MTH745U/P (2013–14)
Further Topics in Algebra
(Fields and Galois Theory)
18th November 2013
Solutions 7
Question 1. Any quadratic extension of F is isomorphic to a quotient of the form
F [X]/(aX 2 +bX +c) where a, b, c ∈ F , a 6= 0 and aX 2 +bX +c is irreducible over
F (which happens if and only if it has no roots in F ). Note that aX 2 + bX + c
will factor into linear factors in either F or F [X]/(aX 2 + bX + c).
Let K be a splitting field over F for aX 2 + bX + c (with a 6= 0 and 2 6= 0). Then
the roots of F in K are α1 and α2 where
√
−b ± b2 − 4ac
.
α1 , α2 =
2a
For i = 1, 2 it is clear√that F (αi ) and F (βi ) are identical subfields of K, where
i −b
βi := b + 2aαi = ± b2 − 4ac (and so αi = β2a
). But β2 = −β1 , and so
√
2
F (β2 ) = F (β1 ). Therefore K = F ( d) where d = b − 4ac (since K is a splitting
field for aX 2 + bX + c), and K has degree 2 over F if and only if d is not a square
in F (otherwise K = F has degree 1 over F ).
√
The solutions of X 2 + X + c = 0 for the case 2 6= 0 are X = 21 (−1 ± 1 − 4c).
In order to prove no such formula can exist in characteristic 2, we can consider
a specific case, so we consider the equation
X2 + X + 1 =
√ 0 over F2 . The roots
√
of this lie in F4 but not in F2 . But x ∈ F2 (and in fact x = x) for all x ∈ F2 .
So no combination of algebraic operations and taking square roots applied to a
collection of elements of F2 can produce an element not in F2 . √
√
Now we let 2 = 0. The two solutions of X 2 + √
X + c = 0 are X√= ∗ c and ∗ c√
+ 1,
∗
∗
∗
and either of these can serve
√ as the√value of c (note that ( c√+ 1) + 1 = c).
This is akin to either + d and − d being able to serve as d when
√ 2 6= 0.
(In some cases, but certainly not universely, we can specify that d denotes
a√ particular square root of d, for instance when d ∈ R+
0 we can specify that
2
d ∈ R+
.)
Now
we
consider
the
equation
aX
+
bX
+
c
= 0 where
0
p a, b, c ∈ F
2
with a 6= 0. If b = 0 then the only root of aX + bX + c = 0 is X = a/c, which
√
has multiplicity 2, and occurs either in F , or in the quadratic extension F ( d)
for d = ac . If b 6= 0 then we let Y = ab X (and so X = ab Y ) and observe that
0 = aX 2 + bX + c =
b2 2
Y
a
+
1
b2
Y
a
+c=
b2
(Y 2
a
+Y +
ac
)
b2
p
p
So Y = ∗ ac
, up to the uncertainty of adding 1. Therefore X = ab ∗ ac
, up to the
b2
b2
b
uncertainty of adding a ∈ F . Thus X either lies in F , or X lies in the quadratic
√
extension F ( ∗ d) where d = ac
.
b2
√
√
√ √
ab
=
a
b
and
a+b =
Note
that
if
F
is
a
field
of
characteristic
2
then
√
√
√
√
√
a + b for√all a, b ∈ F . Also ∗ a + b = ∗ a + ∗ b up to the ambiguity inherent
when using ∗ s. That is, if one adds a root of X 2 + X + a to a root of X 2 + X + b
one obtains a root of X 2 + X + a + b, and either root of X 2 √
+X +a+
b√
can be
√
∗
∗
obtained in this manner. There is no sensible way to relate ab and a ∗ b.
p
−1
Question 2. It is easy to see that for p prime ζp has minimal polynomial XX−1
=
X p−1 + · · · + X 2 + X + 1, which we proved irreducible on Coursework 4, Question
6. Now ζ9 cubes to ω (a primitive cube root of 1), and so is a root of X 6 +X 3 +1,
which is in fact irreducible over Q, which we essentially shall prove here. Our
argument is that Q(ζ9 ) contains subfields of degrees 2 (namely Q(ω)) and 3
(namely Q(γ9 ), and proved here) over Q, so that Q(ζ9 ) cannot have degree less
than 6 over Q by the Tower Law.
