Download Ring Theory (MA 416) 2006-2007 Problem Sheet 2 Solutions 1

Ring Theory (MA 416) 2006-2007
Problem Sheet 2
Solutions
1. Which of the following are subrings of Q[x]? Which (if any) are ideals?
(a) The set consisting of all polynomials of odd degree and the zero polynomial.
This is not a ring since for example it is not closed under addition. To see this
note that x3 + (−x3 + x2 ) = x2 - the sum of two polynomials of odd degree may
have even degree.
(b) The set consisting of all polynomials of even degree and the zero polynomial.
This is not a ring for a similar reason - the sum of two polynomials of even degree
may have odd degree.
(c) The set consisting of all polynomials whose coefficients are all even integers.
This is a subring of Q[x]. However it is not an ideal since for example the element
2 belongs to this subring but 13 × 2 = 23 does not.
(d) The set consisting of all polynomials in which the coefficient of xi is zero whenever
i is odd.
This is a subring of Q[x]; however it is not an ideal since (for example) x2 belongs
to the subring but x × x2 = x3 does not.
2. Decide if each of the following polynomials is irreducible in the indicated ring :
(a) 2x2 − 5x + 5 in R[x]
The roots of this polynomial in C are given by
√
5 ± −15
.
4
These are not real - thus the polynomial is irreducible in R[x] since it is quadratic
and has no real roots
(b) 2x2 − 5x − 5 in R[x]
The roots of this polynomial in C are
5±
√
4
65
.
These are real - thus the polynomial is reducible in R[x].
(c) x3 − 5x2 − 5x − 6 in Q[x]
This polynomial has 6 as a root and is therefore not irreducible in Q[x].
(d) x3 + 2x2 + 2x + 6 in Q[x]
This polynomial is irreducible in Q[x] by Eisenstein’s Criterion with p = 2.
(e) 2x5 − 6x3 + 9x2 − 15x + 12 in Q[x] This polynomial is irreducible in Q[x] by
Eisenstein’s Criterion with p = 3.
3. (a) Let f (x) be a polynomial of degree ≥ 2 in F [x] for some field F . Write x = y + 1
so that f (x) = f (y + 1) = g(y) ∈ F [y]. Show that f (x) is irreducible in F [x] if
and only if g(y) is irreducible in F [y].
Suppose f (x) is reducible in F [x]. Then f (x) = h1 (x)h2 (x) in F [x], where both
h1 (x) and h2 (x) have degree at least 1. Then g(y) = f (y + 1) = h1 (y + 1)h2 (y + 1).
Let h01 (y) = h1 (y + 1) and h02 (y) = h2 (y + 1). Then deg(h01 (y)) = deg(h1 (x))
and deg(h02 (y)) = deg(h2 (x)) and so g(y) = h01 (y)h02 (y) is a proper factorization of
g(y). Since x = y + 1 we have y = x − 1 and a similar argument shows that if g(y)
is reducible in F [y] then f (x) is reducible in F [x].
(b) Show that the polynomial x4 + x3 + x2 + x + 1 is irreducible in Q[x].
Write x = y + 1. Then
x4 + x3 + x2 + x + 1 = y 4 + 5y 3 + 10y 2 + 10y + 5.
This is irreducible by Eisenstein’s criterion with p = 5.
(c) For a prime p ≥ 3, show that xp−1 + xp−2 + · · · + x + 1 is irreducible in Q[x].
Define f (x) = xp−1 + xp−2 + · · · + x + 1 Note that
(x − 1)f (x) = xp − 1.
Replace x with y + 1 to obtain
yf (y + 1) = (y + 1)p − 1.
Thus
p p−1
p p−2
p
g(y) = f (y + 1) = y +
y
+
y
+ ··· +
y + p.
1
2
p−2
p
If 1 ≤ k < p we have
p!
p
=
.
k
k!(p − k)!
