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9.7.8. Each of the polynomials in the previous exercise is irreducible over the rationals. The Eisenstein criterion applies to one of the polynomials, and the others can be checked by computing factorizations of the reductions modulo small primes. For each polynomial, determine which Galois groups are consistent with the computation of the discriminant.
9.7.9. Suppose that f and g are polynomials in KŒx of degrees n and m respectively, and that there exist '.x/ of degree no more than n 1 and .x/ of degree no more than m 1 such that f .x/ .x/ D g.x/'.x/.
Show that f and g have a nonconstant common divisor in KŒx.
9.7.10. Let f and g be polynomials of degree n and m, respectively, with roots i and j .
(a) Show that R.f; g/ D . 1/nCmR.g; f /.
(b) Show that
Y R.f; g/ D am
n
g.i /:
i
(c) Show that
Y R.f; g/ D . 1/nCmbnm
f .j /:
j
9.7.11. Show that
n
m
Y Y .xi yj / i D1 j D1 is a polynomial of total degree m in the elementary symmetric functions i .x1; : : : ; xn/ and of total degree n in the elementary symmetric functions j .y1; : : : ; ym/.
9.7.12. Verify the assertion in the text that det.R.f; g// is divisible by
Q
Q
i
j .i j / D R.f; g/.
9.7.13. Use a computer algebra package (e.g., Maple or Mathematica to find the resultants: (a) R.x3 C 2x C 5; x2 4x C 5/ (b) R.3x4 C 7x3 C 2x 5; 12x3 C 21x2 C 2/ 9.8. Quartic Polynomials
In this section, we determine the Galois groups of quartic polynomials.
Consider a quartic polynomial f .x/ D x4 C ax3 C bx2 C cx C d
(9.8.1)
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9. FIELD EXTENSIONS – SECOND LOOK with coefficients in some field K. It is convenient first to eliminate the cubic term by the linear change of variables x D y a=4. This yields f .x/ D g.y/ D y4 C px2 C qx C r;
(9.8.2) with
3a2
p D
C b;
8
a3
ab
q D
C c;
(9.8.3)
8
2
3a4
a2b
ac
r D
C
C d:
256
16
4
We can suppose, without loss of generality, that g is irreducible, since otherwise we could analyze g by analyzing its factors. We also suppose that g is separable. Let G denote the Galois group of g over K.
Using one of the techniques for computing the discriminant from the previous section, we compute that the discriminant of g (or f ) is 4 p3 q2 27 q4 C 16 p4 r C 144 p q2 r 128 p2 r2 C 256 r3; (9.8.4) or, in terms of a, b, c, and d , a2 b2 c2 4 b3 c2 4 a3 c3 C 18 a b c3 27 c4 4 a2 b3 d C 16 b4 d C 18 a3 b c d 80 a b2 c d 6 a2 c2 d C 144 b c2 d
(9.8.5)
27 a4 d 2 C 144 a2 b d 2 128 b2 d 2 192 a c d 2 C 256 d 3:
We can distinguish two cases, according to whether ı.g/ 2 K: Case 1. ı.g/ 2 K. Then the Galois group is isomorphic to A4 or V Š Z2 Z2, since these are the transitive subgroups of A4, according to Exercise 5.1.20.
Case 2. ı.g/ 62 K. Then the Galois group is isomorphic to S4, D4, or Z4, since these are the transitive subgroups of S4 that are not contained in A4, according to Exercise 5.1.20.
Denote the roots of g by ˛1; ˛2; ˛3; ˛4, and let E denote the splitting field K.˛1; : : : ; ˛4/. The next idea in analyzing the quartic equation is to introduce the elements 1 D .˛1 C ˛2/.˛3 C ˛4/ 2 D .˛1 C ˛3/.˛2 C ˛4/
(9.8.6)
3 D .˛1 C ˛4/.˛2 C ˛3/; and the cubic polynomial, called the resolvent cubic:
h.y/ D .y
1/.y
2/.y
3/:
(9.8.7)
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Expanding h.y/ and identifying symmetric functions in the ˛i with coefficients of g gives h.y/ D y3
2py2 C .p2
4r/y C q2:
(9.8.8)
The discriminant ı2.h/ turns out to be identical with the discriminant ı2.g/ (Exercise 9.8.4). We distinguish cases according to whether the resolvent cubic is irreducible.