We take heed of the hint (omitted from the paper copies of the sheet) that the
minimal polynomial of γn over Q is cubic for each n ∈ {7, 9, 13}. Now for n = 7
or 9, we have γn = ζ + ζ −1 , where ζ = ζn . Therefore, we have:
1 = 1,
γn = ζ + ζ −1 ,
γn2 = ζ 2 + ζ −2 + 2,
γn3 = ζ 3 + ζ −3 + 3(ζ + ζ −1 ).
We now use “symmetrised” forms of the minimal polynomials of ζ7 and ζ9 given
above, namely we observe that ζ 3 + ζ 2 + ζ + 1 + ζ −1 + ζ −2 + ζ −3 = 0 if ζ = ζ7
and that ζ 3 + 1 + ζ −3 = 0 if ζ = ζ9 . (Technically, these are Laurent polynomials
in ζ rather than ordinary ones.) Thus if ζ = ζ9 we find that ζ 3 + ζ −3 = −1, and
so γ93 = ζ 3 + ζ −3 + 3(ζ + ζ −1 ) = −1 + 3(ζ + ζ −1 ) = 3γ9 − 1. Therefore, γ9 is a
root of X 3 − 3X + 1. Similarly, we find that γ7 is a root of X 3 + X 2 − 2X − 1.
The last case is a bit harder. Now we let ζ = ζ13 . We have:
2
γ13
1 = 1, γ13 = ζ + ζ −1 + ζ 5 + ζ −5 ,
= ζ 2 + ζ −2 + ζ 10 + ζ −10 + 4 + 2ζ 6 + 2ζ −6 + 2ζ 4 + 2ζ −4
3
You cay try to calculate γ13
if you want to. Instead we use algebraic conjugacy
(not yet done in lectures) to help us. For 1 6 j 6 12 the element ζ j is a
13 −1
root of g(X) = XX−1
, which we know is irreducible, and thus Q(ζ) ∼
= Q(ζ j ) ∼
=
Q[X]/(g(X)) via an isomorphism φj sending ζ to ζ j and fixing Q. In fact F :=
Q(ζ) = Q(ζ j ) (why?), and so φj is an automorphism of F . For 1 6 j 6 12, we
∗j
∗1
∗5
∗8
∗12
define γ13
:= γ13 φj = ζ j +ζ −j +ζ 5j +ζ −5j . We have γ13
= γ13
= γ13
= γ13
= γ13 ,
together with:
∗2
∗3
∗10
∗11
γ13
= γ13
= γ13
= γ13
= ζ 2 + ζ −2 + ζ 3 + ζ −3
∗4
∗6
∗7
∗9
and γ13
= γ13
= γ13
= γ13
= ζ 4 + ζ −4 + ζ 6 + ζ −6 .
2
2
∗2
∗2
∗4
∗2
Calculating the products γ13
(see above), and γ13 γ13
= γ13 + 2γ13
+ γ13
= γ13
− 1,
and applying various of the maps φj , we get the following relations.
∗4
∗2 2
∗2
∗4 2
∗2
∗4
(γ13 )2 = 3 + γ13
− γ13 , (γ13
) = 3 + γ13 − γ13
, (γ13
) = 3 + γ13
− γ13
,
∗2
∗2
∗4
∗2 ∗4
∗4
γ13 γ13 = γ13 − 1, γ13 γ13 = γ13 − 1 and γ13 γ13 = γ13 − 1.
∗2
∗4
We also note the relations ζ 13 = 1 and γ13 + γ13
+ γ13
= −1 (which follows from
P
6
i
ζ
=
0),
and
we
have
made
some
use
of
them
above,
such as to note that
i=−6 13
2
∗2
∗4
∗4
(γ13 ) = 4 + γ13 + 2γ13 = 3 + γ13 − γ13 . Therefore, we obtain
3
2
∗4
2
2
2
γ13
= γ13 γ13
= 3γ13 + γ13 γ13
− γ13
= 3γ13 + (γ13 − 1) − γ13
= −γ13
+ 4γ13 − 1.
