COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

326 W. D. Maurer Galois theory 327
roots of the polynomial (which are in 2). For each permutation s in the
symmetric group S,, consider it as a permutation of the variables ui and
form the transformed expression se. (For example, if s = (1 2), then s0 =
~luz+azul+cc3u3+er4u4+... +a,~,.) Finally, form the product F of all
the expressions z-s0 for all s in the symmetric group S,,. Now F is a symmetric
function of the CL~, and hence can be expressed in terms of the elementary
symmetric functions of the ai. These are precisely the coefficients
off, and in fact lie in d, so that F is actually in the smaller ring d(ur, . . . ,
u,, z). Decompose F into irreducible factors FI, . . . , F,, in this ring, and
apply the permutations s as above to the resulting equation
F = FI...:F,,
Now: For an arbitrary factor (say FI), those permutations which carry
this factor into itself form a group which is isomorphic to the Galois group
of the given equation.
It is clear that this is a finite method if the associated factorization is a
finite method, and this is shown in [l], vol. 1, p. 77. On the other hand,
van der Waerden’s book first appeared in 1931, a long time before the
first computers, and he pays no attention to considerations of speed.
Some older mathematical algorithms, such as the Todd-Coxeter algorithm,
adapt very well to computers, but it is clear that this is not one of them;
even for a polynomial of degree 4, twenty-four polynomials must be
multiplied and the result decomposed into irreducible factors in five
variables.
Simpler methods are given by van der Waerden in the case in which
the polynomial has degree less than or equal to 4. These methods have been
improved on by Jacobson [2], vol. 3, pp. 94-95. We note first that if the
given polynomial has a linear factor, we may divide by that factor to
obtain a new polynomial with the same Galois group. After all linear
factors have been removed, a polynomial of degree 4 or less is either
irreducible or is a quartic polynomial with two quadratic factors. The
Galois group in this case is the direct sum of two cyclic groups of order 2
unless both polynomials have the same discriminant or unless one discriminant
divides the other and the quotient is a square. Quadratic factors of a
polynomial may easily be found by Kronecker’s method (cf. [l], vol. 1,
p. 77). Therefore we are reduced to the case in which the polynomial is
irreducible. If it is linear, the group is of order 1. If it is quadratic, the
group is of order 2. If it is cubic, the group is of order 3 if the discriminant
is a square, and is otherwise the symmetric group on three letters (of
orderi6). There finally remains the case of an irreducible quartic, and here
Jacobson’s algorithm is as follows:
(1) Calculate the resolvent cubic of the equation. This may be done
pirectly from the coefficients: if the equation is x4-aIx3$a2x2- a3x fad,
then the resolvent cubic is x3- blx2+ bsx- b3, where bI = a2, b2 = a1a3- 4a4,
and 63 = ala4+ai-2aza4.
(2) Calculate the Galois group of the resolvent cubic.
(3) The Galois group of the original equation may now be derived from
the following table :
If the Galois group of the
resolvent cubic is
the identity
a cyclic group of order 2
then the Galois group of the original
equation is
the Klein four-group
a cyclic group of order 4
or
a dihedral group of order 8
the alternating group A3 the alternating group A4
(cyclic of order 3) (of order 12)
the symmetric group Sa the symmetric group S4
(of order 6) (of order 24)
where there is only one ambiguity-the case in which the Galois group
of the resolvent cubic is of order 2. In this case, the Galois group of the
original equation is the cyclic group of order 4, if and only if it is not
irreducible over the field obtained by adjoining the square root of the
discriminant of the resolvent cubic.
This method easily lends itself to calculation. The only apparent difficulty
is in the last step, and this is easily resolved by Kronecker’s method
applied at two levels.
It is of some interest to note that heuristic methods may be used even
in a procedure as obviously combinatorial and manipulative as the one
described above. The heuristics are, however, not of the usual kind. Most
heuristic programs try various approaches, some of which are expected to
fail. In Galois theory, however, we find ourselves faced more than once
with the following situation : A program X solves a problem exhaustively.
A program Y may be run which decreases the number of cases that X must
treat, However, the program Y may take so long to run that no timing
advantage is conferred by running it. Therefore, an estimate of the running
time of X, of the improved X after running Y, and of Y is made, and a
decision made on this basis as to whether to run Y. The result is that, for
different input data, the program will perform the calculations in different
ways, attempting to choose the fastest way as it goes (for the given data).
An example of this occurs in irreducibility test routines. The program
X is the Kronecker’s method program. The program Y finds the factors,
if any, modulo some integer. The number of steps in Kronecker’s method
is the product nln2. . . . ‘IQ, where each ni is twice the number of factors
of some integral polynomial value (including itself and 1). The irreducibility
test modulo p, for prime p, involves checking all the possible factors
CFA 22