Case 1A. ı.g/ 2 K and h is irreducible over K. In this case, ŒK.1; 2; 3/ W K D 6; and 6 divides the order of G. The only possibility is G D A4.
Case 2A. ı.g/ 62 K and h is irreducible over K. Again, 6 divides the order of G. The only possibility is G D S4.
Case 1B. ı.g/ 2 K and h is not irreducible over K. Then G must be V.
(In particular, 3 does not divide the order of G, so the Galois group of h must be trivial, and h factors into linear factors over K. Conversely, if h splits over K, then the i are all fixed by G, which implies that G V.
The only possibility is then that G D V.)
Case 2B. ı.g/ 62 K and h is not irreducible over K. The Galois group must be one of D4 and Z4. By the remark under Case 1B, h does not split over K, so it must factor into a linear factor and an irreducible quadratic (i.e., exactly one of the i lies in K).
We can assume without loss of generality that 1 2 K. Then G is contained in the stabilizer of 1, which is the copy of D4 generated by f.12/.34/; .13/.24/g, D4 D fe; .1324/; .12/.34/; .1423/; .34/; .12/; .13/.24/; .14/.23/g: G is either equal to D4 or to Z4 D fe; .1324/; .12/.34/; .1423/g:
It remains to distinguish between these cases.
Since K.ı/ K.1; 2; 3/ and both are quadratic over K, it follows that K.ı/ D K.1; 2; 3/. K.ı/ is the fixed field of G \ A4, which is either D4 \ A4 D fe; .12/.34/; .13/.24/; .14/.23/g Š V;
or Z4 \ A4 D fe; .12/.34/g Š Z2:
It follows that the degree of the splitting field E over the intermediate field K.ı/ D K.1; 2; 3/ is either 4 or 2, according to whether G is D4 or Z4.
So one possibility for finishing the analysis of the Galois group in Case 2B
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9. FIELD EXTENSIONS – SECOND LOOK is to compute the dimension of E over K.ı/. The following lemma can be used for this.2 Lemma 9.8.1. Let g.x/ D y4 C py2 C qy C r be an irreducible quartic polynomial over a field K. Let h.x/ denote the resolvent cubic of g. Suppose that ı2 D ı2.g/ is not a square in K and that h is not irreducible over K. Let denote the one root of the resolvent cubic which lies in K, and define H.x/ D .x2 C /.x2 C .
p/x C r/:
(9.8.9)
Then the Galois group of g is Z4 if, and only if, H.x/ splits over K.ı/.
Proof. We maintain the notation from the preceding discussion: The roots of g are denoted by ˛i , the splitting field of g by E, and the roots of h by i . Let L D K.ı/ D K.1; 1; 3/. Assume without loss of generality that the one root of h in K is 1. Now consider the polynomial .x ˛1˛2/.x
˛3˛4/.x
.˛1 C ˛2//.x .˛3 C ˛4/
(9.8.10)
D .x2 .˛1˛2 C ˛3˛4/x C r/.x2 C 1/: Compute that p D ˛1˛2 C ˛3˛4. It follows that the preceding poly nomial is none other than H.x/.
Suppose that H.x/ splits over L, so ˛1˛2, ˛3˛4, ˛1C˛2, ˛3C˛4 2 L.
It follows that ˛1 satisfies a quadratic polynomial over L,
.x
˛1/.x
˛2/ D x2 .˛1 C ˛2/x C ˛1˛2 2 LŒx; and ŒL.˛1/ W L D 2. But we can check that L.˛1/ D E. Consequently, ŒE W L D 2 and G D Z4, by the discussion preceding the lemma.