Therefore γ13 is a root of X 3 + X 2 − 4X + 1.
In fact, the minimal polynomials of γ7 , γ9 and γ13 are X 3 + X 2 − 2X − 1,
X 3 − 3X + 1 and X 3 + X 2 − 4X + 1 respectively. In all cases, my claimed
minimal polynomial is in Z[X] and has constant term 1 or −1. Therefore, since
all the cases are cubic, we only have to test (essentially by Gauß’s Lemma) for
1 and −1 being potential roots to prove their irreducibility over Q. Naturally,
none of these polynomials has 1 or −1 as a root.
∗j
Note that since γ13
= γ13 φj for all j with 1 6 j 6 12, the minimal polynomial
∗j
∗2
of γ13
is the same as that of γ13 . Thus X 3 + X 2 − 4X + 1 has roots γ13 , γ13
and
∗4
γ13 . Similar considerations apply to the other two cases.
Question 3. If aX 2 + bX + c2 is an irreducible quadratic (over Q), one of whose
roots is x0 , then the other is − ab + x0 , so that the two roots sum to − ab . So the
splitting field of an irreducible quadratic has degree at most 2 over the ground
field. We showed in lectures that the splitting field of an irreducible cubic has
degree 3 or 6 over the ground field. Note that if aX 3 + bX 2 + cX + d (a 6= 0) is an
irreducible polynomial in F [X] (F a field) having roots x0 , x1 , x2 in a splitting
field then x0 + x1 + x2 = − ab , and so the splitting field is generated over F by
any two of x0 , x1 , x2 .
Not all the polynomials in the question were irreducible, though most were.
Irreducibility of claimed irreducible factors by one or more of: Coursework 4,
Question 6; Eisenstein’s Irreducibility Criterion; Gauß’s Lemma and the fact
that reducible polynomials of degrees 2 or 3 must have a linear factor, and thus
monic reducible polynomials in Z[X] of degrees 2 and 3 have an integer root
dividing the constant term.
√
√
(a) X 2 −2 has splitting field Q( 2); X 2 +7 has splitting field Q( 2); X 2 +X
√ +1
and X 3 −1 = (X −1)(X 2 +X +1) both have splitting field Q(ω) = Q(
√ −3);
3
2
2
and X − 2X + 1 = (X − 1)(X + X − 1) has splitting field Q( 5). So
all the splitting fields have degree 2 over Q.
3
(b) X 4 + X 3 + X 2 + X + 1 has roots ζ5 , ζ52 , ζ53 , ζ54 and splitting field Q(ζ5 ), of
degree 4 over Q. X 6 +X 5 +X 4 +X 3 +X 2 +X +1 has roots ζ7 , ζ72 , ζ73 , ζ74 , ζ75 , ζ76
and splitting field Q(ζ7 ), of degree 6 over Q.
(c) The splitting fields of each of the polynomials f7 , f9 and f13 (square discriminants 49, 81, 169 respectively) each have degree 3 over Q, and in
order to show this it suffices to show we can write all roots of the fi as
Q-polynomials in one of the roots.
We can readily verify that the roots of f7 are ζ7 +ζ7−1 = γ7 , ζ72 +ζ7−2 = γ72 −2
and ζ74 + ζ7−4 = ζ73 + ζ7−3 = −γ72 − γ7 + 2.
In the case of f9 , the roots are ζ9 + ζ9−1 = γ9 , ζ92 + ζ9−2 = γ92 − 2 and
ζ94 + ζ9−4 = −γ92 − γ9 + 1. (Here, ζ94 + ζ9−4 6= ζ93 + ζ9−3 = −1.)
