130 4. Equations
so it makes sense to denote this extension by Q(fi, fi) and to
regard it as the field obtained from Q by adjoining both fi and
6)- It is possible to adjoin to a field F zeros of polynomials of degree
higher than 2. For example, F(21i4) is the set of elements of the form
where a, b, c, d E F.(a) Show that each element of F(21j4) can be written in exactly one
way in the form specified.
(b) Verify that F(21i4) is closed under addition, subtraction and
multiplication.
(c) Show that
(a + 21i4b + 21’,2c + 23’4d)-1
also belongs to F(2114) and deduce that F(21i4) is a field.
(d) In particular, determine (1+23/4)-1 and (1+21/4+21/2+23/4)-‘.
Check your answer.[Remark: Extending an arbitrary field F begs the question of the
existence of the radical in some field larger than F. For example,
in forming Q(d), we know that fi exists as a real number and
so Q(A) is a subset of R. However, F need not be a number field
in order for the extension to be definable. In general, one adjoins a
number making use only of the fact that it is to satisfy a polynomial
equation over F. It is in this spirit that, for example, we need not ask
what J--i is, but just know that it is something which satisfies the
equation x2 + 1 = 0, i.e. something which yields -1 when squared.]- (a) Let F be any field contained in R, and let f(t) be a cubic poly-
nomial over F. Suppose d but not fi is a member of F and that
F(h) contains a zero of f(t). Prove that f(t) is reducible over
F and accordingly F contains a zero of f(t).
(b) Suppose g(t) is a cubic polynomial over Q and that Fs, Fr ,... , F,
is a sequence of fields contained in R such that Fc = Q, and for
i= 1,2,..., n, Fi = Fi-l(&) where di E Fi-1. Prove that, if
g(t) has a zero in F,, then g(t) has a rational zero.
[Remark: This is used to establish the celebrated result that there is
no general ruler-and-compasses construction for trisecting an angle.] - Let f(t)=t3-3t+l.