Conversely, suppose that the Galois group G is Z4. Because 1 D .˛1 C ˛2/.˛3 C ˛4/ is in K and is, therefore, fixed by the generator of G, we have ˙1 D .1324/. The fixed field of 2 D .12/.34/ is the unique intermediate field between K and E, so L equals this fixed field. Each of the roots of H.x/ is fixed by 2 and, hence, an element of L. That is, H splits over L.
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Examples of the use of this lemma will be given shortly. We shall now explain how to solve explicitly for the roots ˛i of the quartic equation in terms of the roots i of the resolvent cubic.
Note that .˛1 C ˛2/.˛3 C ˛4/ D 1 and .˛1 C ˛2/ C .˛3 C ˛4/ D 0; (9.8.11) 2 This result is taken from L. Kappe and B. Warren, “An Elementary Test for the Galois Group of a Quartic Polynomial,”The American Mathematical Monthly 96 (1989) 133-137.
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which means that .˛1 C˛2/ and .˛3 C˛4/ are the two square roots of 1.
Similarly, .˛1 C ˛3/ and .˛2 C ˛4/ are the two square roots of 2, and .˛1 C ˛4/ and .˛2 C ˛3/ are the two square roots of 3. It is possible to choose the signs of the square roots consistently, noting that
q
p
p
p
p
1
2
2 D
123 D q2 D q:
(9.8.12)
That is, it is possible to choose the square roots so that their product is q.
We can check that .˛1 C ˛2/.˛1 C ˛3/.˛1 C ˛4/ D q;
(9.8.13)
and so put
p
p
p
.˛1C˛2/ D
1
.˛1C˛3/ D
2
.˛1C˛4/ D 3: (9.8.14) Using this together with ˛1 C ˛2 C ˛3 C ˛4 D 0, we get
p
p
p
2˛i D ˙
1 ˙
2 ˙ 3;
(9.8.15)
with the four choices of signs giving the four roots ˛i (as long as the characteristic of the ground field is not 2).
Example 9.8.2. Take K D Q and f .x/ D x4 C 3x3
3x
2. Applying the linear change of variables x D y 3=4 gives f .x/ D g.y/ D
179
256 C 3 y
27 y2
8
8
C y4. The reduction of f modulo 5 is irreducible over Z5, and, therefore, f is irreducible over Q. The discriminant of f (or g or h) is 2183, which is not a square in Q. Therefore, the Galois group of f is not contained in the alternating group A4. The resolvent cubic of g is h.y/ D 9
y
y2
64 C 227
16
C 27
4
C y3. The reduction of 64h modulo 7 is irreducible over Z7 and, hence, h is irreducible over Q. It follows that the Galois group is S4.
Example 9.8.3. Take K D Q and f .x/ D 21 C 12 x C 6 x2 C 4 x3 C x4.
Applying the linear change of variables x D y 1 gives f .x/ D g.y/ D 12 C 8 y C y4. The reduction of g modulo 5 factors as Ng.y/ D y4 C 3y C 2 D .1 C y/ 2 C y C 4 y2 C y3 :
By the rational root test, g has no rational root; therefore, if it is not irreducible, it must factor into two irreducible quadratics. But this would be inconsistent with the factorization of the reduction modulo 5. Hence, g is irreducible.
The discriminant of g is 331776 D .576/2, and therefore, the Galois group of g is contained in the alternating group A4. The resolvent cubic of g is h.y/ D 64 48 y C y3. The reduction of h modulo 5 is irreducible over Z5, so h is irreducible over Q. It follows that the Galois group of g is A4.
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9. FIELD EXTENSIONS – SECOND LOOK Example 9.8.4. Take K D Q and f .x/ D x4 C 16x2 1. The reduction of f modulo 3 is irreducible over Z3, so f is irreducible over Q. The discriminant of f is 1081600, so the Galois group G of f is not contained in the alternating group. The resolvent cubic of f is h.x/ D 260 x 32 x2 C x3 D x 260 32 x C x2 :
Because h is reducible, it follows that G is either Z4 or D4. To determine which, we can use the criterion of Lemma 9.8.1.
p
p
p
Note first that ı D 1081600 D 1040 1, so Q.ı/ D Q.