∗2
∗4
In the case of f13 , the roots are γ13 , γ13
and γ13
(noted in the the previous
question). In fact, from the relations given in the previous question, we
∗4
2
∗2
2
easily see that γ13
= γ13
+ γ13 − 3, and so γ13
= −γ13
− 2γ13 + 2.
(d) In this case, the splitting fields of the cubic polynomials contain a quadratic
subfield, and thus have degree 6 (not 3) over Q. Also, note that if a and b
are coprime (that is gcd(a, b) = 1) and if F is generated by subfields having
degrees a and b over Q, then F has degree ab over Q. The cubic polynomials
have non-square discriminants −108 = 62 (−3), −31 and 148 = 22 · 37
respectively (see the hint). So the splitting fields√all have degree
√ 6 in these
cases. The splitting field of X 3 − 4X + 2 is Q(α, 148) = Q(α, 37) where
α is any root of X 3 − 4X + 2. This splitting field is a subfield of R, unlike
√
the splitting fields of X 3 − 2 and X 3 + X +√
1. The roots of X 7 − 3 are ζ7i 7 3
for 0 6 i 6 6, √
so its splitting field is Q(ζ7 , 7 3). It is generated by subfields
Q(ζ7 ) √and Q( 7 3) having degrees (respectively) 6 and 7 over Q, and so
Q(ζ7 , 7 3) has degree 42 over Q. So the splitting fields have respectively
degrees 6, 42, 6 and 6 over Q.
Question 4. For these particular primes we can proceed by trial and error, and
even calculate cubes of all elements in Fp if necessary. For p = 31 we find that
there are three elements cubing to 2, namely 4, 7 and 20 = −11. So over F31 , we
find that X 3 − 2 = (X − 4)(X − 7)(X + 11) has all roots in F31 and thus splitting
field F31 . (The roots have the form 1 · 4, 5 · 4 and 25 · 4, where 1, 5, 25 are the
cube roots of 1 in F31 .)
Now for the case p = 7 we find that the cubes in Fp of 0, 1, 2, 3, 4, 5, 6 are
0, 1, 1, 6, 1, 6, 6 respectively, none of which is 2. So X 3 − 2 has no root in F7 ,
and is therefore irreducible over F7 (as it has no linear factor and degree at most
4
3). Let α(6= 0) be a root of X 3 − 2 in F7 [X]/(X 3 − 2) ∼
= F343 , a field of degree 3
3
over F7 . Then the other two roots of X − 2 are 2α and 4α (note that 2 and 4
are primitive cube roots of 1 in F7 ). Therefore F7 (α) ∼
= F343 is a splitting field
for X 3 − 2 over F7 .
Now we come to the case p = 11, and cubing the elements 0, . . . , 10 in F11 gives
us 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10. So F11 has a unique element (7) that cubes to 2,
and we can show that this is not a repeated root, for if it were, we would have
X 3 − 2 = (X − 7)3 , which is not the case. (There is also only one cube root of 1
in F11 .) So over F11 we have X 3 − 2 = (X + 4)(X 2 − 4X + 5), where X 2 − 4X + 5
2
is irreducible over F11 as its discrimant is (−4)2 − 4 · 5 = −4 =
√ 2 (−1) which
3
is a non-square in F11 . So the roots of X − 2, namely 7, 2 ± −1, lie in and
generate an extension of degree 2 over F11 . This extension is the splitting
√ field,
and is F121 . (Note that elements of Fp2 can be written in the form a + b −1 for
a, b ∈ Fp if and only if p ≡ 3 (mod 4). You should
√ consider what Question 1 says
about writing elements of Fp2 in the form a + b d for a, b, d ∈ Fp .)
Let us consider what happens for general p. So let fp = f = X 3 − 2 ∈ Fp [X]. We
have Df = 3X 2 , and so if the characteristic is not 2 or 3 we have −2 = f − 13 (Df ),
and so 1 = − 12 f + 16 (Df ), and thus f does not have a repeated roots except
possibly when p = 2 or 3 (since f and Df have a g.c.d. of 1). In fact, for p = 2
and 3, there is a single root with multiplicity 3, for we have f (X) = f2 (X) = X 3
(case p = 2) and f (X) = f3 (X) = X 3 − 2 = X 3 + 1 = (X 3 + 1)3 (case p = 3).
