Newton polygon

{{Short description|Tool for solving polynomial equations}}

{{Use British English|date=October 2024}}

In mathematics, the Newton polygon is a tool for understanding the behaviour of polynomials over local fields, or more generally, over ultrametric fields.

In the original case, the ultrametric field of interest was essentially the field of formal Laurent series in the indeterminate X, i.e. the field of fractions of the formal power series ring KX,

over K, where K was the real number or complex number field. This is still of considerable utility with respect to Puiseux expansions. The Newton polygon is an effective device for understanding the leading terms aX^r

of the power series expansion solutions to equations P(F(X)) = 0

where P is a polynomial with coefficients in K[X], the polynomial ring; that is, implicitly defined algebraic functions. The exponents r here are certain rational numbers, depending on the branch chosen; and the solutions themselves are power series in KY

with Y = X^{\frac{1}{d}} for a denominator d corresponding to the branch. The Newton polygon gives an effective, algorithmic approach to calculating d.

After the introduction of the p-adic numbers, it was shown that the Newton polygon is just as useful in questions of ramification for local fields, and hence in algebraic number theory. Newton polygons have also been useful in the study of elliptic curves.

Definition

Image:Newton-polygon.gif

A priori, given a polynomial over a field, the behaviour of the roots (assuming it has roots) will be unknown. Newton polygons provide one technique for the study of the behaviour of the roots.

Let K be a field endowed with a non-archimedean valuation v_K: K \to \mathbb R\cup \{ \infty \}, and let

:f(x) = a_nx^n + \cdots + a_1x + a_0 \in K[x],

with a_0 a_n \ne 0. Then the Newton polygon of f is defined to be the lower boundary of the convex hull of the set of points P_i=\left(i,v_K(a_i)\right),

ignoring the points with a_i = 0.

Restated geometrically, plot all of these points Pi on the xy-plane. Let's assume that the points indices increase from left to right (P0 is the leftmost point, Pn is the rightmost point). Then, starting at P0, draw a ray straight down parallel with the y-axis, and rotate this ray counter-clockwise until it hits the point Pk1 (not necessarily P1). Break the ray here. Now draw a second ray from Pk1 straight down parallel with the y-axis, and rotate this ray counter-clockwise until it hits the point Pk2. Continue until the process reaches the point Pn; the resulting polygon (containing the points P0, Pk1, Pk2, ..., Pkm, Pn) is the Newton polygon.

Another, perhaps more intuitive way to view this process is this : consider a rubber band surrounding all the points P0, ..., Pn. Stretch the band upwards, such that the band is stuck on its lower side by some of the points (the points act like nails, partially hammered into the xy plane). The vertices of the Newton polygon are exactly those points.

For a neat diagram of this see Ch6 §3 of "Local Fields" by JWS Cassels, LMS Student Texts 3, CUP 1986. It is on p99 of the 1986 paperback edition.

Main theorem

With the notations in the previous section, the main result concerning the Newton polygon is the following theorem,For an interesting demonstration based on hyperfields, see Matthew Baker, Oliver Lorscheid, (2021). Descartes' rule of signs, Newton polygons, and polynomials over hyperfields.Journal of Algebra, Volume 569, p. 416-441. which states that the valuation of the roots of f are entirely determined by its Newton polygon:

Let \mu_1, \mu_2, \ldots, \mu_r

be the slopes of the line segments of the Newton polygon of f(x) (as defined above) arranged in increasing order, and let

\lambda_1, \lambda_2, \ldots, \lambda_r

be the corresponding lengths of the line segments projected onto the x-axis (i.e. if we have a line segment stretching between the points P_i and P_j then the length is j-i).

  • The \mu_i are distinct;
  • \sum_i \lambda_i = n;
  • if \alpha is a root of f in K, v(\alpha) \in \{-\mu_1, \ldots , -\mu_r\};
  • for every i, the number of roots of f whose valuations are equal to -\mu_i (counting multiplicities) is at most \lambda_i, with equality if f splits into the product of linear factors over K.

Corollaries and applications<span class="anchor" id="Applications"></span>

With the notation of the previous sections, we denote, in what follows, by L the splitting field of f over K, and by v_L an extension of v_K to L.

Newton polygon theorem is often used to show the irreducibility of polynomials, as in the next corollary for example:

  • Suppose that the valuation v is discrete and normalized, and that the Newton polynomial of f contains only one segment whose slope is \mu and projection on the x-axis is \lambda. If \mu = a/n, with a coprime to n, then f is irreducible over K. In particular, since the Newton polygon of an Eisenstein polynomial consists of a single segment of slope -\frac{1}{n} connecting (0,1) and (n,0), Eisenstein criterion follows.

Indeed, by the main theorem, if \alpha is a root of f, v_L(\alpha) = -a/n.

If f were not irreducible over K, then the degree d of \alpha would be < n , and there would hold v_L(\alpha) \in {1\over d}\mathbb Z. But this is impossible since v_L(\alpha) = -a/n with a coprime {{nobr|to n}}.

