Zachary Scherr
1
Math 503 HW 10
Due Monday, Apr 25
Reading
1. Read sections 13.1, 13.2, 13.4, 13.5.
2
Problems
1. Show that g(x) = x3 + x + 1 is irreducible in F2 [x] and let γ be a root. Compute
1+γ
1 + γ + γ2
in F2 (γ) = {a + bγ + cγ 2 | a, b, c ∈ F2 }.
Solution: Since deg(g(x)) = 3, it suffices to check that g(x) has no roots in F2 . This follows since
g(0) = g(1) = 1. Under the isomorphism F2 (γ) ∼
= F2 [x]/(g(x)) we need to compute a multiplicative
inverse of 1 + x + x2 (mod 1 + x + x3 ). Using long division we have:
1 + x + x3 = (1 + x + x2 )(x + 1) + x
1 + x + x2 = x(x + 1) + 1.
We can back-solve to obtain:
1 = (1 + x + x2 ) + x(x + 1)
= (1 + x + x2 ) + ((1 + x + x3 ) + (1 + x + x2 )(x + 1))(x + 1)
= (1 + x + x2 )x2 + (1 + x + x3 )(x + 1),
showing that
1
≡ x2
1 + x + x2
(mod 1 + x + x3 ).
Thus in F2 (γ) we have
γ+1
= (γ + 1)γ 2
1 + γ + γ2
= γ3 + γ2
= 1 + γ + γ2.
2. Find all monic irreducible polynomials of degrees 1 and 2 in F3 [x]. Show that the product of all of these
polynomials is x9 − x.
Solution: All linear polynomials are irreducible, so the degree 1 monic irreducibles are:
{x, x + 1, x + 2}.
The pairwise products of these polynomials, allowing for repetition, are:
{x2 , x2 + x, x2 + 2x, x2 + 2x + 1, x2 + 2, x2 + x + 1}.
Thus the irreducible monic degree 2 polynomials are thus
{x2 + 1, x2 + x + 2, x2 + 2x + 2}.
Zachary Scherr
Math 503 HW 10
Due Apr 25
The product is
x(x + 1)(x + 2)(x2 + 1)(x2 + x + 2)(x2 + 2x + 2) = (x3 + 2x)(x2 + 1)(x2 + x + 2)(x2 + 2x + 2)
= (x5 + 2x)(x2 + x + 2)(x2 + 2x + 2)
= (x7 + x6 + 2x5 + 2x3 + 2x2 + x)(x2 + 2x + 2)
= (x9 + 2x)
= x9 − x.
REMARK: It’s true more generally that the product of all polynomials of degree d for all d dividing a
n
fixed n is xp − x over Fp . This is not hard to prove given what we know about finite fields. The finite
n
n
field Fpn is the splitting field of xp − x over Fp . Let p(x) be an irreducible factor of xp − x. If γ is a root
of p(x) then Fp (γ) is a subfield of Fpn . Thus deg(p(x)) = [Fp (γ) : Fp ] divides [Fpn : Fp ] = n. Conversely,
let q(x) be any irreducible polynomial of degree d a divisor of n. If one knows that Fpn contains Fpd
as a subfield, then by uniqueness of finite fields we must have a root of q(x) in Fpn as any root of q(x)
generates a degree d extension of Fp . But then all roots of q(x) are contained in Fpn and since the roots
n
n
of xp − x are precisely the elements of Fpn , we must have q(x) dividing xp − x. The last piece of the
n
proof is to know that xp − x is separable, which is obvious by taking derivatives.
3. Determine the splitting field and its degree over Q for the polynomials
(a) x4 − 2,
(b) x6 − 4,
(c) x4 + x2 + 1.
Solution:
(a) The polynomial x4 − 2 is irreducible by Eisenstein. If i denotes a square root of −1 and
denotes a fourth root of 2, then all the roots of x4 − 2 are given by
√
√
√
√
4
4
4
4
2, i 2, − 2, −i 2.
√
4
2
Thus the splitting field is
√
√
√
√
√
4
4
4
4
4
Q( 2, i 2, − 2, −i 2) = Q( 2, i).
√
It is not asked of you, but the Galois group of the extension Q( 4 2, i)/Q is generated by two
automorphisms, σ and τ where
√
√
√
√
4
4
4
4
σ( 2) = i 2
τ ( 2) = 2
τ (i) = −i.
σ(i) = i
√
It’s easy to see that |σ| = 4 and |τ | = 2. Check that στ = τ σ 3 and hence Gal(Q( 4 2, i)/Q) = D8 .
