A famous result of David Hilbert asserts that there exist irreducible polynomials of every degree over having the largest possible Galois group . However, Hilbert’s proof, based on his famous irreducibility theorem, is non-constructive. Issai Schur proved a constructive (and explicit) version of this result: the Laguerre polynomial is irreducible and has Galois group over .

In this post, we give a simple proof of Schur’s result using the theory of Newton polygons. The ideas behind this proof are due to Robert Coleman and are taken from his elegant paper *On the Galois Groups of the Exponential Taylor Polynomials*. (Thanks to Farshid Hajir for pointing out to me that Coleman’s method applies equally well to the Laguerre polynomials.) Before we begin, here is a quote from Ken Ribet taken from the comments section of this post:

Coleman was light years ahead of Steve Jobs: he was the original guy to think different. This was especially true where Newton polygons were concerned. (Steve never got into those.) I remember how obvious it was to Robert how to prove Schur’s theorem concerning the Galois groups of the polynomials obtained by truncating the Taylor series for exp(x). At first, no one could follow Robert’s reasoning. Fortunately, Robert was patient; he was always happy to provide further details.

In order to help the reader understand Coleman’s beautiful ideas, we begin with a quick crash course on valued fields and Newton polygons. A **valued field** is a field together with a valuation , which is a function satisfying and . A valuation induces a metric on by the rule , so one can talk about completeness and completions. For example, the completion of with respect to the -adic valuation is the field of -adic numbers. If is a complete valued field and is a finite extension of , there is a unique valuation on extending the given one on , which can be defined by setting , where is the determinant of multiplication by viewed as an endomorphism of the -vector space . In particular, there is a unique extension of the valuation on to a fixed algebraic closure , and every root of an irreducible polynomial of degree over has valuation belonging to . It follows that every root of a polynomial of degree over has valuation belonging to for some .

The theory of Newton polygons asserts that if is a polynomial with coefficients in a complete valued field , then the valuations of the roots of in some fixed algebraic closure of can be determined in a purely combinatorial way from the valuations of the coefficients of . More concretely, suppose with and nonzero, and consider the points in the Cartesian plane (where we either omit points with or take ). The **Newton polygon** of is defined to be the lower convex hull of these points. Let denote the successive vertices of the Newton polygon, and for let be the slope of the segment. The theorem of the Newton Polygon asserts that there are exactly roots of in having valuation .

More vividly, imagine the points as nails sticking out from the plane and attach a long piece of string with one end nailed to and the other end free. Rotate the string counter-clockwise until it meets one of the nails; this will be the point . As we continue rotating, the segment of string between and (whose slope is ) will be fixed. Continuing to rotate the string in this manner until the string catches on the point yields the Newton polygon. (Click here for a java applet which draws Newton polygons.)

In the figure above, the vertices of the Newton polygon for the truncated exponential polynomial over are , with corresponding slopes . Thus has four roots with valuation , two roots with valuation , and one root with valuation .

Newton polygons are very useful for proving irreducibility of polynomials; for example, they yield an illuminating proof of Eisenstein’s criterion. Indeed, the Newton polygon of a degree Eisenstein polynomial consists of a single segment of slope connecting and , hence all roots of have valuation . However, as mentioned above every root of a polynomial of degree over a field with value group has valuation belonging to for some . Thus is irreducible over .

We now prove, following Coleman, that the exponential Taylor polynomials and the Laguerre polynomials are irreducible over for all . We will then compute their Galois groups using a classical group-theoretic result of Jordan and some elementary number theory.

Fix a prime number . To compute the -adic Newton polygons of the above polynomials, one uses the following two well-known facts:

(1) Let be a positive integer, and write in base as with and let be the sum of the base digits of . Then .

(2) [Lucas' Theorem] Let be positive integers with , written in base as and . (We add extra zeros to the base expansion of if necessary so that the two expansions have the same length.) Then .

It is an exercise using (1) to show that if we write with and , then the vertices of the Newton polygon of are and for , where , and the corresponding slopes of are .