1/.
The root of the resolvent cubic in Q is zero, so the polynomial H.x/ of
p
Lemma 9.8.1 is x2.x2
16x
1/. The nonzero roots of this are 8 ˙ 65, so H does not split over Q.ı/. Therefore, the Galois group is D4.
The following lemma also implies that the Galois group is D4 rather than Z4: Lemma 9.8.5. Let K be a subfield of R and let f be an irreducible quartic polynomial over K whose discriminant is negative. Then the Galois group of f over K is not Z4.
Proof. Complex conjugation, which we denote here by , is an automorphism of C that leaves invariant the coefficients of f . The splitting field E of f can be taken to be a subfield of C, and induces a K-automorphism of E, since E is Galois over K.
The square root ı of the discriminant ı2 is always contained in the splitting field. As ı2 is assumed to be negative, ı is pure imaginary. In particular, E is not contained in the reals, and the restriction of to E has order 2.
Suppose that the Galois group G of f is cyclic of order 4. Then G contains a unique subgroup of index 2, which must be the subgroup generated by . By by the Galois correspondence, there is a unique intermediate field K L E of dimension 2 over K, which is the fixed field of .
But K.ı/ is a quadratic extension of K which is not fixed pointwise by .
This is a contradiction.
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Example 9.8.6. Take the ground field to be Q and f .x/ D x4 C 5x C 5.
The irreducibility of f follows from the Eisenstein criterion. The discriminant of f is 15125, which is not a square in Q; therefore, the Galois group G of f is not contained in the alternating group. The resolvent cubic of f is 25 20 x C x3 D .5 C x/ 5 5 x C x2. Because h is reducible, it follows that the G is either Z4 or D4, but this time the discriminant is positive, so the criterion of Lemma 9.8.5 fails.
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p
Note that the splitting field of the resolvent cubic is Q.ı/ D Q. 5/,
p
p since ı D 55 5. The one root of the resolvent cubic in Q is 5. There fore, the polynomial H.x/ discussed in Lemma 9.8.1 is H.x/ D 5 C x2 5 5 x C x2 :
p
Because H.x/ splits over Q. 5/, the Galois group is Z4.
Example 9.8.7. Set f .x/ D x4 C 6x2 C 4. This is a so-called biquadratic
p
polynomial, whose roots are ˙ ˛; ˙pˇ, where ˛ and ˇ are the roots of the quadratic polynomial x2 C 6x C 4. Because the quadratic polynomial is irreducible over the rationals, so is the quartic f . The discriminant of f is 25600 D 1602, so the Galois group is contained in the alternating group.
But the resolvent cubic 20 x 12 x2 C x3 D . 10 C x/ . 2 C x/ x is reducible over Q, so the Galois group is V .
A Mathematica notebook Quartic.nb for investigation of Galois groups of quartic polynomials can be found on my Web site. This notebook can be used to verify the computations in the examples as well as to help with the Exercises.
Exercises 9.8 9.8.1. Verify Equation (9.8.8).
9.8.2. Show that a linear change of variables y D x C c does not alter the discriminant of a polynomial.
9.8.3. Show how to modify the treatment in this section to deal with a nonmonic quartic polynomial.
9.8.4. Show that the resolvent cubic h of an irreducible quartic polynomial f has the same discriminant as f , ı2.h/ D ı2.f /.
9.8.5. Let f .x/ be an irreducible quartic polynomial over a field K, and let ı2 denote the discriminant of f . Show that the splitting field of the resolvent cubic of f equals K.ı/ if, and only if, the resolvent cubic is not irreducible over K.
9.8.6. Show that x4 C 3x C 3 is irreducible over Q, and determine the Galois group.
9.8.7. For p a prime other than 3 and 5, show that x4 C px C p is irreducible with Galois group S4. The case p D 3 is treated in the previous exercise, and the case p D 5 is treated in Example 9.8.6.