(b) The polynomial x6 − 4 is a difference of squares, so we may factor to get
x6 − 4 = (x3 − 2)(x3 + 2).
√
Both of the factors on the right hand side are irreducible by Eisenstein. Let 3 2 be a cube-root
of 2 and let ω be a primitive third root of unity ω 3 = 1. Then the six roots of this polynomial
are:
√
√
√
√
√
√
3
3
3
2, ω 2, ω 2 2, i 3 −2, ωi 3 −2, ω 2 i 3 −2.
Page 2
Zachary Scherr
Math 503 HW 10
Due Apr 25
Notice that (−1)3 = −1 so in fact
√
√
√
√
√
√
√
3
3
3
3
Q( 2, ω 2, ω 2 2, 3 −2, ω 3 −2, ω 2 3 −2) = Q( 2, ω).
We proved in class that the Galois group of this extension is D6 = S3 .
(c) We can factor
x4 + x2 + 1 = (x2 − x + 1)(x2 + x + 1).
Both factors on the right are irreducible since they don’t have roots. By the quadratic formula
their roots are
√
√
1 ± −3 −1 ± −3
,
.
2
2
√
Notice that these four elements are
√contained in Q( −3), but
√ any field containing any four of
those elements must also contain −3. This shows that Q(√ −3) is the
√ splitting field. It is a
degree two extension with Galois group Z/2Z generated by −3 7→ − −3.
How might you try to solve this if you couldn’t factor x4 + x2 + 1? One way is the following.
Let t = x2 , then the roots of x4 + x2 + 1 are roots of t2 + t + 1. If ω denotes a primitive cube
root of unity, then the roots of t2 + t + 1 are
t = ω, ω 2 .
Then
√
√
√
x = ± t = ± ω, ± ω 2 .
Notice that (ω 2 )2 = ω 4 = ω and hence these values are
x = ±ω 2 , ±ω.
Thus the splitting field is
Q(ω) = Q
√ √
−1 + −3
= Q( −3).
2
4. Let β1 , β2 , . . . , βn ∈ C be complex numbers for which βi2 ∈ Q, and let L = Q(β1 , β2 , . . . , βn ). Prove that
the polynomial x3 − 3 is irreducible in L[x].
Solution: Let F be a field, and let K/F be an extension of F . If β ∈ K is such that β 2 ∈ F , then I
claim that [F (β) : F ] = 1 or 2. To see this, note that β is a root of x2 − β 2 ∈ F [x]. If this polynomial
is irreducible then we know that [F (β) : F ] = 2. Otherwise, since a quadratic reducible polynomial
must factor into linears we have β ∈ F already, and hence [F (β) : F ] = 1. For this problem, let
F0 = Q and recursively define Fi = Fi−1 (βi ). Since βi2 ∈ Q ⊆ Fi−1 , our above lemma shows that
[Fi : Fi−1 ] = 1 or 2. Thus
[Fn : F0 ] = [Fn : Fn−1 ][Fn−1 : Fn−2 ] · · · [F2 : F1 ][F1 : F0 ]
shows that [Fn : F0 ] = 2k for some nonnegative integer k. Since L = Fn , we have [L : Q] = 2k .
Now, assume for the sake of contradiction that x3 − 3 is reducible in L[x]. The only way a cubic
polynomial can be reducible in L[x] is if it already has a root α ∈ L. But then Q(α) is a subfield of
L, and hence 3 = [Q(α) : Q] divides [L : Q] = 2k which is a contradiction.
Page 3
Zachary Scherr
Math 503 HW 10
Due Apr 25
5. Let F be a field of characteristic p > 0, and let a ∈ F be an element for which the polynomial f (x) =
xp − x − a has no roots in F . Let K/F be the splitting field of f (x).
(a) Let α ∈ K be a root of f (x). Show that α + k is a root of f (x) for every k ∈ Fp .
(b) Prove that every root of f (x) in K has the same degree over F .
(c) Deduce that f (x) is irreducible, that K = F (α), and that [K : F ] = p.
Solution:
(a) Let α ∈ K be a root of f (x). Then for any k ∈ Fp we have
f (α + k) = (α + k)p − (α + k) − a = (αp + k p ) − (α + k) − a = f (α) + (k p − k).
We know that f (α) = 0 and k p = k since k ∈ Fp . Therefore
f (α + k) = f (α) + (k p − k) = 0
showing that α + k is a root of f (x).