(It is no coincidence that 2 and 3 are the only prime factors of −108, which is
the discriminant X 3 − 2.)
For general p, and working over a splitting field for X 3 − 2, we find (much like
the situation over C) that the roots are α, ωα and ω 2 α, where α3 = 2 and ω is
a primitive cube root of 1 (take ω = 1 in characteristic 3). Now ω is a root of
X 2 + X + 1, and if 3 6= 0 then X 2 + X + 1 does not have 1 as a root, and so (away
∼
from the case 3 = 0) ω lives in Fp if and only if (F×
p , ·) = Cp−1 has an element
of order 3. This happens if and only if p ≡ 1 (mod 3). Alternatively, for p 6= 2
or 3, we can√argue that ω lies in Fp if and only if −3 is a square in
√ Fp , since
1
1
ω = 2 (−1 + −3) (which we cannot really distinguish from 2 (−1 − −3)).
If p ≡ 1 (mod 3) then (since ω ∈ Fp ) all of the roots α, ωα, ω 2 α of X 3 − 2 lie
in Fp , or none do. In the former case (such as p = 31), X 3 − 2 factors into
linear factors over Fp , and so the splitting field is Fp . In the latter case (such
as p = 7), the polynomial X 3 − 2 has no root in Fp , and so is irreducible over
Fp . The splitting field of X 3 − 2 is then Fp (α, ωα, ω 2 α) = Fp (α) = Fp3 . In this
case, α ∈ Fp if and only if 2 is a cube in (F×
p , ·), which happens if and only if
(p−1)/3
2
≡ 1 (mod p).
If p ≡ 2 (mod 3), we argue that the cubing map on Fp is injective, and hence
surjective since Fp is finite. For if x3 = y 3 (with x, y ∈ Fp ), then 0 = x3 − y 3 =
5
(x − y)(x2 + xy + y 2 ) = (x − y)(x − ωy)(x − ω 2 y) [now working in Fp (ω)], thus
showing that x = y, as x−ωy = 0 implies that x = y = 0 and x−ω 2 y = 0 implies
∼
that x = y = 0. Alternatively, we note that (F×
p , ·) = Cp−1 has order coprime
to 3. From the Euclidean Algorithm in Z, we find that 1 = (p − 1) − 3( p−2
)=
3
2p−1
2p−1
−2(p − 1) + ( 3 )3. We let m := 3 , and observe that m > 0 is an integer,
and that xm cubes to x, and so is the unique cube root of x in Fp , as we have
now shown that the cubing map is surjective. (If x = 0 we observe directly that
xm = 0m = 0 cubes to x, while if x 6= 0, we get 3m = 2p − 1 = 2(p − 1) + 1
and so (xm )3 = x3m = (x(p−1) )2 x = x.) So we may pick α = 2m to be the unique
cube root of 2 in Fp . If α 6= 0 (that is p 6= 2), then the splitting field of X 3 − 2 is
Fp (α, ωα, ω 2 α) = Fp (ω) = Fp2 , while if α = 0 (that is p = 2), then the splitting
field of X 3 − 2 is Fp (α, ωα, ω 2 α) = Fp .
To summarise, the full results are as follows.
• If p = 2 or 3 then X 3 − 2 has a single root with multiplicity 3 in Fp , and
so the splitting field is Fp .
• If 2 6= p ≡ 2 (mod 3) then X 3 − 2 has 3 distinct roots in its splitting field,
which is always Fp2 . Precisely one of these roots lies in Fp
• If p ≡ 1 (mod 3) then X 3 − 2 has splitting field Fp if 2(p−1)/3 ≡ 1 (mod p),
and splitting field Fp3 if 2(p−1)/3 6≡ 1 (mod p). In this case, all roots of
X 3 − 2 generate the same field when appended to Fp .
