274 Answers to Exercises and Solutions to ProblemsSince plcc and p2rt)cg, p must divide exactly one of a0 and bo, say plac,
p$bo. Then
plcl plalbo =$plal
P I c2 P I do - P I U2.
Continue on to find in succession that plug,... ,~[a,. But then plc, = unb,,
yielding a contradiction.
1.13. Apply the Eisenstein Criterion with prime 3.
1.14.t + 1, 3t+4.
1.15. t2 + t + 1 has nonreal zeros and so is irreducible over Z.
1.16. Assume that h(t) is irreducible; use the notation of Exercise 12. If p
fails to divide one of uc or bo, say the latter, the argument of Exercise 12
leads to a contradiction. The remaining case is that each of ue and bo are
divisible by p but not by p2.
We show by induction that plai, pjbi for i = 0, 1,... , m. This is true for
i = 0. Suppose that it has been shown for i = 1,2,... , R - 1, where k < m.
Consider
c2k = uob2k + albak-1 + * * * + akbk + uk+lbk-l +. * * + uxkbo.Since p is a divisor of cgk, ac,... , a&i, bo,... , bk-1, p alSO divides Uk bk,
and thus at least one of ak and bk.
Now, consider
ck =aobk +albk-1 + .**+ak-lbl +akbo.Sincep2isadivisorofck,aibk-I,..., ak,lbl, p2 also divides aobk + akbo.
Now, suppose plak. Then, p21ukbo, so p21aobk. But p21fue, so plbk. Similarly,
if p]bk, then pjak. Hence p divides both ak and bk.
Therefore, plai, plbi for 0 < i 5 m, and so p divides ~a,,,+1 = uOb2m+l +
.. e + amb,+l + um+lbm +... + az,,,+l bo, yielding a contradiction.
1.17. Apply the Eisenstein Criterion for p = 2 to show irreducibility over Z,
hence over Q. The polynomial is reducible over R by the Factor Theorem
since it has a real root 2i/‘, when r > 1.
1.18. Construct a polynomial of the form t” +... + 2, where r - 2 terms
apart from the leading and constant ones have even nonzero coefficients.
Such a polynomial is irreducible by the Eisenstein Criterion with prime 2.
1.19. By Exercise 6, P-l + tPv2 +a.. + t + 1 is irreducible as a polynomial
in t if and only if (1 + s)P-’ + (1 + s)Ps2 +... + (1 + s) + 1 is irreducible
as a polynomial in s. The latter polynomial can be written(l+s)‘-1
S