On the structure of the classical discriminant Текст научной статьи по специальности «Математика»

УДК 517.55

On the Structure of the Classical Discriminant

Evgeny N. Mikhalkin* Avgust К. Tsikh

Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041

Russia

Received 18.09.2015, received in revised form 18.10.2015, accepted 25.10.2015 Consider a general polynomial of degree n with variable coefficients. It is known that the Newton polytope of its discriminant is combinatorially equivalent to an (n — 1)-dimensional cube. We show that two facets of this Newton polytope are prisms, and that truncations of the discriminant with respect to facets factor into discriminants of polynomials of smaller degree.

Keywords: general algebraic equation, discriminant, Newton polytope. DOI: 10.17516/1997-1397-2015-8-4-426-436

We consider a general polynomial of degree n

f (y) = ao + aiV + ••• + anyn. (1)

By its discriminant we call an irreducible polynomial A(a0, ai,... ,an) with integer coefficients that vanishes if and only if the polynomial (1) has multiple roots. The purpose of this article is to refine some classical results on the structure of facets of the Newton polytope of the discriminant A, as well as to study factorization of truncations of the discriminant with respect to the facets. The knowledge of this structure is important in the study of a general algebraic function y = y(a) of roots of the polynomial (1) ( [1,2]).

1. The Newton polytope for the classical discriminant

Recall that the Newton polytope N (A) of the polynomial A(a0,..., an) is the convex hull in Rn+i of exponents k = (k0, ki,... kn) of all monomials participating in A.

Note that the Newton polytope of the polynomial (1) is the segment [0, n] C R. The following theorem shows that each vertex of the Newton polytope N (A) of the discriminant of f corresponds to an appropriate triangulation (i.e. a partition into segments) of the segment [0, n]. Each partition is given by a set of integer points

0 = io < ii < . . . < is < is+i = n.

It is clear that such a set is identified by a subset I C {1, 2,...,n — 1} of the type I = {ii < i2 < ... < is}, 0 < s < n — 1. The number of all such subsets equals 2n-i, since we include in the list the empty set too, which corresponds to s = 0.

Theorem 1 ( [3], p. 412). The Newton polytope N (A) of the discriminant A is combinatorially equivalent to an (n — 1)-dimensional cube; its 2n-i vertices are in a bijective correspondence with all possible subsets

I C {1, 2,...,n — 1}.

* [email protected]

1 [email protected] © Siberian Federal University. All rights reserved

The vertex vj corresponding to subset I = {ii < i2 < ... < is} has the coordinates

ko = ii - io - 1, kn = is+i - is - 1 (2)

kiq = ig+1 - iq— 1 for iq € I,

ki = 0 for i €I U {0, n}. Let lq = iq+1 - iq (0 ^ q ^ s). Then the monomial

is in A with the coefficient

H-l^q (3)

q=0

Let us illustrate the theorem by an example of a cubic polynomial

f (y) = ao + aiy + a^y2 + a3y3.

Its discriminant is

* 0 0 0 0 A = -27a0a2 - 4a-La3 - 4aoa2 + a-a- + 18aoaia-a2.

In this case there are 4 subsets I C {1,2}:

Io = 0, Ii = {1}, 12 = {2}, I3 = {1, 2}.

The corresponding monomials are the following

-270^^, -4a2a3, -4aoa2, a-a-.

As for the monomial 18aoaia2a2, it corresponds to an interior integer point (1,1,1,1) € N(A), and the theorem says nothing about it.

Further on we consider (1) with ao = an = 1, i.e. a reduced polynomial

fred(y) = 1 + ai y + ... + an—iyn—i + yn. (4)

The Newton polytope of the discriminant of this polynomial lies in Rn—i (the coordinates ko and kn in the expression (2) are missing). The discriminant of the polynomial (4) we call a reduced discriminant. For example, for a cubic equation the reduced discriminant equals

A = 27 + 4a2 + 4a- - 18aia2 - a\a\.

