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4.3. Solving Special Polynomial Equations 133

The polynomial t4 - 4t3 + 6t2 - 4t - 1 = (t - 1)4 - 2 is irreducible

over Q, and therefore cannot be factored as a product of polynomials of

strictly lower. degree with rational coefficients. Consequently, there are no

linear, quadratic or cubic polynomials over Q with some of the ti as zeros.

Furthermore, any polynomial over Q of degree at least 4 which has any ti

as a zero must be divisible by t4 - 4t3 + 6t2 - 4t - 1 and therefore have all

the ti among its zeros. Accordingly, there is no polynomial in one variable

over Q which is satisfied by some but not all of the ti.

Let us take the ti a pair at a time. Verify that tl + t2 and t3 + t4 are

both equal to 2 while tl + t3, tl + t4, t2 + t3, t2 + t4 all differ from 2. Thus

the equation

u+v-2=0

is satisfied by (u, v) = (tl, t2), (t3, t4), but not by other pairs of roots. This
shows that there are polynomial equations over Q which are satisfied by
some but not all of the pairs of the roots. Intuitively, this suggests that the
four roots are not as far from being in Q as they might be.
We now introduce the group of the quartic equation. To do so, we describe
first the permutations associated with the elements tl, t2, t3, t4: there are
the 4! = 24 ways of reordering the four elements by replacing each by
another.
The identity permutation, denoted by c, leaves each ti fixed.
There are six permutations which interchange two of the roots and leave
the remaining two alone. These will be denoted by (12), (13), (14), (23),
(24), (34); the numbers between the parentheses indicate the indices of the
roots to be interchanged.
There are eight permutations which cyclically interchange three of the
roots and leave the remaining one alone. These are (123), (124), (132),
(134), (142), (143), (234), (243). F or example, (142) corresponds to replac-
ing tl by t4, replacing t4 by t2, and replacing t2 by tl.
There are three permutations which interchange pairs of the four roots.
These are (12)(34), (13)(24), (14)(23). For example, (13)(24) interchanges
tl and t3, and interchanges t2 and t4.
There are six permutations which cyclically interchange all four of the
roots. These are (1234), (1243), (1324), (1342), (1423), (1432). Thus, (1342)
replaces tl by t3, t3 by t4, t4 by t2 and t2 by tl.
A product of two permutations is defined by applying the permutations
in succession. For example the product of (123) and (1243) is effected as
follows:

replace tl by t2 (by (123)), then t2 by t4 (by (1243)), for a net replacement
of tl by t4;
replace t2 by t3, then t3 by tl, for a net replacement of t2 by tl;
replace t3 by 21, then tl by t2, for a net replacement of t3 by t2;
replace t4 by t3.