Abstract Algebra II
Spring 2014
Homework 1
Problem 1. Let p ∈ Z be an odd prime. Show that x2 + 1 is reducible in Zp [x] if and only if
p is of the form 4k + 1. (Hint: Use the fact that Z×
p is cyclic to show that Zp has an element
of order 4 if and only if p is of the form 4k + 1.)
Since Zp is a field, a polynomial of degree two, x2 + 1 is reducible in Zp [x] if and only if
it has a root in Zp , which is equivalent to having an element of order 4 in Z×
p (why?). We
×
proved that the multiplicative group of a field is cyclic, so Zp is cyclic, so it has an element
of order 4 if and only if its order p − 1 is divisible by 4, that is, p is of the form 4k + 1.
Problem 2. Find all the values of a in Z5 such that the quotient ring
Z5 [x]/hx3 + 2x2 + ax + 3i
is a field.
The quotient ring is a field if and only if the ideal I = hx3 + 2x2 + ax + 3i is maximal.
Since Z5 is a field, Z5 [x] is a PID. Hence the ideal I is maximal if and only if it is prime,
which happens if and only if the polynomial f (x) = x3 + 2x2 + ax + 3 is irreducible in Z5 [x]
(since Z5 [x] is a UFD). A polynomial of degree 3 with coefficients in a field is irreducible if
and only if it has no roots in the field. Plugging in 0, 1, 2, 3, 4 for x we see that the values of
a for which f has no roots in Z5 are 0, 1, and 2.
Problem 3. Determine whether the following polynomials are irreducible in the rings indicated.
1. x4 + 1 in Z5 [x].
Reducible: x4 + 1 = (2x2 + 1)(3x2 + 1).
2. x4 + 10x2 + 1 in Z[x].
Irreducible. The only possible rational roots are ±1, and they don’t work, so f has no
linear factors. If
x4 + 10x2 + 1 = (x2 + ax + 1)(x2 + bx + 1) or x4 + 10x2 + 1 = (x2 + ax − 1)(x2 + bx − 1)
The product of the leading terms is 1, so either both of them are 1 or −1. If it’s the
latter we can multiply both quadratics by −1. We get a + b = 0 in both cases. Also,
2 + ab = 10 in the first case and ab − 2 = 10 in the second. We get a2 = −8 or −12,
neither of which has integer solutions.
3. x4 + 4x3 + 6x2 + 2x + 1 in Z[x] (substitute x − 1 for x).
Irreducible. Following the hint, f (x − 1) = x4 − 2x + 2, which is irreducible by
Eisenstein’s Criterion. Hence f (x) is irreducible as a factorization for f (x) would give
a factorization for f (x − 1).
4.
xp −2p
x−2
in Z[x], where p is prime.
Irreducible. For p = 2 the polynomial is x + 2, so it is irreducible. Assume next that
p 6= 2.
p (x + 2)p − 2p X p p−k k−1
ϕ(x + 2) =
=
2 x .
x
k
k=1
Here all the coefficients except the first one are divisible by p and the constant term
p2p−1 is not divisible by p2 . By Eisenstein’s Criterion we conclude that f is irreducible.
Problem 4. Let F be a field and suppose that I is an ideal in F [x1 . . . xn ] generated by
a (possibly infinite) set S of polynomials. Prove that there exists a finite subset of S that
generates I. (Use Hilbert’s basis theorem.)
By Hilbert’s basis theorem, I has a finite generating set {f1 , · · · , fm }. Express each fi
in this generating set as a polynomial combination of elements in S. Let the polynomials
involved in expressing fi be the elements of a finite subset Si of S. Then the finite set of
elements ∪m
i=1 Si in S generates I.
2
© Copyright 2024 ExpyDoc