The paper [4] gives the inequalities defining the Newton polytope N (A) for the discriminant of the polynomial (4). There are 2 • (n - 1) such inequalities:

n—1

tk ^ 0, ^^ min(j, k)[n - max(j, k)]tj ^ nk(n - k), k = 1, 2,...,n - 1. (5)

j=i

Thus, the Newton polytope of the reduced discriminant of a cubic equation is given by the system of inequalities

ti > 0, t2 > 0, 2ti +12 < 6, ti +2t2 < 6

(see Fig. 1).

c

(0,0) (3,0)

Fig. 1. The Newton polytope of the reduced discriminant of a cubic equation

2. The prism facets of the Newton polytope N (A)

Recall that we consider a reduced discriminant A. Further on it will be convenient for us to denote the facets in (5) that are not in the coordinate hyperplanes as

( n-1 ^

9k := \t € N (A) : ^^ min(j, k)[n — max(j, k)]tj = nk(n — k), (6)

1 j=1

k = 1, 2,...,n — 1.

We shall call a polytope G of dimension d a d-prism, if it is a Minkowski sum of a (d — 1)-dimensional polytope and a segment. This (d — 1)-dimensional polytope and its translate we call the prism's bases.

Theorem 2. The facets g2 and 9n-2 of the Newton polytope N (A) of the reduced discriminant A are (n — 2) -prisms.

To prove Theorem 2 we need two lemmas. Recall that according to Theorem 1 each vertex vI = of the Newton polytope N(A) of the reduced discriminant of the polynomial (4) is

written down as follows

ii \ [Q, . . . . . . - il,...,lA - 12,...,is— is-2, .. .,n — is-1, . .., 0j (7)

(there are zeroes on vacant places).

Lemma 1. Each vertex vI = of the Newton polytope N (A) belongs to all s facets gk,

k = i1,... ,is of the list (6) and doesn't lie in any remaining n — 1 — s facets of this list.

Proof of Lemma 1. First we show that the vertex lies in the facets gk with k =

i1,...,is. According to (6), if it belongs to gip, p =1, 2,... ,s then

p s

(n — ip)Y, ivtiv + ip XI (n — iv )tiv = n(n — ip)ip, (8)

v=1 v=p+1

where tiv are nonzero coordinates of , which in accordance with (7) can be written down

as tiv = iv+1 — iv-1, v = 1, 2,..., s, with i0 = 0, is+1 = n.

p

Transform both sums in the left hand side of the equality (8). The first sum ^ iv(iv+1 —iv-1)

v=1

s

equals ipip+1. As for the second sum ^ (n — iv)(iv+1 — iv-1), here all the terms except four

v=p+1

of them vanish, and as a result we get

s

53 (n - iv )(iv+1 - iv-1) = -nip - nip+1 + ipip+1 + n2 = ip(ip+l - n) + n(n - ip+1) = v=p+1

= (n - ip)(n - ip+l). Then for the left hand side of (8) we have

ps

(n - ip) 53 iv tiv + ip 53 (n - iv )tiv = (n - ip)ipip+1 + ip(n - ip)(n - ip+1) = n(n - ip)ip,

v=1 v=p+1

i.e. the required equality (8).

Further on, for an arbitrary partition of the segment [0, n] by i1,... ,is we prove the following fact. If i' £ {i1,... ,is} then the vertex vi1,..,is doesn't lie in the facet gi'. Let i' lie between ip and ip+1, then we need to prove that

ps

(n - i') iv tiv + i' (n - iv )tiv = ni' (n - i'),

v=1 v=p+1

that is that if i' does not coincide with the points i1,... ,is of the segment partition, then the function

ps

h(i') = (n - i' )^2 iv (iv+1 - iv-1) + i' "^2 (n - iv )(iv+1 - iv-1) - ni' (n - i') v=1 v=p+1

does not vanish. It is not hard to see that h(i') = 0 for i' = ip and i' = ip+1. Indeed, using the

equalities proved in this lemma

pp

ivtiv =53 iv (iv+1 - iv-1) = ipip+1,

v=1 v=1

ss

53 (n - iv )tiv = 53 (n - iv )(iv+1 - iv-1) = (n - ip)(n - ip+1),

v=p+1 v=p+1

we get

h(i') = (n - i')ipip+1 + i'(n - ip)(n - ip+1) - ni'(n - i'). Then for h(ip) we have

h(ip) = (n - ip)ipip+1 + ip(n - ip)(n - ip+1) - nip(n - ip) =

= ip(n - ip)(ip+1 + n - ip+1 - n) = 0.