Another simple corollary is the following:

  • Assume that (K, v_K) is Henselian. If the Newton polygon of f fulfills \lambda_i = 1 for some i, then f has a root in K.

Proof: By the main theorem, f must have a single root \alpha whose valuation is v_L(\alpha) = -\mu_i. In particular, \alpha is separable over K.

If \alpha does not belong to K, \alpha has a distinct Galois conjugate \alpha' over K, with v_L(\alpha') = v_L(\alpha),Recall that in Henselian rings, any valuation extends uniquely to every algebraic extension of the base field. Hence v_K extends uniquely to v_L. But v_L\circ \sigma is an extension of v_K for every automorphism \sigma of L, therefore v_L(\alpha') = v_L\circ \sigma (\alpha) = v_L(\alpha). and \alpha' is a root of f, a contradiction.

More generally, the following factorization theorem holds:

  • Assume that (K, v_K) is Henselian. Then f = A\,f_1\, f_2\cdots f_r,, where A\in K, f_i\in K[X] is monic for every i, the roots of f_i are of valuation -\mu_i , and \deg(f_i) = \lambda_i. J. W. S. Cassels, Local Fields, Chap. 6, thm. 3.1.

:Moreover, \mu_i = v_K(f_i(0))/\lambda_i, and if v_K(f_i(0)) is coprime to \lambda_i, f_i is irreducible over K.

Proof:

For every i, denote by f_i the product of the monomials (X - \alpha) such that \alpha is a root of f and v_L(\alpha) = -\mu_i. We also denote f = A P_1^{k_1}P_2^{k_2}\cdots P_s^{k_s} the factorization of f in K[X] into prime monic factors (A\in K).

Let \alpha be a root of f_i. We can assume that P_1 is the minimal polynomial of \alpha over K.

If \alpha' is a root of P_1, there exists a K-automorphism \sigma of L that sends \alpha to \alpha', and we have v_L(\sigma \alpha) = v_L(\alpha) since K is Henselian. Therefore \alpha' is also a root of f_i.

Moreover, every root of P_1 of multiplicity \nu is clearly a root of f_i of multiplicity k_1\nu, since repeated roots share obviously the same valuation. This shows that P_1^{k_1} divides f_i.

Let g_i = f_i/P_1^{k_1}. Choose a root \beta of g_i. Notice that the roots of g_i are distinct from the roots of P_1. Repeat the previous argument with the minimal polynomial of \beta over K, assumed w.l.g. to be P_2, to show that P_2^{k_2} divides g_i.

Continuing this process until all the roots of f_i are exhausted, one eventually arrives to

f_i = P_1^{k_1}\cdots P_m^{k_m}, with m \leq s. This shows that f_i\in K[X], f_i monic.

But the f_i are coprime since their roots have distinct valuations. Hence clearly f = A f_1\cdot f_2\cdots f_r, showing the main contention.

The fact that \lambda_i = \deg(f_i) follows from the main theorem, and so does the fact that \mu_i = v_K(f_i(0))/\lambda_i, by remarking that the Newton polygon of f_i can have only one segment joining (0, v_K(f_i(0)) to (\lambda_i, 0 = v_K(1)). The condition for the irreducibility of f_i follows from the corollary above. (q.e.d.)

The following is an immediate corollary of the factorization above, and constitutes a test for the reducibility of polynomials over Henselian fields:

  • Assume that (K, v_K) is Henselian. If the Newton polygon does not reduce to a single segment (\mu, \lambda), then f is reducible over K.

[[Image:Diagram_of_a_Newton_Polygon_Convex_hull.svg|thumb|right| The Newton polygon for {{nowrap|1= 3x2 y3xy2 + 2x2y2x3y}} with positive monomials in red and negative monomials in cyan. Faces are labelled with their limiting terms.

]]

Other applications of the Newton polygon comes from the fact that a Newton Polygon is sometimes a special case of a Newton polytope, and can be used to construct asymptotic solutions of two-variable polynomial equations like

3 x^2 y^3 - x y^2 + 2 x^2 y^2 - x^3 y = 0.

Symmetric function explanation

In the context of a valuation, we are given certain information in the form of the valuations of elementary symmetric functions of the roots of a polynomial, and require information on the valuations of the actual roots, in an algebraic closure. This has aspects both of ramification theory and singularity theory. The valid inferences possible are to the valuations of power sums, by means of Newton's identities.

History

Newton polygons are named after Isaac Newton, who first described them and some of their uses in correspondence from the year 1676 addressed to Henry Oldenburg.Egbert Brieskorn, Horst Knörrer (1986). Plane Algebraic Curves, pp. 370–383.

See also

References

{{reflist}}

  • {{Citation | last=Goss | first=David | title=Basic structures of function field arithmetic | publisher=Springer-Verlag | location=Berlin, New York | series=Ergebnisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas (3)] | isbn=978-3-540-61087-8 |mr=1423131 | year=1996 | volume=35 | doi=10.1007/978-3-642-61480-4}}
  • Gouvêa, Fernando: p-adic numbers: An introduction. Springer Verlag 1993. p. 199.