Moreover, using (2) we see that the -adic Newton polygon for is **equal** to the Newton polygon for . Indeed, each coefficient of has valuation at least as big as the corresponding coefficient of , and it follows from Lucas’ theorem that , so in particular . (Instead of (2), one could also apply Kummer’s theorem that is equal to the number of carries when is added to in base .)

Let be any polynomial of degree with coefficients in having the same Newton polygon as and for all primes . We draw the following global conclusions from the local considerations above:

(A) is irreducible over .

Indeed, if divides then divides the denominator of each in lowest terms, hence the denominator of the valuation of each root of in lowest terms. This implies that divides the degree of every irreducible factor of over , hence over as well. Thus every irreducible factor of over has degree divisible by .

(B) If is a prime number, then the Galois group of contains a -cycle.

To see this, first note that implies that , and thus divides the denominator of . As above, this implies that divides the degree of any extension of formed by adjoining a root of with valuation . Thus divides the degree of the splitting field of over , and hence over as well. This means that divides the order of , so by Cauchy’s theorem contains an element of order . Since , the only elements of order in are -cycles.

(C) contains the alternating group for all . If contains a 3-cycle then this holds for as well.

For this, we first observe that by (the proof of) Bertrand’s Postulate, for each integer there is a prime number with . On the other hand, a well-known group-theoretic result due to Camille Jordan says that a transitive subgroup of which contains a -cycle for some prime with must contain . Combined with (A) and (B), this proves (C) for . For , we can instead use the fact that a transitive subgroup of which contains a 3-cycle and a -cycle for some prime with must contain (cf. Theorem 2.2 here).

(D) (Schur) The Galois group of over is for all . The Galois group of is if and if .

First, we can use Dedekind’s theorem (Theorem 4.14 in these notes) to show that the Galois group contains for all ; this amounts to finding a prime for such that factors into an irreducible cubic times linear factors. (Exercise!) It remains to distinguish between the possibilities and for , which can be done by looking at the discriminant: the Galois group will be if the discriminant is a square in and otherwise (see Theorem 4.7 here). It so happens that there are beautiful explicit formulas due to Schur for the discriminant of and : the discriminant of is and the discriminant of is . (For the latter, see Coleman’s paper referenced above, and for the former see Theorem 6.71 of Szego’s book *Orthogonal Polynomials.*) It follows easily from these formulas and Bertrand’s postulate that the discriminant of is never a square, and the discriminant of is a square iff .

** Concluding Remarks:**

1. For more on p-adic numbers, valuations, and local fields, including proofs of many of the assertions made above, see for example Chapter 5 of my course notes on Algebraic Number Theory, these notes by Jack Thorne, or Serre’s book *Local Fields*. A reference for the theory of Newton polygons is the book *An Introduction to G-Functions* by Dwork, Gerotto, and Sullivan.

2.Schur’s proof of the irreducibility of exponential Taylor polynomials was quite different from what we’ve presented here; it relies on a stronger version of Bertrand’s postulate and applies to a much larger class of polynomials; see this handout by Keith Conrad.

3. There is a one-parameter family of generalized Laguerre polynomials which includes both the classical Laguerre polynomials (when ) and the exponential Taylor polynomials considered in this post. One can use the Newton polygon method described here to prove that for fixed , the Galois group of contains for all but finitely many . See this paper and the references therein.

4. Another explicit family of polynomials over having Galois group for all is given by .

The irreducibility of these polynomials was proved by Ernst Selmer in this paper; the argument is tricky but elementary.

To see that the Galois group is , we follow the exposition of Serre from his book *Topics in Galois Theory* (p.42). A prime divides the discriminant of if and only if and its derivative have a common root modulo . Substituting into , we get . Hence there can be at most one double root modulo for each prime which ramifies in a splitting field for . This implies that the inertia subgroup at in is either trivial or of order 2 and generated by a transposition. Since is simply connected (i.e., has no nontrivial unramified extensions), is generated by its inertia subgroups. By Selmer’s theorem, is transitive, and we have just shown that is generated by transpositions. But a transitive subgroup of generated by transpositions must be equal to .

I hadn’t seen that argument of Serre before — it’s quite slick.