Similarly, we get h(ip+1) = 0.

Since h is a polynomial of degree two in i', we conclude that

h(i') = n(i' - ip)(i' - ip+1).

Taking into account the fact that i' is not a point of the partition, i.e. i' = ip, i' = ip+1, we see that h(i') =0. □

Lemma 2. Let i' ,i'' be two different integer points of the segment [0, n] that are not in {i1,..., is}. Then in l"-1 the following vector equalities hold:

Vi' Vi'i'' = Vi

is < i < i ;

Vi'' Vi'i'' = Vi'',i1,...,is Vi'i''iu... is , i <i <is.

Proof of Lemma 2. Calculate the coordinates of vertices in the equalities:

(0,...,n,..., 0), vi' i" = (0,...,i>-',..., "",..., 0),

il i2 is — 1 i'

(Q, . . . ,i2, . . . ,h - i1,. . .,is - is-2, . . . ,n- is,. . . , 0^ ,

i1 i2 is—1 i i

Vi1, ,.,is, i', i'' =(0,. . . ,i2, . . . ,'h - i1,... ,is - is-2, .. .,i" - is,... ,n - i',.

(there are zeroes on vacant places). Hence

Vi' = 0,

■0)

Similarly,

Vi'i'' — Vi' = Vi

Vi'i''— Vi'' = Vi'i''

i', i'', il, ... , is

vi ,i1,

i i (o,...,i" — n,.. . ,n — i', .. ., 0^ .

^0,...,l?/,..., -i',..., 0).

Thus, the lemma is proved. □

Now we can turn to the proof of Theorem 2.

Proof of Theorem 2. First, we prove the theorem for the facet g2. According to Lemma 1, it contains all vertices of the polytope N(A) of the type (i.e. corresponding to partitions of

[0, n] having the point 2 in the defining set of integer points) and only of this type. Let us show that these points constitute a prism. In order to do this we divide all the mentioned partitions into two groups. The first group includes all the partitions which do not employ the point 1, the vertices of N (A) corresponding to them are of the type v2,...; the second group includes all partitions employing the point 1, the corresponding vertices of N (A) are of the type v1,2,....

Show that vertices of each group lie in two parallel planes of dimension n - 3. Namely, the vertices of the first group lie in the plane given by

ti =0, (n — 2)t2 + (n — 3)ts + ... + 1tn-i = n(n — 2).

(9)

The coordinates of the vertices v2can be written down as (7) with i1 = 2. Substituting these coordinates into the left hand side of the second equation of (9) we get

ss

y^(n - iv )(i v+1 - iv-1) =53 n(i v+1 - iv-1 - iv iv+1 + iv-1iv) = v=1 v=1

= nis + nis+1 - nio - ni1 - isis+1 + ioi1 = n(n - 2).

Here we take into account that is+;t = n, i0 = 0, i1 = 2. Thus the second equality in (9) holds.

Now, according to Lemma 2, each vertex of N(A) corresponding to a partition from the second group, i.e. each point of the form v1}2}i1}..,is, s = 0,...,n - 3, is a translate of the point v2,i1,...,is corresponding to a partition from the first group by the vector (2, -1,0,..., 0). It follows from here that two faces, one formed by the points of the form v1,2,i1,...,is and another

//

V

i1,...,is,i

//

formed by the points v2,ilt...,is, s = 0,... ,n — 3, are conguent. Also, it is clear that the other edges of the facet are parallel. So, for the facet g2 the theorem is proved.

The bases are parallel, and the base formed by the vertices of N (A) of the type v1i2,... passes through the point v12...,n-1 = (2,..., 2) of the polytope, from these facts it follows that the equation of this base is

t1 = 2, (n — 2)t2 + (n — 3)t3 + ... + 1 • tn-1 = (n — 1)(n — 2).