Since p appears
once in the numerator and not at all in the denominator, the
p
integer k is a multiple of p. Thus g(y) is irreducible in Q[y] by Eisenstein’s
criterion applied to the prime p, and so f (x) is irreducible in Q[x] by part (a).
4. Note that the set of nth roots of unity in C is precisely the set of roots of the polynomial
xn − 1 in C[x]. Express each of the following polynomials as a product of irreducible
factors in Q[x] :
(a) x4 − 1
x4 − 1 = (x − 1)(x + 1)(x2 + 1).
(b) x5 − 1
x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1).
(c) x6 − 1
x6 − 1 = (x − 1)(x + 1)(x2 + x + 1)(x2 − x = 1).
(d) x8 − 1
x8 − 1 = (x − 1)(x + 1)(x2 + 1)(x4 + 1).
5. An element ω of C is called a primitive nth root of unity if ω n = 1 but ω k is not equal
to 1 for any positive integer k < n. For example −1 is a fourth root of unity but it is
not a primitive fourth root of unity, since (−1)2 = 1; the primitive fourth roots of unity
in C are i and −i.
For n = 4, 5, 6, 8 identify a connection between the primitive kth roots of unity for each
k dividing n and the irreducible factors in Q[x] of xn − 1 found above.
(a) x4 − 1 x4 − 1 = (x − 1)(x + 1)(x2 + 1).
The 4th roots of unity in C are either primitive 1st roots of unity (1, the root in
C of x − 1), primitive 2nd roots of unity (−1, the root in C of x + 1), or primitive
4th roots of unity (i and −i, the roots in C of x2 + 1). So the irreducible factors
of x4 − 1 in Q[x] correspond to the primitive roots of unity in C of the different
orders dividing 4.
(b) x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1).
The 5th roots of unity in C are either primitive 1st roots of unity (1, the root in
C of x − 1), or primitive 5th roots of unity (the roots in C of x4 + x3 + x2 + x + 1).
So the irreducible factors of x5 − 1 in Q[x] correspond to the primitive roots of
unity in C of the different orders dividing 5.
(c) x6 − 1 = (x − 1)(x + 1)(x2 + x + 1)(x2 − x = 1).
The 6th roots of unity in C are either primitive 1st roots of unity (1, the root in C
of x − 1), primitive 2nd roots of unity (−1, the root in C of x + 1), primitive 3rd
roots of unity (the roots in C of x2 + x + 1), or primitive 6th roots if unity (the
roots in C of x2 − x + 1). So the irreducible factors of x6 − 1 in Q[x] correspond
to the primitive roots of unity in C of the different orders dividing 6.
(d) x8 − 1 = (x − 1)(x + 1)(x2 + 1)(x4 + 1).
The 8th roots of unity in C are either primitive 1st roots of unity (1, the root in C
of x − 1), primitive 2nd roots of unity (−1, the root in C of x + 1), primitive 4th
roots of unity (i and −i, the roots in C of x2 + 1), or primitive 8th roots if unity
(the roots in C of x4 + 1). So the irreducible factors of x8 − 1 in Q[x] correspond
to the primitive roots of unity in C of the different orders dividing 8.
6. Let R be a an integral domain (with at least two elements). Prove that R is a field if
and only if the only ideals of R are R and {0R }.
First assume that R is a field, and let I be a non-zero ideal of R. Choose a non-zero
element a of I. Since R is a field, a has an inverse a−1 for multiplication in R. Then
a−1 a = 1R belongs to I. Now if r is any element of R4 we have r1R = r belonging to
I. Thus I = R.
On the other hand suppose that the only ideals of R are R and {0R }. We need to show
that every non-zero element of R has an inverse with respect to multiplication. So let
a ∈ R, a 6= 0. The principal ideal of R generated by a contains all elements of the form
ar, r ∈ R. This is not the zero ideal of R since it contains a, so it must be equal to R.
Then 1R ∈ hai =⇒ 1r = ab for some b ∈ R. Thus b is an inverse for a in R, and R is a
field.