Theorem:

Let M be a torsion-free R-module, and d > 0. Let

f = ∑_{n_{1},...,n_{d}} a_{n_{1},...,n_{d}} x_{1}^{n_{1}} ⋯ x_{d}^{n_{d}} ∈ M[[x_{1}, … x_{d}]]

be a rational function, i.e. there are P ∈ M[x_{1}, …, x_{d}] and Q ∈ R[x_{1}, …, x_{d}] with Q(0, …, 0) = 1 and Q · f = P. Then the full diagonal of f

g = ∑_{n=0}^{∞} a_{n,...,n} x_{1}^{n}

is a D-finite element of M[[x_{1}]], w.r.t. R[x_{1}] and {∂_{x_{1}}}. Proof:

From the hypotheses, M[[x_{1}, …, x_{d}]] is a torsion-free differential module over R[x_{1}, …, x_{d}] w.r.t. the derivatives {∂_{x_{1}}, …, ∂_{x_{d}}}, and f is a D-finite element of M[[x_{1}, …, x_{d}]] over R[x_{1}, …, x_{d}] w.r.t. {∂_{x_{1}}, …, ∂_{x_{d}}}. Now apply the general diagonal theorem ([1], section 2.18) d − 1 times. Theorem:

Let R be an integral domain of characteristic 0 and M simultaneously a torsion-free R-module and a commutative R-algebra without zero divisors. Let

f = ∑_{m,n≥0} a_{m,n}x^{m}y^{n} ∈ M[[x, y]]

be a rational function. Then the diagonal of f

g = ∑_{n=0}^{∞} a_{n,n} x^{n}

is algebraic over R[x]. Motivation of proof:

The usual proof ([2]) uses complex analysis and works only for R = M = C. The idea is to compute

g(x^{2}) = \frac{1}{2πi} ∮_{|z|=1} f(xz, \frac{x}{z})\frac{dz}{z}

This integral, whose integrand is a rational function in x and z, is calculated using the residue theorem. Since f(x, y) is continuous at (0, 0), there is a δ > 0 such that f(x, y) ≠ ∞ for |x| < δ, |y| < δ. It follows that for all ε > 0 and |x| < δε all the poles of f(xz, \frac{x}{z}) are contained in \{z : |z| < ε\}∪\{z : |z| > \frac{1}{ε}\}. Thus the poles of f(xz, \frac{x}{z}), all algebraic functions of x – let’s call them ζ_{1}(x), … ζ_{s}(x) –, can be divided up into those for which |ζ_{i}(x)| = O(|x|) as x → 0 and those for which \frac{1}{|ζ_{i}(x)|} = O(|x|) as x → 0. It follows from the residue theorem that for |x| < δ

g(x^{2}) = ∑_{ζ=0∨ζ=O(|x|)} Res_{z=ζ} f(xz, \frac{x}{z})

This is algebraic over C(x). Hence g(x) is algebraic over C(x^{1/2}), hence also algebraic over C(x).

Proof:

Let

h(x, z) := f(xz, \frac{x}{z}) = ∑_{m,n=0}^{∞} a_{m,n}x^{m+n}z^{m−n} ∈ M[[xz, xz^{−1}]]

Then g(x^{2}) is the coefficient of z^{0} in h(x, z). Let N(x, z) := z^{d}Q(xz, \frac{x}{z}) (with d := max(deg_{y} P, deg_{y} Q)) be “the denominator” of h(x, z). We have N(x, z) ∈ R[x, z] and N ≠ 0 (because N(0, z) = z^{d}). Let K be the quotient field of R. Thus N(x, z) ∈ K[x][z] \ {0}.

It is well-known (see [3], p.64, or [4], chap. IV, §2, prop. 8, or [5], chap. III, §1) that the splitting field of N(x, z) over K(x) can be embedded into a field L((x^{1/r})), where r is a positive integer and L is a finite-algebraic extension field of K, i.e. a simple algebraic extension L = K(α) = Kα^{0} + ⋯ + Kα^{u−1}.

M̃ := (R \ {0})^{−1} · M is a K-vector space and a commutative K-algebra without zero divisors. M̂ := M̃α^{0} + ⋯ + M̃α^{u−1} is an L-vector space and a commutative L-algebra without zero divisors.