Similar arguments are applicable to the facet gn-2, whose each point looks like v..,n-2,.... For this facet each vertex of N (A) of the form vilt...,is,n-2,n-1, s = 0,... ,n — 3, belonging to one of the prism's bases is a translate by the vector (0,..., —1,2) of a point of the form vil..,is,n-2, belonging to another base. Herewith, the equations of the bases of the prism gn-2 are

tn-1 =0, 1t1 +2t2 + ... + (n — 2)tn-2 = n(n — 2),

tn-1 = 2, 1t1 +2t2 + ... + (n — 2)tn-2 = (n — 1)(n — 2). The theorem is proved. □

3. Factorability of the discriminant's truncations A|

9k

A discriminant truncation with respect to a facet gk of the Newton polytope N(A) is the polynomial AI consisting of all monomials A with exponents from gk. The main observation of

' 9k

this section is the following statement about discriminants of polynomials of degree not greater than 6.

Theorem 3. For n = 2, 3, 4, 5, 6 the truncation An | of the reduced discriminant is a product(up to a monomial factor) Ak(1, a1,..., ak )An-k(1, an-1,..., ak) of the discriminants of polynomials of degrees k and n — k.

Conjecture. The statement of Theorem 3 seems to be valid for any degree n.

Before we begin to prove Theorem 3, remark that A1 = 1 since Ak (1, a1) is the discriminant of the polynomial f (y) = a1y + 1. Therefore the statement of Theorem 3 for n = 2, 3 is trivial. So, we consider only the cases when n = 4, 5, 6. We begin our proof for facets that are prisms.

3.1. Truncations of the discriminant with respect to prism facets

In this section we study truncations of the reduced discriminant with respect to facets g2 and 9n-2, which according to Theorem 2 are prisms. Consider a reduced polynomial of degree 4:

fred(y) = 1 + aYy + a2y2 + a3 y3 + yA.

In this case prisms g2 and gn-2 coincide, therefore only one of the facets of N (A) is a 2-prism, i.e. a parallelogram. According to (6) the facet g2 is given by

3t1 +2t2 +13 < 12, t1 +2t2 +13 = 8, t1 + 2t2 + 3t3 < 12

in the positive orthant. Integer solutions of this system are the four points

V2 = (0,4, 0), V12 = (2, 3, 0), V123 = (2, 2, 2), V23 = (0, 3, 2),

which are the vertices of the parallelogram. Using formula (3) we compute the truncation

A|92 = 16a4 — 4a\a2 + a21a2ia^ — It turns out that it factors into

A|g2 = a2(4a2 — o2)(4O2 — a2), (10)

where binomial factors are discriminants of quadratic polynomials

a2y2 + a3y + 1 and a2y2 + a1y + 1.

For n = 5 the polynomial (4) takes the form

fred(y) = 1 + a1y + a2y2 + a3y3 + a^yA + y5. (11)

In this case according to Theorem 2 the facets g2 and g3 are 3-prisms. The facet g3 is defined by the system

4t1 + 3t2 + 2t3 + t4 < 20, 3t1 + 6t2 + 4t3 + 2t4 < 30, 2t1 + 4t2 + 6t3 + 3t4 = 30, t1 + 2t2 + 3t3 + 4t4 < 20.

Calculations show that a solution of this system which defines the truncation A1 of the dis-

93

criminant of (11) consists of the following 10 points

v3 = (0,0, 5, 0), v13 = (3, 0, 4,0), v23 = (0, 3, 3, 0), v123 = (2, 2, 3, 0), V34 = (0, 0, 4, 2), V134 = (3, 0, 3, 2), V234 = (0, 3, 2, 2), V1234 = (2, 2, 2, 2),

(1,1,4, 0), (1,1, 3, 2).