(b) We just found p distinct roots of f (x) and since deg(f (x)) = p we know that we’ve found all of
them. For every k ∈ Fp we have k ∈ F , and hence
F (α + k) ⊆ F (α).
Conversely, we have α = (α + k) − k showing that F (α) ⊆ F (α + k) and therefore we get
equality F (α) = F (α + k). Since both of these extensions are simple and algebraic, we get
deg(α) = [F (α) : F ] = [F (α + k) : F ] = deg(α + k).
(c) Suppose we factor f (x) into irreducible f (x) = p1 (x)p2 (x) · · · pk (x) in F [x]. Since we have
found all roots of f (x), we know that for each i there is some k for which
pi (α + k) = 0.
Since the pi (x)’s are irreducible, we therefore have that
deg(pi (x)) = deg(α + k) = deg(α)
for each i. Thus all pi (x) have the same degree d, and hence
X
p = deg(f (x)) =
deg(pi (x)) = kd.
This shows that either d = 1 or d = p. By assumption the polynomial f (x) has no roots in F
and hence d 6= 1. Thus d = p and k = 1 proving that f (x) is irreducible.
We can thus conclude that [F (α) : F ] = deg(f (x)) = p. Moreover, we’ve seen that this field
contains all the roots of f (x) and so K = F (α).
REMARK: From what we’ve done recently, we know that K/F is a Galois extension of degree p. This
forces Gal(K/F ) = Z/pZ since there is only one group of order p. This group is very easy to understand
given what we’ve done in this problem. For each k ∈ Fp = Z/pZ we have an automorphism σk which
acts of K = F (α) via
σk (α) = α + k.
These are all automorphisms since they permute the roots of f (x), and we’ve found p of them, hence we
have them all.
Page 4
Zachary Scherr
6. Let α = 1 +
Math 503 HW 10
√
2+
√
3
Due Apr 25
√
2 ∈ Q( 6 2).
(a) Find a degree 6 polynomial f (x) ∈ Q[x] for which f (α) = 0.
√
(b) Show that 6 2 ∈ Q(α). Deduce that the polynomial f (x) you found in part (a) is irreducible.
Solution:
(a) There √
are several√ways to do this. The easiest way is probably the following. Let β =
Then 3 2 = β − 2. Cubing both sides gives
√
2 = (β 3 + 6β) − (3β 2 + 2) 2.
Rewrite this as
√
2=
√
2+
√
3
2.
β 3 + 6β − 2
3β 2 + 2
and square both sides to get
2=
β 6 + 12β 4 − 4β 3 + 36β 2 − 24β + 4
.
9β 4 + 12β 2 + 4
Thus
β 6 + 12β 4 − 4β 3 + 36β 2 − 24β + 4 = 18β 4 + 24β 2 + 8
which rewrites as
β 6 − 6β 4 − 4β 3 + 12β 2 − 24β − 4 = 0.
Thus β is a root of
g(x) = x6 − 6x4 − 4x3 + 12x2 − 24x − 4.
Then α = β + 1 is a root of
f (x) = g(x − 1) = x6 − 6x5 + 9x4 + 3x2 − 42x + 31.
There are two more computational
yet algorithmic
√
√ ways to attack this problem. Both of them
rely on the fact that α ∈ Q( 6 2). A basis for Q( 6 2) over Q is
B = {1, 21/6 , 21/3 , 21/2 , 22/3 , 25/6 }.
Thus the seven elements
1, α, α2 , α3 , α4 , α5 , α6
√
6
must be linearly dependent in Q( 2). You can express these seven elements in terms of the
basis B and then use row reduction to find a dependence.
√
√
You can also look at the linear transformation T : Q( 6 2) → Q( 6 2) given by T (v) = αv. If you
represent T as M = MBB , then α will have to be a root of the characteristic polynomial of M .
See homework 8, problem 4 for why this works.
√
(b) It’s clear that α ∈ Q( 6 2) since
√
√
6
6
α = 1 + 21/2 + 21/3 = 1 + ( 2)3 + ( 2)2 .
Now, consider β = α − 1 from above. Then certainly Q(β) = Q(α). In part a) we showed
√
2=
β 3 + 6β − 2
3β 2 + 2
Page 5
Zachary Scherr
Math 503 HW 10
which implies that
√
Thus
√
3
In particular,
2 ∈ Q(β) = Q(α).
2=β−
√
6
Due Apr 25
√
2= √
2 ∈ Q(β) = Q(α).
2
√ ∈ Q(α).