9.8.8. Find the Galois group of f .x/ D x4 C 5x2 C 3. Compare Example 9.8.7.
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9. FIELD EXTENSIONS – SECOND LOOK 9.8.9. Show that the biquadratic f .x/ D x4 C px2 C r has Galois group V precisely when r is a square in the ground field K. Show that in general
p K.ı/ D K. r/.
9.8.10. Find examples of the biquadratic f .x/ D x4 C px2 C r for which the Galois group is Z4 and examples for which the Galois group is D4.
Find conditions for the Galois group to be cyclic and for the Galois group to be the dihedral group.
9.8.11. Determine the Galois group over Q for each of the following polynomials. You may need to do computer aided calculations, for example, using the Mathematica notebook Quartic.nb on my Web site.
(a) 21 C 6 x 17 x2 C 3 x3 C 21 x4 (b)
1
12 x C 36 x2 19 x4
(c) 33 C 16 x 39 x2 C 26 x3 9 x4
(d)
17
50 x
43 x2
2 x3
33 x4
(e)
8
18 x2
49 x4
(f) 40 C 48 x C 44 x2 C 12 x3 C x4 (g) 82 C 59 x C 24 x2 C 3 x3 C x4 9.9. Galois Groups of Higher Degree Polynomials
In this section, we shall discuss the computation of Galois groups of polynomials in QŒx of degree 5 or more.
If a polynomial f .x/ 2 KŒx of degree n has irreducible factors of degrees m1 m2 mr , where Pi mi D n, we say that .m1; m2; : : : ; mr / is the degree partition of f . Let f .x/ 2 ZŒx and let
Q f .x/ 2 ZpŒx be the reduction of f modulo a prime p; if Q f has degree partition ˛, we say that f has degree partition ˛ modulo p.
The following theorem is fundamental for the computation of Galois groups over Q.
Theorem 9.9.1. Let f be an irreducible polynomial of degree n with coefficients in Z. Let p be a prime that does not divide the leading coefficient of f nor the discriminant of f . Suppose that f has degree partition ˛ modulo p. Then the Galois group of f contains a permutation of cycle type ˛.
If f is an irreducible polynomial of degree n whose Galois group does not contain an n-cycle, then the reduction of f modulo a prime p is never irreducible of degree n. If the prime p does not divide the leading coefficient of f or the discriminant, then the reduction of f modulo p factors, according to the theorem. If the prime divides the leading coefficient, then
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the reduction has degree less than n. Finally, if the prime divides the discriminant, then the discriminant of the reduction is zero; but an irreducible polynomial over Zp can never have multiple roots and, therefore, cannot have zero discriminant.
We shall not prove Theorem 9.9.1 here. You can find a proof in B. L. van der Waerden, Algebra, Volume I, Frederick Ungar Publishing Co., 1970, Section 8.10 (translation of the 7th German edition, Springer– Verlag, 1966).
This theorem, together with the computation of the discriminant, often suffices to determine the Galois group.
Example 9.9.2. Consider f .x/ D x5 C 5x4 C 3x C 2 over the ground field Q. The discriminant of f is 4557333, which is not a square in Q and which has prime factors 3, 11, and 138101. The reduction of f modulo 7 is irreducible, and the reduction modulo 41 has degree partition .2; 1; 1; 1/.
It follows that f is irreducible over Q and that its Galois group over Q contains a 5-cycle and a 2-cycle. But a 5-cycle and a 2-cycle generate S5, so the Galois group of f is S5.
Example 9.9.3. Consider f .x/ D 4 4 x C 9 x3 5 x4 C x5 over the ground field Q. The discriminant of f is 15649936 D 39562, so the Galois group G of f is contained in the alternating group A5. The prime factors of the discriminant are 2, 23, and 43. The reduction of f modulo 3 is irreducible and the reduction modulo 5 has degree partition .3; 1; 1/.
Therefore, f is irreducible over Q, and its Galois group over Q contains a 5-cycle and a 3-cycle. But a 5-cycle and a 3-cycle generate the alternating group A5, so the Galois group is A5.
