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Math 417 – Homework VIII: July 10, 2007
3.84(i). If f(x)∈R[x], show that f(x)has no repeated roots in Cif and only
if (f,f0)=1 in R.
Exercise 3.67 says that if kis a field, then f(x)∈k[x]has no repeated roots
if and only if (f,f0)=1 in k[x]. Thus, the hypothesis here says that f(x)has
no repeated roots in R, and the question is whether it can have repeated roots
in C. But Corollary 3.75 says that the gcd (f,f0)=1 in C[x]if and only if
(f,f0)=1 in R[x]. To elaborate, even though there may be more complex
roots of f(x)than real roots [for example, f(x)may have no real roots at all],
the question whether a root is repeated in the larger field Ccan be decided in R.
(ii). Prove that if p(x)∈Q[x]is an irreducible polynomial, then p(x)has no
repeated roots.
If p(x)=Pn
i=0aixi, then p0(x)=Pn
i=1iaixi−1. It follows that deg(p0)=
n−1, and so p0is not the zero polynomial. Since p(x)is irreducible, the gcd
(p,p0)=1, and so part (i) applies to show that p(x)has no repeated roots.
(Note that this argument may fail if Qis replaced by Fp, where pis prime.)
3.87.Determine whether the following polynomials are irreducible in Q[x].
(i). f(x)=3x2−7x−5.
There are no rational roots: the candidates are
±1,±5,±1
3,±5
3.
Therefore, f(x)is irreducible, by Proposition 3.65.
(iii). f(x)=2x3−x−6.
There are no rational roots: the candidates are
±1
2,±1,±3
2,±2,±3,±6.
Therefore, f(x)is irreducible, by Proposition 3.65.
(v). f(x)=x3+6x2+5x+25.
f(x)≡x3+x+1 mod 2. Since there are no roots over F2, the cubic f(x)
is irreducible over F2, by Proposition 3.65. It now follows from Theorem 3.97
that f(x)is irreducible in Q[x].
