In this post I discuss a funny observation about algebraic Laurent series. In a future post I will explain why it is also an interesting fact. Thanks to Will Sawin and Raymond Cheng for helping me figure this out! As usual, all mistakes are mine.

Let A be a complete discrete valuation ring with uniformizer t and fraction field K. Let x be a variable. Consider an element f in the t-adic completion of A[x, 1/x]. Then we can write

f = sum a_{n} x^{n}

where the sum is over all integers n (postive and negative) and where a_{n} is in A and tends to zero t-adically as the absolute value of n goes to infinity. We say f is *algebraic* if there exists a relation

sum P_{i} f^{i} = 0

where the sum is finite and the P_{i} are rational functions of x not all zero. Of course by clearing denominators we may assume P_{i} is in A[x] and not all of them equal to zero. Finally, given f we can of course write

f = fplus + fminus

where fplus is the sum of a_{n} x^{n} for n ≥ 0 and fminus is the sum of a_{n} x^{n} for n < 0.

**Question.** If f is algebraic, must fplus and fminus be algebraic?

It turns out that if K has positive characteristic, then the answer is yes and if K has characteristic zero, then the answer isn’t yes in general. Let’s get to work.

**Char p.** If f is algebraic, then we can find a relation of the form

sum P_{i} f^{pi} = 0

for some polynomials P_{i} not all zero. Then looking at powers of x we observe that

sum P_{i} fplus^{pi}

is a polynomial in x! This of course means that fplus is algebraic. QED.

**Char 0.** This is more difficult. We claim that the square root f of (1 + tx)(1 + t/x) which starts as 1 + 1/2t(x + 1/x) + … is a counter example. To see this we consider the differential operator

L = 2x(1 + tx)(t + x)D – t(x^2 – 1)I

where D = d/dx and I is the identity operator. An easy computation shows that L(f) = 0. Looking at powers of x the reader easily shows that

L(fplus) = a + bx

for some nonzero a, b in A. In fact, a computation shows that we have

a = t * sum_{i = 0, 1, 2, …} binomial(1/2, i)^2 * t^(2*i)

and

b = sum_{i = 1, 2, 3, …} -t^(2*i) * binomial(2*i – 2, i – 1) * 4 * 3 * (3 – 8*i + 4*i^2)^(-1) * 2^(-i*4) * binomial(2*i, i)

(It isn’t necessary to verify this for what we’re going to say next.)

**Lemma.** If fplus is algebraic, then fplus is the sum of a rational function and a K-multiple of f.

**Proof.** Observe that the operator L’ = (a + bx)D – bI annihilates a + bx. Hence we see that fplus is a solution of L’L. Choose an embedding of K into the complex numbers C. On some affine open U of P^1_C we see that L defines a local system V of rank 1 with finite monodromy; in fact the monodromy group is of order 2 because L is of order 1 with 4 regular singular points whose residues are each a half integer (you can also see this without computation by considering how we chose f in the first place). On the other hand, L’ defines a local system V’ of rank 1 with trivial monodromy (as it has a solution which is a rational function). Finally, L’L defines an extension W of V’ by V, i.e., we have a short exact sequence 0 → V → W → V’ → 0. The monodromy of W is either finite and then the extension is split, or the monodromy is infinite and the extension is nonsplit. Of course fplus defines a horizontal section of W over some small disc. Since L(fplus) is nonzero, we conclude that fplus isn’t in V. If fplus is algebraic (before embedding into C then of course also after embedding K into C), then we conclude that the monodromy of W is finite and the extension is split over all of U. This means there is a rational function h over C with L’L(h) = 0. By a change of fields argument (omitted), this implies we can find a rational function h over K with L’L(h) = 0. Then any solution, and in particular fplus, is a complex linear combination of h and f. QED

OK, so now we know that we have fplus = P/Q + c f for some constant c and some P, Q in A[x] with Q not zero. Multiplying through by Q we obtain Q * fplus = P + c * Q * f. Looking at very negative powers of x we conclude that c = 0 but looking at very positive powers of x we conclude that c = 1. This contradiction finishes the proof.