(Note that if p is a prime such that p ≡ 1 (mod 3) then it is reasonably easy to
show that there are unique integers a, b > 0 such that p = a2 + 3b2 . It is much
harder to show that those primes p ≡ 1 (mod 3) for which 2 is a cubic residue
modulo p are exactly those that have the form p = a2 + 27b2 for some integers
a, b > 0.)
√
Question 5.
/ Q, √
and √
from Coursework 2, Question 3, we
√ that √2 ∈
√ We know
/ Q( 2) and
5∈
/√
Q( 2, 3). Therefore repeated application
know that 3 ∈
√ √
of the Tower Law gives |Q( 2, 3, 5) : Q| = 8.
Time permitting, we shall prove a general result that will imply that any finite extension of Q is simple.
Without
√ √
√ that result, and working
√ √from√first principles, we
shall prove that Q( 2, 3, 5) = Q(α) where α = 2+ 3+ √5, though
√ √ there is
a lot of lattitude in the choice of α. It is clear that Q(α) 6 Q( 2, √3, √5), √
so we
seek to prove this inclusion the other way round by showing that 2, 3, 5 ∈
Q(α). We can calculate that
√
√
√
√
√
√
α2 = ( 2 + 3 + 5)2 = 10 + 2 6 + 2 10 + 2 15,
6
a result that does not seem to help much (we still
square roots). Even
√ three
√
√ have
2
the fact that α is in the proper subfield Q( 6, 10, 15) of degree 4 over Q
does not seem to help much.
√
√
√
2
However, we have shown that β := 6 + 10 + 15 = α −10
is in Q(α). Now
2
√
√
√
1
2
we find that β = 31 + 10 6 + 6 10 + 4 15, and so γ := 2 (β 2 − 4β − 31) =
√
√
√
√
2
3 6 + 10 ∈ Q(α).
thus√giving 15 ∈ Q(α). So
√ γ = 64 + 12 15, √
√ So√we get
6+ √
10 ∈ Q(α), and so 6, 10 ∈ Q(α)
now we get β − 15 = √
√ (as
√ we have
already
established
that
3
6
+
10
∈
Q(α)).
From
calculating
6α,
10α and
√
√
√
√
√
15α, we may √
observe√that
√ 3α + √6α −1 √15α
√ = 6 2, and so 2 ∈ Q(α). Thus
1
Q(α) contains 3 = 2 2 6 and 5 = 2 2 10.
√
(If you follow through the above, you can obtain 15 as a degree
√ 4 polynomial
√
in β, and thus
a
degree
8
polynomial
in
α.
The
same
applies
to
6√and √
10.√We
√
√ √
then obtain 2 as a degree 9 polynomial in α, and 3, 5 and 30 = 2 15
as degree 17 polynomials. But the minimal polynomial of √
α has √
degree√8, and is
X 8 − 40X 6 + 352X 4 − 960X 2 + 576, which has 8 roots ± 2 ± 3 ± 5. So in
fact all the above square roots can be written as [now unique] polynomials in α
of degree at most 7. These polynomials are:
√
1
(α7 − 28α5 − 56α3 + 960α),
√2 = 576
1
(−α7 + 37α5 − 244α3 + 360α),
√3 = 96
1
(−α7 + 37α5 − 244α3 + 360α),
√5 = 96
1
(−α6 + 40α4 − 328α2 + 480),
√ 6 = 96
1
(α6 − 36α4 + 216α2 − 256),
√10 = 32
1
6
4
2
√15 = 481 (−α7 + 34α5 − 136α3 − 96)
and 30 = 192 (α − 40α + 376α − 1440α).
Note that I used a computer to help me calculate these polynomials. But I do not
7
recommend
√ √ √calculating
√ √ by
√ hand
√ either 1, α, . . . , α as a Q-linear combination of
1, 2, 3,
√ 5, 6, 10, 15, 30 or (by7 hand) doing the linear algebra required
to write 2, etc, in terms of 1, α, . . . , α .)
Dr John N. Bray, 18th November 2013
7