2014 Algebra SEP
∼
1. (Jan-10.3) Let
Fields Day 2 Solutions, by E. Dummit
F ⊆E
be nite elds where
|F | = q
(a) Show that every monic irreducible polynomial in
over
F
of some element of
[E : F ] = n.
and
F [x]
(b) Compute the product of all the monic irreducible polynomials in
(c) If
|F | = 2,
n
of degree dividing
is the minimal polynomial
E.
F [x]
of degree dividing
nd the number of monic irreducible polynomials of degree 10 in
n.
F [x].
Solution:
a)
If
[F (α) : F ] divides [E : F ] = n. We also know the Frobenius map ϕ : x → xq is an
automorphism of E/F , so it is therefore a generator for the Galois group: the degree of the extension
q n−1
hence the order of the Galois group is n and σ is also order n (because x
− x only has q n−1 roots in
E ). In particular, we see that the intermediate elds between E and F are the xed elds of powers of
σ , and are thus the elds whose orders are q k where k divides n. Then if p(x) is any monic irreducible
deg(p)
polynomial over F of degree dividing n, the extension F [x]/p(x) is a eld of q
elements, hence is
isomorphic to F (α) for α ∈ E .
α ∈ E
Note
then
This argument assumes the well-known fact that there is a unique nite eld (up to isomorphism) of
any prime-power order. This in fact follows from the argument in part (b), if one wishes to be especially
pedantic.
b)
The product is
|E × | = q n − 1
n
xq − x.
To see this observe that every element of
x ∈ E×
xq
n
−1
satises
= 1;
E
is a root of this polynomial, since
x adds 0 as a root. Since
q n , we have found all the roots. Now we observe that every monic irreducible
in F [x] of degree dividing n must divide this polynomial (since by part (a) all of its roots lie in E ), and
conversely this polynomial can have no other type of irreducible factors over F (because it is separable,
all its roots lie in E , and the minimal polynomial of every element of E has degree dividing n).
so by Lagrange every
multiplying by
this polynomial is degree
c)
The product of all of the irreducibles of degree dividing
n
is
n
x2 − x.
By inclusion-exclusion (or Mobius
inversion) the degree of the product of the irreducibles of exact degree
10
is
210 − 25 − 22 + 21 = 990,
and since each has degree 10, there are 99 of them.
2. (Jan-09.3): Suppose
f (x) = xm + 1
(a) Show that every root of
(b) Show that
2m
(c) Show that
m 6= 4.
divides
f
m
is irreducible over
in a splitting eld of
p −1
Fp [x]
f
but does not divide
where
p
is an odd prime.
has multiplicative order
n
p −1
for any
n
with
2m.
0 < n < m.
Solution:
a)
If
b)
An irreducible polynomial of degree
αm + 1 = 0 then α2m − 1 = 0 so α2m = 1. Thus the order of any root divides 2m. It cannot be equal to
m since then αm − 1 = 0 whence 2 = 0 and the characteristic is not 2. It also cannot be less than m since
m
then α would satisfy a polynomial of degree less than that of x + 1, and hence taking the polynomial
m
gcd would give a nontrivial factor of x + 1.
m over Fp has splitting eld Fpm . Since by part (a) every root of
m
xm + 1 has multiplicative order 2m, we see that 2m must divide F×
pm = p − 1. If 2m were to divide
n
p − 1 for 0 < n < m, then since over Fpm the polynomial factors as (x − α)(x − α3 ) · · · (x − α2m−1 )
m
where α is any root (since if α
= −1 then α(2k+1)m = −1 for any integer k ), we see that Fpn would
m
contain an element of multiplicative order 2m hence contain a root of x + 1 hence contain all roots of
m
m
x + 1. But this would contradict the irreducibility of x + 1, since it would only generate an extension
of degree n < m.
c) Suppose m = 4. Then if p = 2k + 1 we have p2 − 1 = (2k + 1)2 − 1 = 8 · k2 , so p2 − 1 is divisible by
2m = 8, violating the condition of (b).
3. (Jan-13.2): Let
k
be a eld. We say a polynomial
(not necessarily in
k)
such that
f (x) ∈ k[x]
is consecutive-root if it has two roots
(a) Show that there is no irreducible consecutive-root polynomial in
(b) Let
p
be a prime. Show that
p
x −x−1
Q[x].
is consecutive-root and irreducible in
(c) Characterize all irreducible monic consecutive-root polynomials in
Note
Fp [x]
Fp [x].
of degree
≤ p.
Compare to Jan-92.3.
