit in $\mathbb Z_p[[X]]$ that all non-leading coefficients of $Q$ are divisible by $p$. One then checks that $A= BQ +R $ for $B \in \mathbb Z_p[[X]]$ and $R \in \mathbb Z_p[X]$ of degree at most $n$, one checks that the natural map $\mathbb Z_p[X]/Q \to \mathbb Z_p[[x]]/Q$ is an isomorphism by subtracting a suitable multiple of $Q$ to cancel the degree $n$ term mod $p$, then the degree $n+1$ term mod $p$, and so on, and then the degree $n$ term mod $p^2$, and the degree $n+1$ term mod $p^2$, and so on, etc.

Answer 3

The condition that only finitely many primes divide the denominators is not sufficient. One needs to add at least the additional condition that their growth is polynomial. I suspect there is a more standard reference, but see for Example 2.3(c) in Algebraic patching over complete domains by Elad Paran. It says there that ${\rm Quot}(\mathbb{Z}[[X]])$ is contained in the field $$ \left\{\frac{1}{c}\sum_{i=m}^\infty\frac{b_i}{a^i}X^i:a,b_i,c,m\in\mathbb{Z},ac\neq 0\right\}. $$ This does of course not yet answer the question how to describe ${\rm Quot}(\mathbb{Z}[[X]])$ more precisely.

Answer 4

If $R$ is a UFD with field of quotients $K$, then the group of units $K^*$ is the direct product of $R^*$ with the free abelian group generated by the irreducible/prime elements of $R$ (one for each class of associates). This at least gives you a description of the group $K^* = K \backslash \{0\}$. In the case of $\mathbb{Z}[[X]]$, which is a UFD, its prime elements are described here (see Theorem 1.4): https://www.ams.org/journals/tran/2014-366-08/S0002-9947-2014-05903-5/S0002-9947-2014-05903-5.pdf . Moreover, $\mathbb{Z}[[X]]^*$ is isomorphic to $\{1,-1\} \times (1+X\mathbb{Z}[[X]])$, where also $1+X\mathbb{Z}[[X]]$ is the additive group of the universal lambda ring $\Lambda(\mathbb{Z})$ over $\mathbb{Z}$. Perhaps more is known about the structure of that group.

Answer 5

This is actually a remark on Arno Fehm's answer, but too long for a comment (the remark is due to P. Samuel). Let $p$ be a prime, and let us look at power series $u(X)=\sum_{n\geq 1}a_nX^n$ in $\mathbb{Q}((X))$ satisfying $u^2-pu+X=0$. This is equivalent to $a_1=p^{-1}$, $a_n=a_1a_{n-1}+\ldots +a_{n-1}a_1$. So the $a_i$ are uniquely determined, and of the form $\dfrac{b_i}{p^i} $ with $b_i\in\mathbb{Z}$. Still $u$ does not belong to $K$, because it is integral over $\mathbb{Z}[[X]]$ which is integrally closed. If it were in $K$, it would belong to $\mathbb{Z}[[X]]$, and this is clearly not the case.