The situation is quite a bit more difficult if the Galois group is not Sn or An. In this case, we have to show that certain cycle types do not appear.
The rest of this section is devoted to a (by no means definitive) discussion of this question.
Example 9.9.4. Consider f .x/ D 3 C 7 x C 9 x2 C 8 x3 C 3 x4 C x5 over the ground field Q. The discriminant of f is 1306449 D 11432, so the Galois group G of f is contained in the alternating group A5. The prime factors of the discriminant are 3 and 127. The transitive subgroups of the alternating group A5 are A5, D5, and Z5, according to Exercise 5.1.20. The reduction of f modulo 2 is irreducible, and the reduction modulo 5 has degree partition .2; 2; 1/. Therefore, G contains a 5-cycle and an element of cycle type .2; 2; 1/. This eliminates Z5 as a possibility for the Galois group.
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9. FIELD EXTENSIONS – SECOND LOOK
If we compute the factorization of the reduction of f modulo a num ber of primes, we find no instances of factorizations with degree partition .3; 1; 1/. In fact, for the first 1000 primes, the frequencies of degree partitions modulo p are
15
2; 14 22; 1 3; 12 3; 2 4; 1
5
:093
0
:5
0
0
0
:4
These data certainly suggest strongly that the Galois group has no 3 cycles and, therefore, must be D5 rather than A5, but the empirical evidence does not yet constitute a proof! I would now like to present some facts that nearly, but not quite, constitute a method for determining that the Galois group is D5 rather than A5.
Let us compare our frequency data with the frequencies of various cycle types in the transitive subgroups of S5. The first table shows the number of elements of each cycle type in the various transitive subgroups of S5, and the second table displays the frequencies of the cycle types, that is, the number of elements of a given cycle type divided by the order of the group.
15
2; 14 22; 1 3; 12 3; 2 4; 1
5
Z5
1
0
0
0
0
0
4
D5
1
0
5
0
0
0
4
A5
1
0
15
20
0
0
24
Z4 Ë Z5
1
0
5
0
0
10
4
S5
1
10
15
20
0
30
24
Numbers of elements of various cycle types
15
2; 14 22; 1 3; 12 3; 2 4; 1
5
Z5
:2
0
0
0
0
0
:8
D5
:1
0
:5
0
0
0
:4
A5
:017
0
:25
:333
0
0
:4
Z4 Ë Z5
:05
0
:25
0
0
:5
:2
S5
:0083
:083
:125
:167
0
:25
:2
Frequencies of various cycle types Notice that our empirical data for frequencies of degree partitions mod ulo p for f .x/ D 3 C 7 x C 9 x2 C 8 x3 C 3 x4 C x5 are remarkably close to the frequencies of cycle types for the group D5. Let’s go back to our previous example, the polynomial f .x/ D 4 4 x C 9 x3 5 x4 C x5,
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whose Galois group is known to be A5. The frequencies of degree partitions modulo the first 1000 primes are
15
2; 14 22; 13 3; 12 3; 2 4; 1
5
:019
0
:25
:32
0
0
:41
These data are, again, remarkably close to the data for the distribution of cycle types in the group A5. The following theorem (conjectured by Frobenius, and proved by Chebotarev in 1926) asserts that the frequencies of degree partitions modulo primes inevitably approximate the frequencies of cycle types in the Galois group: Theorem 9.9.5. (Chebotarev). Let f 2 ZŒx be an irreducible polynomial of degree n, and let G denote the Galois group of f over Q. For each partition ˛ of n let d˛ be the fraction of elements of G of cycle type ˛. For each N 2 N, let d˛;N be the fraction of primes p in the interval Œ1; N such that f has degree partition ˛ modulo p. Then lim d˛;N D d˛: N !1
Because the distribution of degree partitions modulo primes of the polynomial f .x/ D 3 C 7 x C 9 x2 C 8 x3 C 3 x4 C x5 for the first 1000 primes is quite close to the distribution of cycle types for the group D5 and quite far from the distribution of cycle types for A5, our belief that the Galois group of f is D5 is encouraged, but still not rigorously confirmed by this theorem. The difficulty is that we do not know for certain that, if we examine yet more primes, the distribution of degree partitions will not shift toward the distribution of cycle types for A5.