As we see, the first five points lie on the prism's base in the hyperplane t4 = 0, and the rest of these points lie on the base in the hyperplane t4 = 2. The expression for A|g is

A| = 108a3 + 16a\a4 + 16a^a^ — 4a21a2a^ — 27a^a24 — 4a3a3a4 — 4a\ o'^o,4+ +a1a2a2a^ — 72a10,20,3 + 18a1 a2a3a It turns out that this truncation also factors:

A|93 = a^(4a3 — a24)(27a^ + 4O3O,3 + 4a3 — a\a2 — 18CJ,10,20,3) . (12)

Note that the factor 4a3 — o?4 in this expression is the discriminant A2;3, 4(a3, a4) of a quadratic polynomial f2;3,4(y) = a3y2 + a4y + 1 and the last factor is the discriminant A3;3 , 2,1(a1, a2, a3) of a cubic polynomial

f3;3,2,1(y) = a3y3 + a2y2 + a1y + 1.

Consider now the polynomial (4) for n = 6:

fred(y) = 1 + aiy + a2y2 + a3y3 + a4y4 + a5y5 + y6. (13)

Compute A|s4. According to (6) the integer points of N(A) that belong to g4 are defined by the following conditions

5ti + 4t2 + 3t3 + 2t4 + t5 < 30, 4ti + 8t2 + 6t3 + 4t4 + 2t5 < 48, 3t1 + 6t2 + 9t3 + 6t4 + 3t5 < 54, 2ti + 4t2 + 6t3 + 8t4 + 4t5 = 48, t1 + 2t2 + 3t3 + 4t4 + 5t5 < 30.

It follows from the equations for the prisms' bases obtained in the previous section that for the level t5 the values 0,1,2 are possible. For the base in t5 =0 we get the system:

t4 > 3, t3 + 2t4 > 8, t2 + 2t3 + 3t4 > 15, ti + 2t2 + 3t3 + 4t4 = 24,

(14)

the solution of which are 16 points from Z>0, and 8 of them are vertices

v1234 = (2, 2, 2, 3, 0), v234 = (0, 3, 2, 3,0), v134 = (3,0, 3, 3, 0),

v34 = (0,0,4, 3, 0), v124 = (2, 3, 0, 4,0), v24 = (0,4,0, 4, 0), v14 = (4,0,0, 5, 0), v4 = (0,0,0, 6, 0).

The remaining 8 points

(1,1, 3, 3,0) (3,1,1, 4,0), (1, 2,1,4,0), (2, 0, 2,4,0), (0,1, 2,4,0),

(2,1, 0, 5, 0), (0, 2, 0, 5,0) , (1, 0,1, 5,0)

are not vertices but correspond to some monomials of the discriminant of the equation under consideration.

For t5 = 1 for each integer solution of (14) from Z>0 there is no corresponding monomial participating in the discriminant of the polynomial of degree 6. As for the plane t5 = 2, here we use the fact that each point of this base is a translate from the base in t5 = 0 by the vector (0,0,0, —1,2). As a result we get the following points in the remaining base (in t5 = 2):

v12345 = (2, 2, 2, 2, 2), v2345 = (0, 3, 2, 2, 2), v^5 = (3, 0, 3, 2, 2),

v345 = (0, 0, 4, 2, 2), v1245 = (2, 3, 0, 3, 2), v245 = (0,4,0, 3, 2), v145 = (4, 0,0,4, 2), v45 = (0, 0, 0, 5, 2); (1,1, 3, 2, 2) (3,1,1, 3, 2), (1, 2,1, 3, 2), (2, 0, 2, 3, 2), (0,1, 2, 3, 2), (2,1, 0,4, 2), (0, 2,0,4, 2) , (1,0,1,4, 2). Then for A|g we get the following representation:

A|g4 = — (4a2a2a| a| — 16a|a3 — 16a3a|a4 — 108a3 a| — 16a2a| a4 + 64a|a4 —

— 108a4a4 + 1024a6) + (72a1a2a3a3 + 72afa2a3a4 — 320a1 a|a3a4 — 24a1a3a4+ +576a2a3a4 + 576a2 a2a5 — 512a2a5 — 768a1a3a5) — a5 ((—a!a2a^a^+