Solution:
x1 ),
Suppose
f (x)
is consecutive-root. By hypothesis,
so their gcd has positive degree. If
f (x) and f (x + 1) have the same
f is clearly consecutive-root.
since
then
a)
x0 , x1
x1 − x0 = 1.
f
f (x)
and
f (x + 1)
have a common root (namely,
f , but
f (x) = f (x + 1),
is irreducible, then this forces the gcd to be equal to
degree, we see
f (x) = f (x + 1).
Conversely, if
From the above observation we see immediately that there are no irreducible consecutive-root polynomials
in
Q[x],
since it is not possible for
f (x)
to equal
f (x + 1),
as the second-highest-degree terms are not
equal.
b)
For
c)
From the criterion at the beginning, we see that
xp − x − 1 we see that f (x) = f (x + 1), since (x + 1)p − (x + 1) − 1 = (xp + 1) − (x + 1) − 1 = xp − x − 1
in characteristic p. Hence f is consecutive-root. To see it is irreducible, observe that the Galois action
p
of Fp [x]/(x − x − 1) over Fp in this extension of nite elds is Frobenius, the pth power, but since
p
x 7→ x = x + 1 is the same as addition by 1. But now the Galois action is transitive on the roots, so the
p
polynomial is irreducible. (Alternatively, if q(x) is a divisor of x − x − 1, and α is any root, then the
sum of the roots of q(x) is deg(q)α + r where r ∈ Fp ; since this term is also in Fp we see that α would
necessarily be in Fp but it is easy to see that f has no roots in Fp .)
f (x) = f (x + 1) = · · · = f (x + p − 1), and so we see
f (0) = f (1) = · · · = f (p − 1) = a. Hence f (x) − a is identically zero on Fp , so it is divisible by
x(x − 1) · · · (x − (p − 1)), which by standard nite eld facts is xp − x. Hence since f has degree ≤ p and
p
p
is monic, we necessarily have f (x) − a = x − x, so f (x) = x − x + a for some a ∈ Fp . We see that, by
the same argument in part (b), this polynomial is irreducible as long as a 6= 0.
Remark
These extensions are called Artin-Schreier extensions, and they characterize all (cyclic) degree-p
Galois extensions of
4. (Jan-89.3) Prove that
Solution:
Fp .
x9 − 2
is an irreducible factor of
x27 − 1
over
F7 .
(x9 − 2)(x18 + 2x9 + 4) = x27 − 8 (over Z, even). For the irreducibility, let E be the
9
splitting eld of x − 2 over F7 . Each root has multiplicative order 27 in E , since none of them has order
9
9
9 (as α = 2, not 1, for any root α of x − 2). Therefore, if d = [E : F7 ], then it must be the case that
×
d
|E | = 7 − 1 is divisible by 27. Reducing mod 9 gives (−2)d = −1 mod 9, so d ≡ 3 mod 6. We also
3
see that d = 3 does not work since 7 − 1 = 342 is not divisible by 27, so it must be true that d ≥ 9.
9
However, obviously d = 9 does work because every root of x − 2 lives in an extension of degree at most
9
9: thus we see that the splitting eld of x − 2 over F7 is degree 9, hence the polynomial is irreducible.
Remark
over
We have
Using an analysis like the above, one can show that the full factorization of
F7
is
3
3
9
9
(x − 1)(x − 2)(x − 4)(x − 2)(x − 4)(x − 2)(x − 4).
x27 − 1 into irreducibles
E⊆C
5. (Aug-08.3): Let
(a) Show that
(b) If
α∈E
a)
over
Q.
[E : Q] = 6.
and
α5 ∈ Q
α ∈ Q.
show that
β∈E
(c) Show that there exists
Solution:
x3 − 2
be the splitting eld of
with
√
E = Q( 3 2, ζ3 ) where ζ3
β2 ∈ Q
but
β 6∈ Q.
x3 −2 is irreducible over Q by Eisenstein's
√
3
2
criterion or the fact that it has no roots, so Q[ 2] : Q = 3. Similarly, [Q(ζ3 ) : Q] = 2 since x + x + 1
has no roots. Then [E : Q] ≤ 6 since [EF : Q] ≤ [E : Q] · [F : Q] but it is also divisible by 2 and 3, hence
Clearly,
is a nonreal cube root of unity.
must be 6.
b)
x5 − α5 ∈ Q[x]
First observe that
cannot be irreducible, since otherwise
Q[α]
would have degree 5 over
Q
so it must have a linear or quadratic factor. If it has a quadratic factor, the constant term must be
αζ5−2d = α5 (α2 ζ5d )−2 is in Q, but this is one of the roots of x5 − α5 . Thus, x5 − α5
d
has a linear factor over Q, so αζ5 ∈ Q for some 0 ≤ d ≤ 4. If d 6= 0 then Q[α] = Q[ζ5 ] has degree 4 over
Q, which is impossible since Q[α] is a subeld of E . We conclude α ∈ Q.