2· 32
√
√
Therefore Q(α) = Q( 6 2). Notice that 6 2 is a root of x6 − 2 which is irreducible by Eisenstein
at 2. Thus
√
6
[Q(α) : Q] = [Q( 2) : Q] = 6
implying that the minimal polynomial of α has degree 6. Since we found a degree 6 polynomial
f (x) in part (a) for which α is a root, it follows that f (x) must be irreducible!
7. Let F = Fp (t). Show that f (x) = xp −t is irreducible in F [x] yet is inseparable. Conclude that Frobenius
is not surjective on F .
Solution: Let R = Fp [t] and let F = Fp (t), the fraction field of R. The ring R is a UFD, so a
monic polynomial g(x) ∈ R[x] is irreducible in F [x] if and only if it is irreducible in R[x] already.
The ideal (t) ⊆ R is prime, and so Eisenstein applies immediately to show that xp − t is irreducible
in R[x] and hence F [x] as well. Notice that f 0 (x) = 0 and so
gcd(f (x), f 0 (x)) = f (x) 6= 1
shows that f (x) is inseparable. Frobenius cannot be surjective on F . This is because we proved that
if π is surjective on a field K then every irreducible polynomial in K[x] is automatically separable.
3
Challenge Problems
Challenge Problems tend to be harder than the rest of the problems (and sometimes more interesting). You
do not need to turn these in, but you should get something out of thinking about these.
1. Let F be a field of characteristic p > 0. Show that if a ∈ F is not a p’th power, then the polynomial
k
xp − a is irreducible for every k ≥ 1.
k
Solution: Assume for the sake of contradiction that the polynomial f (x) = xp − a is reducible for
some k ≥ 1. Factor f (x) as f (x) = g(x)h(x) where g(x) and h(x) are non-constant polynomials.
k
Let K be any splitting field of f (x). Then in K there exists α so that 0 = f (α) = αp − a. Thus in
K[x] we have
k
k
k
k
f (x) = xp − a = xp − αp = (x − α)p .
By unique factorization in K[x] we must have
g(x) = (x − α)r
for some integer r in the range 0 < r < pk . Write r = p` · s for some p not dividing s, then since
r < pk we know that ` < k. Then
g(x) = (x − α)r = (x − α)p
Page 6
`
·s
`
`
= (xp − αp )s .
Zachary Scherr
Math 503 HW 10
Due Apr 25
`
`
Using binomial expansion, we see that the coefficient of xp (s−1) in g(x) is −sαp . Since g(x) ∈ F [x],
`
`
we get −sαp ∈ F which shows that αp ∈ F as well. But then
β = αp
k−1
`
= (αp )p
k−`
∈F
and therefore a = β p gives a contradiction.
2. Let a > 1 be an integer, and let n and d be positive integers. Prove that ad − 1 divides an − 1 if and
only if d divides n. Use this to show that Fpd ⊆ Fpn if and only if d divides n.
Solution: Write n = dq + r where 0 ≤ r < d. Then d divides n if and only if r = 0. We have
an − 1 = adq+r − 1
= (adq+r − ar ) + (ar − 1)
= (adq − 1)ar + (ar − 1)
= (ad − 1)[(a(q−1)d + a(q−2)d + · · · + ad + 1)(ar )] + (ar − 1).
Since 0 ≤ r < d we get 0 ≤ ar − 1 < ad − 1. Thus we have found the quotient and remainder when
dividing an − 1 by ad − 1. Thus ad − 1 divides an − 1 if and only if
ar − 1 = 0
which happens if and only if r = 0 or equivalently d divides n.
We shall use this to prove that Fpd ⊆ Fpn if and only if d n (a trivial proof can be given using
Galois theory by looking at the subgroups of Gal(Fpn /Fp ) = Z/nZ.) We know that Fpd ⊆ Fpn if
d
d
n
and only if xp − x = x(xp −1 − 1) has all its roots in Fpn . The group F×
pn is cyclic of order p − 1.
d
d
n
This cyclic group has a subgroup of order p − 1 if and only if p − 1 divides p − 1, which we’ve just
seen happens if and only if d divides n. If d does divide n then by the structure of cyclic groups,
there will be exactly pd − 1 elements of order pd − 1. These elements, along with 0, will be precisely
d
d
the roots of xp − x. Conversely, if d does not divide n then F×
pn contains no elements of order p − 1
d
and hence xp − x does not split. Therefore Fpd ⊆ Fpn if and only if d divides n.
Page 7