M̂⟨⟨x, z⟩⟩ := M̂[[x^{1/r} · z, x^{1/r} · z^{−1}, x^{1/r}]][x^{−1/r}]
= {∑_{m,n} c_{m,n}x^{m/r}z^{n} : c_{m,n} ≠ 0 ⇒ |n| ≤ m + O(1)}

is an L-algebra which contains M̂((x^{1/r})).

Since N(x, z) splits into linear factors in L((x^{1/r}))[z], N(x, z) = l ∏_{i=1}^{s}(z− ζ_{i}(x))^{k_{i}}, there exists a partial fraction decomposition of h(x, z) = \frac{P (xz, \frac{x}{z})}{Q(xz, \frac{x}{z})} =

\frac{z^{d}P (xz, \frac{x}{z})}{N(x,z)} in M̂⟨⟨x, z⟩⟩:

h(x, z) = ∑_{j=0}^{l} P_{j}(x)z^{j} + ∑_{i=1}^{s} ∑_{k=1}^{k_{i}} \frac{P_{i,k}(x)}{(z − ζ_{i}(x))^{k}}

with P_{j}(x), P_{i,k}(x) ∈ M̂((x^{1/r})).

Recall that we are looking for the coefficient of z^{0} in h(x, z). We compute it separately for each summand.

If ζ_{i}(x) = ax^{m/r} + ... with a ∈ L \ {0}, m > 0, or ζ_{i}(x) = 0, we have

\frac{1}{(z − ζ_{i}(x))^{k}} = \frac{1}{z^{k}} · \frac{1}{(1 − \frac{ζ_{i}(x)}{z})^{k}}

= \frac{1}{z^{k}} · ∑_{j=0}^{∞} \binom{k − 1 + j}{k − 1} (\frac{ζ_{i}(x)}{z})^{j}

= ∑_{j=0}^{∞} \binom{k − 1 + j}{k − 1}\frac{ζ_{i}(x)^{j}}{z^{k+j}}

hence the coefficient of z^{0} in \frac{P_{i,k}(x)}{(z−ζ_{i}(x))^{k}} is 0.

If ζ_{i}(x) = ax^{m/r} + ... with a ∈ L \ {0}, m < 0, we have

\frac{1}{(z − ζ_{i}(x))^{k}} = \frac{1}{(−ζ_{i}(x))^{k}} ·\frac{1}{(1 − \frac{z}{ζ_{i}(x)})^{k}} = \frac{1}{(−ζ_{i}(x))^{k}} ·∑_{j=0}^{∞} \binom{k − 1 + j}{k − 1} (\frac{z}{ζ_{i}(x)})^{j}

hence the coefficient of z^{0} in \frac{P_{i,k}(x)}{(z−ζ_{i}(x))^{k}} is \frac{P_{i,k}(x)}{(−ζ_{i}(x))^{k}}.

The case ζ_{i}(x) = ax^{m/r} + ... with a ∈ L \ {0}, m = 0, cannot occur, because it would imply 0 = N(0, ζ_{i}(0)) = N(0, a) = a^{d}.

Altogether we have

g(x^{2}) = [z^{0}]h(x, z) = P_{0}(x) + ∑_{\frac{1}{ζ_{i}(x)} =o(x)} ∑_{k=1}^{k_{i}} \frac{P_{i,k}(x)}{(−ζ_{i}(x))^{k}} ∈ M̂((x^{1/r}))

Since all ζ_{i}(x)(in L((x^{1/r}))) and all P_{j}(x), P_{i,k}(x) (in M̂((x^{1/r}))) are algebraic over K(x), the same holds also for g(x^{2}). Hence g(x) is algebraic over K(x^{1/2}), hence also over K(x). After clearing denominators, we finally conclude that g(x) is algebraic over R[x].

References

1 Bruno Haible: D-finite power series in several variables. Diploma thesis, University of Karlsruhe, June 1989. Sections 2.18 and 2.20.

2 M. L. J. Hautus, D. A. Klarner: The diagonal of a double power series. Duke Math. J. 38 (1971), 229-235.

3 C. Chevalley: Introduction to the theory of algebraic functions of one variable. Mathematical Surveys VI. American Mathematical Society.

4 Jean-Pierre Serre: Corps locaux. Hermann. Paris 1968.

5 Martin Eichler: Introduction to the theory of algebraic numbers and functions. Academic Press. New York, London 1966.