+4a3a|a2 + 4afa3a4 + 27a'^a24 + 4a2a2a3 — 16a^a4 + 27a4a4 — 256a4)+

O Q o O Q O Q Q O 0 Q

+( — 18o,10,20,3 a4 — 18ala2a3a4 + 800,10,20,3 a3 + 6a1a3a3 — 144020,3a3 — — 144a\a2a4 + 128a2a44 + 192a1 a3a44)),

which admits a factorization

*| 0/0 \ / 0 0 AO 0 00

A|g = —a4 (O,5 — 4O,4)(6O,10,30,4 + 27a1a4 — 18a1a2a3a4 + 80a1a2a3a4 — 144a1a2 0,4+

oo o o o 0 0 0 A

+4o,3o,3 — 18o1o2o,3 + 192a1a3a\ — 256o,4 — a1a^a^ — 16a,4a4+

+128a2a4 + 27a4 + 4a[o32a4 — 144a2a^a4 + 4a3 a2).

In the obtained product the factor a2 — 4a4 is the discriminant A2;4,5(a4,a5) of a quadratic polynomial

f2;4,5(y) = a4y2 + a5 y + 1, and the last factor is the discriminant A4;4,3i2i1 (a1, a2, a3, a4) of a polynomial of degree 4:

f4;4,3,2,1(y) = a4y4 + a3y3 + a2y2 + a1y + 1. Then for A|94 we get the following representation:

A^i = a2 A2;4i5(a4, a5 ) A4-t4t3t2,1(a1, a2, a3, a4).

3.2. Truncations of the discriminant with respect to the remaining facets

Now we turn again to the polynomial (11) of degree 5. Consider the truncation A| of its

gk

discriminant with respect to the remaining facets gk, i.e. g1 and g4. The facet g4 is given by the system

{4t1 + 3t2 + 2t3 + t4 < 20, 3t1 + 6t2 + 4t3 + 2t4 < 30, 2t1 + 4t2 + 6t3 + 3t4 < 30, t1 + 2t2 + 3t3 + 4t4 = 20.

Calculations show that among integer solutions of this system only the following points are exponents of monomials of the discriminant A

(1,1, 3, 2), (3,1,1, 3), (2, 2, 2, 2), (3,0, 3, 2), (2,0, 2, 3), (1,0,1,4),

(0,1, 2, 3), (2,1, 0, 4), (0, 3, 2, 2), (2, 3,0, 3), (1, 2,1, 3), (4,0,0, 4), (0,0, 0, 5), (0, 2,0, 4), (0,4,0, 3), (0,0,4, 2). The truncation of A with respect to 94 equals

*| 0 ^ 0 0 0 0 0 0 0 A

A^ = 18a1a2a3a4 + 18a1a2a3a4 + a1a2a3a4 — 4a1a3a4 — 6a1a3a4 — 192o,1 a3a4+

OQ OA Q O O OQQ OQ A A ^

+144o,2a:io,4 + 144a1a2a4 — 4O,20,30,4 — 4a1a2a4 — 80a1a2a3a4 — 27a4 O,4 + 256a4 —

OA A ^ AO 0 0 0 0

— 128O,20,4 + 16a2a4 — 27a3 O^ = 0,4(180,10,2a,3 + 18a2a2aзa4 + O^O^O,2 — 40,30,3 —

0 0 0 0 0 0 ^00^ 0 —6a2a^a4 — 192010301 + 144020304 + 1440,20,20,4 — 40203 — 4a2a.3a4 — 800,10,20,30,4 —

—27a\a\ + 256a3 — 128o\o\ + 16a2a4 — 27a4) = a\A4;4,3i2,1(a1, 02, 03, 04),

where A4.4i3,2i1(a1, a2, a3, a4) is the discriminant of a polynomial

f4;1,2,3(y) = 04y4 + 03y3 + 02y2 + 0,1 y + 1.

It is known that this discriminant is irreducible [4]. Since Ai = 1, we can write as desired:

A|fl4 = a2A4;4 , 3 , 2 ,i(ai,a2,a3,a4)Ai;4(a4).