α2 ζ5d
for some
Remark
d:
then
E chosen here: the argument given above actually
α ∈ Q for any prime p such that neither p nor p − 1 divides [E : Q].
p
p
To see this, consider the minimal polynomial of α over Q: by hypothesis it must divide x − α , so its
d
r
constant term is some product of Galois conjugates of α hence is of the form α · ζp for some d and r . If
d < p then d is invertible mod p: then for a, b with ad − bp = 1, we have (αd ζpr )a · (αp )−b = α · ζpra is in
Q. Hence [Q[α] : Q] is either 1 or p − 1 (depending on whether ra is divisible by p or not). Otherwise, if
d = p, we see xp − αp is irreducible over Q, so [Q[α] : Q] = p. Thus, since α ∈ E , we require one of p − 1
or p to divide [E : Q].
√
√
c) β = −3 is in E , since −3 = 1 + 2ζ3 , and ζ3 ∈ E .
The result of (b) has very little to do with the eld
shows that
αp ∈ Q and α ∈ E
implies
f (x) = x6 + 3 ∈ Q[x],
6. (Aug-96.3): Let
let
α
be a root of
(a) Show that
E
contains a primitive 6th root of unity.
(b) Show that
E
is Galois over
f
over
C,
and set
Q.
(c) Find the number of intermediate elds
F
with
Q⊂F ⊂E
with
[F : Q] = 3.
Solution:
a)
If
6
α = −3,
thenα
√
3
= ± −3,
so (in either case)
√
−3 ∈ E .
Then
sixth root of unity.
b)
and
c)
f
The other roots of
ζ6
are both in
are
E
αζ6k
E = Q[α].
1
+
2
√
−3
∈ E,
2
and this is a primitive
k = 1, · · · , 5 and ζ6 = eiπ/3 is a primitive sixth root
E is the splitting eld of f (x), hence it is Galois.
where
we see that
of unity. Since
α
The Galois group has order 6 so the question is whether it is abelian or not. Complex conjugation and
multiplication by
−1
are both elements of the Galois group (since
order 2, so the Galois group must be nonabelian, hence
to subgroup of index 3 = order 2 in
c-alt)
If the extension
E/Q
3
S3 ,
hence
not
3
A3
we see that
and
S3
1/3
Q.
and have
Then the extensions of degree 3 correspond
of which there are 3 (generated by the 3 transpositions).
were abelian then by the Kronecker-Weber theorem
some cyclotomic extension of
1/6
S3 ,
S3 .
ᾱ6 + 3 = (−α)6 + 3 = 0)
E
would be contained in
By adjoining additional roots of unity we then see that
would also be contained in a cyclotomic extension. But because
3
x −3
(−3)1/6
hence
has Galois group
(complex conjugation is an element of order 2, or because its discriminant is not a square),
31/3
is not).
is not contained in any cyclotomic extension (since cyclotomic extensions are abelian,
7. (Aug-09.3): Let
F
(a) Show that
(b) If
f (x) ∈ F [x] irreducible with splitting eld E .
g(x) ∈ F [x] irreducible with g(αn ) = 0.
be a eld and
and a positive integer
n
deg(g)
and let
divides
deg(f )/ deg(g) = n
deg(f )
and
Choose
α∈E
with
f (α) = 0
deg(f )/ deg(g) ≤ n.
and the characteristic of
F
does not divide
n,
show
E
contains a primitive
nth
root of unity.
Solution:
E 0 = F [αn ]. By the assumptions, g(x) is the minimal polynomial of αn , so [E 0 : F ] = deg(g) and
[E : F ] = deg(f ). Then deg(g) · [E 0 : E] = deg(f ) and [E 0 : E] ≤ n since α satises the polynomial
xn − αn ∈ E 0 [x].
a)
Let
b)
If equality holds, the polynomial
xn − α n
must be irreducible over
E0.
Now since
E
is Galois over
F
(it
is a splitting eld), it is also Galois over any intermediate extension hence in particular it is Galois over
E0.
E is the splitting eld of xn − αn over E 0 , and so since the polynomial is separable (its
n−1
n
n
derivative is nx
which is relatively prime to x − α by the assumption on the characteristic) all of
k
its roots lie in E . In particular, since its roots are α · ζn for 0 ≤ k ≤ n − 1 where ζn is a primitive nth
root of unity, E contains α and αζn hence contains ζn .
Therefore,