Unknown

4.3. Solving Special Polynomial Equations 135

similar way, eliminate (132), (142), (1342), (1432), (13), (134), (234), (14),

(23), (24) (143) (243). Th us, G consists of the permutations

We enlarge the base field (the smallest containing the coefficients) by

adjoining a radical to Q: the hope is that by adjoining enough radicals, we

will eventually obtain a large enough field to contain not only the coeffi-

cients, but also the roots as well. In the present example, we can achieve

this by adjoining first the radical i = fl, and then the radical 2114.

The smallest field Q( i ) containing Q and the radical i is

{a + bi : a, b E Ql.

Just as we did for Q, we can now define the group of our polynomial over
Q(i)ath e se t o f a 11 permutations which preserve the validity of polynomial
equations of the roots whose coefficients are allowed to lie in Q(i). The
family of admissible equations will now be larger than before; for example,
it will include

(k) (tl - t2)(t3 - t4)3 = -32i

(1) (tl - t2)3(t3 - t4) = 32i.

Accordingly, the group over Q(i) will be smaller; the equations (k) and
(1) can be used to argue that it will not contain (12), (34), (13)(24) and
(14)(23). However, it turns out that each of the permutations c, (1423),
(W(W, (1324) converts the two equations above as well as every other
valid equation over Q(i) into a valid equation.
The field Q(i) is a splitting field for the polynomial t2 + 1, being the
smallest field that contains both the coefficients and the roots of t2 + 1,
or alternatively, being the smallest field in which the polynomial can be
“split” into linear factors. Show that the group of t2 + 1 over Q consists
of the two permutations of its zeros, namely the identity permutation and
the permutation which interchanges them. Call this group G1. This group
can be tied in with the groups of the original polynomial over Q and Q(i)
in the following way. Partition the group G into two subsets:

H1 = {E., (1423), (12)(34), (1324))

Hz = ((1% (34)s (13X24), (I4)(23)).

Observe that HI is the group over Q(i). Verify that, if each of the permu-
tations in Hr is applied to (tl - t2)3(t3 - t4)/32, we obtain a polynomial
in the ti equal to i. However, verify also that each of the permutations in
Hs gives a polynomial equal to -i. This actually holds for other polyno-
mials in the roots which are equal to i as well. Thus, we can think of the
sets HI and H2 as corresponding to the two elements of Gr, and we can