An introduction to the modern theory of equations, by Florian Cajori.

GALOIS THEORY OF ALGEBRAIC NUMBERS 141

Ex. 2. Are the following reducible in U(1)?
(1) x3+ 2 2 + 3 x-6. (5) X4 + 1 -100 x2 - x + 1.
(2) X3 +- 3 x2 + 8 -2. (6) X5 + X3 + x2 4- X 4- 7.
(2) x _+3x2+8x-2. (6) x5+ 3+z2+x+ 7.
(3) X4 + X3 + 2 + x - 4. (7) x5 - 2 x 4- 3 3 + 4 x2+ 3 x + 2.
(4) X4 + 9X3 + 25 2 + 22 x 6. (8) X5 + x + 1.
129. Eisenstein's Theorem. If p is a prime member, and
ao, aC,..., a, integers, all (except ao) divisible by p, but t,, not
divisible by p2, then is f(x) _ x + alx"-' + *.* + a, irreducible.
For, if f(x) could be resolved into factors, the coefficients of
the factors could be integers. We could have
f(x) _ (co, + c1,h- +... + c,)(doX0 + d1xk-1...+ d0),
where 7h + k = n.
Since c,, is divisible by p, but not by p2, and (t, = c, * dk, it
follows that one of the factors c,, d,, is divisible by p, but not
the other. Let c0 be the factor divisible by p. Then not all
the coefficients c are divisible by p, else a0 would be divisible
by p. Let c, be a coefficient not divisible by ), while c,,v, c,,,,
- c,, are each divisible by p. The coefficient of x"-", in the
product of the two factors of f(x), is then
(dkC + Clk7lCv+ + Cdk_2C,,+2+
Since ever?, term in this polynomial is divisible by p, except
the first term, the polynomial is not divisible by p. But, by
assumption, the only coefficient of f(x) which is not divisible
by p is co. Hence xA^- =_ ax', which is impossible, since h must
be less than n.
Ex. 1. Show by ~ 129 the irreducibility of
2 x3 + 9 x2 + 6 x -- 12,
4 x5 + 14 X4 + 21 x + -35.