At the end of the section we show factorability of V|g3 for the polynomial of degree 6. The facet g3 is given by the system

5t1 + 4t3 + 3t3 + 2t4 + t5 < 30, 4t1 + 8t3 + 6t3 + 4t4 + 2t5 < 48, 3t1 + 6t2 + 9t3 + 6t4 + 3t5 = 54, 2t1 + 4t2 + 6t3 + 8t4 + 4t5 < 48, t1 + 2t2 + 3t3 + 4t4 + 5t5 < 30.

Its solution consists of 25 points

(1,1, 3, 3,0), (1,1, 4,0, 3), (2, 2, 2, 3,0), (2, 2, 3,0, 3), (0, 3, 3,1,1),

(0, 3, 2, 2, 2), (3, 0, 4,1,1), (3, 0, 3, 2, 2), (0, 0,4, 3, 0), (0,0, 5,0, 3),

(0, 3, 4, 0,0), (3, 0, 5,0,0), (2, 2, 3,1,1), (1,1, 3, 2, 2), (0,0, 5,1,1),

(0,0, 4, 2, 2), (1,1, 5,0,0), (2, 2,4,0,0), (0, 3, 2, 3, 0), (0, 3, 3,0, 3),

(3,0, 3, 3,0), (3, 0, 4,0, 3), (0, 0, 6,0,0), (1,1,4,1,1), (2, 2, 2, 2, 2). Accordingly, the truncation V|g3 is

I QQ /1 Q OOOQ OOQQ QQ

V|g = 72aia2 a+ 72ai a2^a5 + + 0,30,5 + 72a2 030405+

+4a20,2G,4 0,5 —+720^^405 + 40103 a^3 — 108a4a3 — 108a3a3 — 108a,a4—

O c O O Q QOO ^ /10 0

— 108a2a5 — 18a1a203a4a5 — 180102030405 +486030405 + 27040305+

E* OO/l QOQ Q Q Q QQQ

+486aia2a3 + 27a,la,2 a4 — 16a2aз a3 — 16a2aзa5 — 16a2aзa4 —

Q/1Q P. A OOOOO

— 16a10403 — 729a3 — 3240102030405 — a1a2a3a405. Calculations show that this expression admits the following factorization

0/0 000 —a,з(27a,з — 180102 0,3 — ajO + 4a3 + 4a1a3) x

/0 0*3 0 \

x (27 03 — 18030405 — 90305 + 4a3 — 30405 — a,5).

Thus, we get |

V|g3 = —a3A3;3 , 3 , 1 A3;3 ,4,5,

where A3;3 , 3 ,1(a1, a3, a3), A3;3 , 4 ,5(a3,a4,a5) are the discriminants of polynomials

f3;3,3,1 (y) = 03y3 + 03y2 + 01y + 1, f3;3,4,5(y) = 03y3 + 04y3 + 05y + 1.

The first author was supported by the RFBR grant 14-01-31265 mol-a, and the second by RFBR grant 14-01-00544-a.

References

[1] I.A.Antipova, A.K.Tsikh. The discriminant locus of a sistem of n Laurent polynomials in n variables, Izv. Math., 76(2012), no. 5, 881-906.

[2] E.N.Mikhalkin, A.K.Tsikh. Singular strata of cuspidal type for the classical discriminant, Sb. Math., 206(2015), no. 2, 282-310.

[3] I.Gelfand, M.Kapranov, A.Zelevinsky, Discriminants, resultants and multidimensional determinants, Birkhauser, Boston, 1994.

[4] M.Passare, A.Tsikh, Algebraic equations and hypergeometric series, The legacy of Niels Henrik Abel, Springer, Berlin-Heidelberg-New York, 2004, 653-672.

О структуре классического дискриминанта

Евгений Н. Михалкин Август К. Цих

Рассматривается полином степени п с переменными коэффициентами. Известно, что многогранник Ньютона дискриминанта такого многочлена комбинаторно эквивалентен n — 1-мерному кубу. В статье показано, что две гиперграни многогранника Ньютона являются призмами, а срезки дискриминанта на грани факторизуются на дискриминанты полиномов степени меньше п.

Ключевые слова: общее алгебраическое уравнение, дискриминант, многогранник Ньютона.