MA 3419: Galois theory
Selected answers/solutions to the assignment due October 20, 2015
1. (a) Consider the quaternion group Q8 . Its centre Z consists of ±1, and
Q8 /Z ≡ Z/2Z × Z/2Z
(it is a group of order four where each element is of order 2).
(b) This question contained a misprint, it was meant to be about Z/6Z, in which case the
polynomial x2 − x works, as it has roots 0, 1, 3, 4. In case of Z/4Z, every monic polynomial
has at most two distinct roots. (For the record, the polynomial 2x(x + 1) (not of the type
considered in the question has four roots in Z/4Z). In any case, this question was not marked
due to the misprint.
(c) Consider any quaternion ai + bj + ck where a2 + b2 + c2 = 1. This quaternion solves
the equation x2 + 1 = 0 (direct inspection).
2. (a) We have e = (ab)2 = abab. Also, a2 = e and b2 = e, so
ba = bae = baabab = bbab = ab.
This holds for any a, b ∈ G, so ab = ba for any a, b ∈ G, and G is Abelian.
(b) By Lagrange’s theorem, the order of any element is a divisor of 6. If there is an
element a of order 6, then the group is cyclic generated by a and therefore Abelian. If there
are no elements of order 3 or 6, the order of each element different from e must be two, so
the group is Abelian by previous question. Therefore, there must be an element of order
3. If g ∈ G is of order 3, and h ∈ G is distinct from e, g, g2 , the elements h, hg, hg2 are
different from e, g, g2 , hence G = {e, g, g2 , h, hg, hg2 }. We cannot have h2 ∈ {h, hg, hg2 }, so
either h2 = e or h2 = g or h2 = g2 . In case h2 = g or h2 = g2 , the element h is of order 6,
a contradiction. Also, we note that we cannot have gh ∈ {e, g, g2 , h}, so either gh = hg or
gh = hg2 . If hg = gh, the group is Abelian, since g and h generate it. Hence gh = hg2 , and
this determines the multiplication table uniquely, so we must obtain S3 .
3. (a) This polynomial does not have roots in F2 , hence irreducible (it is of degree two).
The quotient F2 [x]/(x2 + x + 1) hence consists of 4 elements and is a field. It has elements
0, 1, x, 1 + x, and the multiplication table is


0
1
x
1+x
 0
0
0
0
0 


 1
0
1
x
1 + x


 x
0
x
1+x
1 
1+x 0 1+x
1
x
(b) This polynomial does not have roots in F3 , hence irreducible (it is of degree two).
The quotient F3 [x]/(x2 + 1) hence consists of 9 elements and is a field. The representatives
of cosets are elements a + bx with a, b ∈ F3 , and x2 = −1 in the quotient, hence the product
rule
(a + bi)(c + di) = ac − bd + (ad + bc)i.
x2
4. (a) Degree 1: x and x + 1. Degree 2: x2 + x + 1 (the others x2 , x2 + x = x(x + 1) and
+ 1 = (x + 1)2 are clearly reducible). Degree 3: reducibility for degree 3 is still equivalent
to having a root, so we just need to avoid the polynomials with root 0 (constant term 0) and
root 1 (sum of coefficients zero). We obtain the polynomials x3 + x + 1 and x3 + x2 + 1. Degree
4: a reducible polynomial of degree 4 either has a root, or is a product of two irreducibles
of degree 2, which we already know. This gives the polynomials x4 + x + 1, x4 + x3 + 1,
x4 + x3 + x2 + x + 1.
(b) The rings F2 [x]/(x3 + x + 1) and F2 [x]/(x4 + x + 1) are fields of 8 and 16 elements
respectively.
5. Suppose that f1 (x), . . . , fn (x) are all irredicible polynomials. Consider the polynomial
f1 (x)f2 (x) · · · fn (x) + 1. It must have an irreducible divisor, but neither of fi (x) is its divisor,
a contradiction.
6. (a) Eisenstein’s criterion for p = 2.
(b) We have a3 − 2a − 2 = 0, so a(a2 − 2) = 2, and 1/a = (a2 − 2)/2. Also, we have
x3 − 2x − 2 = (x2 − x − 1)(x + 1) − 1 (long division), so we have 1/(a + 1) = a2 − a − 1. Finally,
we have a3 − 2a − 2 = a(a2 + 1) − 3a − 2, so 1/(a2 + 1) = a/(3a + 2), so we need to compute
27
1/(3a + 2), which we do as above, obtaining 26
(1/3a2 − 2/9a − 14/27). Multiplying by a and
1
3
reducing using the equation a − 2a − 2 = 0, we obtain 1/(a2 + 1) = 26
(−6a2 + 4a + 18).
√
7. The roots of this polynomial are 105 9e2πki/105 , 0 6 k 6 104. If x105 − 9 = f(x)g(x),
where f(x), g(x) ∈ Z[x], the roots of f(x) are some of the roots of x105 − 9, and hence the
absolute value of the constant term of f(x) is 9d/105 , where 1 6 d 6 104 is the degree of f(x).
However, for these values of d this number is not an integer: 9d/105 = 32d/105 , and 2d/105 is
not an integer for 0 < d < 105.√ √
√ √
√ √
√ √
4
2
8. It is the polynomial (x− 2− 3)(x+
√
√2− 3)(x− 2+ 3)(x+ 2+ 3) = x −10x +1.
Note that the minimal polynomial of 2 + 3 must divide this one, so it is sufficient to show
that x√4 − 10x2 + 1 is irreducible.
√
√ The
√ roots
√ of√this polynomial are not rational,√since if, e.g.,
2+ 3 is rational, then 1/( 2+ 3) = 3− 2 is rational, and consequently 2 is rational.
Thus, if this polynomial factorises, it is a product of two quadratic polynomial with integer
coefficients, x4 − 10x2 + 1 = f(x)g(x). The sum of roots of f(x), which is, up to a sign, one of
4
2
its coefficients, is√obtained
adding
√ by √
√ two of the roots of x − 10x + 1; the possible values of
this sum is 0, −2 2, 2 2, −2 3, 2 3. Hence, for√the coefficients
of f(x)√to be integers, that
√
√
sum √
must √
be zero, so,
without
loss
of
generality,
2
+
3
and
−
2 − 3 are roots of f(x),
√
√
and 2√− 3 and − 2 + 3 are roots of g(x). But then the product of the roots of f(x) is
−5 + 2 6 which is not an integer.