What we would need in order to turn these observations into a method is a practical error estimate in Chebotarev’s theorem. In fact, a number of error estimates were published in the 1970s and 1980s, but I have not able to find any description in the literature of a practical method based on these estimates.
The situation is annoying but intriguing. Empirically, the distribution of degree partitions converge rapidly to the distribution of cycle types for the Galois group, so that one can usually identify the Galois group by this method, without having a proof that it is in fact the Galois group. (By the way, there exist pairs of nonisomorphic transitive subgroups of Sn for n 12 that have the same distribution of cycle types, so that frequency of degree partitions cannot always determine the Galois group of a polynomial.)
A practical method of computing Galois groups of polynomials of small degree is described in L. Soicher and J. McKay, “Computing Galois
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9. FIELD EXTENSIONS – SECOND LOOK Groups over the Rationals,” Journal of Number Theory, vol. 20 (1985) pp. 273–281. The method is based on the computation and factorization of certain resolvent polynomials. This method has two great advantages:
First, it works, and second, it is based on mathematics that you now know.
I am not going to describe the method in full here, however. This method, or a similar one, has been implemented in the computer algebra package Maple; the command galois( ) in Maple will compute Galois groups for polynomials of degree no more than 7.
Both the nonmethod of Chebotarev and the method of Soicher and McKay are based on having at hand a catalog of potential Galois groups (i.e., transitive subgroups of Sn) together with certain identifying data for these groups. Transitive subgroups of Sn have been cataloged at least for n 11; see G. Butler and J. McKay, “The Transitive Subgroups of Degree up to 11,” Communications in Algebra, vol. 11 (1983), pp. 863–911.
By the way, if you write down a polynomial with integer coefficients at random, the polynomial will probably be irreducible, will probably have Galois group Sn, and you will probably be able to show that the Galois group is Sn by examining the degree partition modulo p for only a few primes. In fact, just writing down polynomials at random, you will have a hard time finding one whose Galois group is not Sn. Apparently, not all that much is known about the probability distribution of various Galois groups, aside from the predominant occurrence of the symmetric group.
A major unsolved problem is the so-called inverse Galois problem: Which groups can occur as Galois groups of polynomials over the rational numbers? A great deal is known about this problem; for example, it is known that all solvable groups occur as Galois groups over Q. (See Chapter 10 for a discussion of solvability.) However, the definitive solution to the problem is still out of reach.
Exercises 9.9 Find the probable Galois groups for each of the following quintic polynomials, by examining reductions of the polynomials modulo primes. In Mathematica, the command Factor[f, Modulus ! p] will compute the factorization of the reduction of a polynomial f modulo a prime p. Using this, you can do a certain amount of computation “by hand.” By writing a simple loop in Mathematica, you can examine the degree partition modulo p for the first N primes, say for N D 100. This will already give you a good idea of the Galois group in most cases. You can find a program for the computation of frequencies of degree partitions on my Web site; consult the notebook Galois-Groups.nb.
9.9.1. x5 C 2
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9.9. GALOIS GROUPS OF HIGHER DEGREE POLYNOMIALS
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9.9.2. x5 C 20x C 16 9.9.3. x5
5x C 12
9.9.4. x5 C x4
4x3
3x2 C 3x C 1 9.9.5. x5 x C 1
9.9.6. Write down a few random polynomials of various degrees, and show that the Galois group is Sn.
9.9.7. Project: Read the article of Soicher and McKay. Write a program for determining the Galois group of polynomials over Q of degree 5 based on the method of Soicher and McKay.
9.9.8. Project: Investigate the probability distribution of Galois groups for polynomials of degree 4 over the integers. If you have done the previous exercise, extend your investigation to polynomials of degree 5.
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