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Math 121 Homework 2 Solutions
Problem 13.2 #16. Let K/F be an algebraic extension and let R be a ring
contained in K that contains F . Prove that R is a subfield of K containing
F.
We will give two proofs. The first is more explicit, but the second contains
important ideas that underlie the Theorems in Section 3.1.
First Solution. Let t ∈ R, t 6= 0. We must show that t is invertible in R.
Since t is algebraic over F , it satisfies a polynomial equation f (t) = 0,
an xn + an−1 xn−1 + . . . + a0 = 0,
ai ∈ F.
We may assume that this polynomial is irreducible, and so it is not divisible
by x. This means that a0 6= 0. Since R contains F , a0 is invertible in R, and
so
n−1
t(−a−1
+ . . . + a−1
0 an t
0 a1 ) = 1.
n−1
This means that t−1 = −a−1
+. . .+a−1
0 an t
0 a1 ∈ R.a0 6= 0. Since R contains
F , a0 is invertible in R, and so
n−1
t(−a−1
+ . . . + a−1
0 an t
0 a1 ) = 1.
n−1
This means that t−1 = −a−1
+ . . . + a−1
0 an t
0 a1 ∈ R.
Second Solution. Let t ∈ R, t 6= 0. We give an alternate proof show
that t is invertible in R. It is sufficient to show that t is contained in a
subfield of K that is contained in R. Consider the ring homomorphism
εt : F [x] −→ R that is evaluation at t: thus εt (g(x)) = g(t). The image
of this homomorphism is an integral domain, and so its kernel p is a prime
ideal of F [x]. The ideal p is nonzero because t is algebraic, and if f (x) is
a polynomial it satisfies, then f (x) ∈ p. Since a nonzero prime ideal of a
principal ideal domain is maximal, p is a maximal ideal. Therefore F [x]/p is
a field. By the First Isomorphism Theorem for Rings (Theorem 7 in Chapter
1
7 of Dummit and Foote, page 243), F [x]/p is isomorphic to the image of εt ,
which is therefore a subfield of K. The image of εt contains t = εt (x), and
so we are done.
Problem 13.2 #19. Let K be an extension of F of degree n.
(a) For any α ∈ F prove that multiplication by α is an F -linear transformation.
(b) Prove that K is isomorphic to a subfield of the ring Matn (F ) of n × n
integers over F . Thus Matn (F ) contains a copy of every extension of F of
degree 6 n.
Solution. Part (a) is obvious. Indeed, define Mα : K −→ K to be the map
Mα (x) = αx. Then certainly Mα (x + y) = Mα (x) + Mα (y) and Mα (cx) =
cMα (x) for c ∈ F , so Mα is a linear transformation.
Review of relevant Linear Algebra. If V is an F -vector space, let
EndF (V ) be the space of F -linear transformations of V . (Linear transformations are also called vector space endomorphisms, hence the notation EndF .)
Proposition 1. If V is an n-dimensional vector space then EndF (V ) ∼
=
Matn (F ).
Proof. Choose a basis v1 , · · · , vn of V . If T : V −→ V is a linear transformation, let MT be the n × n matrix (tij ) whose entries are defined by the
equation
X
T (vj ) =
tij vi .
i
Then T 7−→ MT is a ring homomorphism. Indeed, let us check thatP
MT U =
MT · MU (matrix multiplication). Write MU = (uij ) so that U (vj ) = uij vi .
Then
!
X
X
XX
T U (vj ) = T
ukj vk =
ukj T (vk ) =
ukj tik vi .
k
k
k
i
P
Therefore the i, j-th entry in MT U is k tik ukj , proving that MT U = MT MU .
We have defined a ring homomorphism T 7−→ MT from EndF (V ) to Matn (F ).
It is easy to see that it is bijective.
Now we discuss (b). The map α 7−→ Mα is a ring homomorphism K −→
End(V ) ∼
= Matn (F ). It is injective since if α is in then kernel, then Mα is
2
the zero map, so α = Mα (1) = 0. This proves that Matn (F ) contains a copy
of K. But we are asked to show that it contains a copy of every extension of
F of degree 6 n. So we have to consider what happens if [K : F ] is strictly
< n. Let d = [K : F ]. Then we embed K into Matd (F ) as before, then
embed Matd (F ) into Matn (F ) by sending a d × d block X to the matrix
X
,
In−d
where In−d is the (n − d) × (n − d) identity matrix.
Problem 13.2 #20. Show that if the matrix of the linear transformation
“multiplication by α” considered in the previous exercise is A, then α is a root
of the characteristic polynomial of A. This gives an effective procedure for
determining the equation of degree n satisfied by an element of an extension
of F of degree n. Use
obtain the monic polynomials of
√ to3 √
√ this procedure
3
3
2 and 1 + 2 + 4.
degree 3 satisfied by
√
Solution. Let α = 3 2. We choose the basis 1, α, α2 for K = Q(α).
Multiplication by α has this effect on the basis:
1 −→ α
α −→ α2
α2 −→ 2 · 1
and the matrix of this linear transformation is


0 0 2
 1 0 0 .
0 1 0
√
√
On the other hand, multiplication by 1 + 3 2 + 3 4 = 1 + α + α2 has the
matrix


1 2 2
 1 1 2 .
1 1 1
The characteristic polynomials of these two matrices are:
x3 − 1,
x3 − 3x2 − 3x − 1
You can probably compute the characteristic polynomials of these matrices
by hand, or we can use sage as follows.
3
sage: M1=Matrix([[0,0,2],[1,0,0],[0,1,0]])
sage: M1
[0 0 2]
[1 0 0]
[0 1 0]
sage: M1.characteristic_polynomial()
x^3 - 2
sage: M2=Matrix([[1,2,2],[1,1,2],[1,1,1]])
sage: M2
[1 2 2]
[1 1 2]
[1 1 1]
sage: M2.characteristic_polynomial()
x^3 - 3*x^2 - 3*x - 1
Problem 13.4 #1. Determine the splitting field and its degree over Q for
x4 − 2.
√
Solution. Let α = 4 2 be the positive real fourth root of 2. We will prove
that the splitting field K of x4 − 2 is Q(α, i) and that it has degree 8 over Q.
The polynomial x4 − 2 is irreducible over Z by Eisenstein’s criterion (By
Proposition 13 on page 309 of Dummit and Foote) and then over Q by Gauss’
Lemma (Proposition 5 on page 303).
By Proposition 13 on page 309 of Dummit and Foote it is irreducible in
the polynomial ring Z[x]. (Take P to be the prime ideal (3) of Z.) But it
follows from Gauss’ Lemma (Proposition 5 on page 303) that if it can be
factored in Q[x] then it can be factored in Z[x], which it can’t be, so it is
irreducible in Q[x].
The roots of x4 − 2 = 0 are α, iα, −α, −iα since 1, i, −1, −i are the fourth
roots of unity. Since the splitting field K of x4 −2 over Q is the field generated
by all the roots of the polynomial, we see that K = Q(α, iα, −α, −iα). The
last two roots −α and −iα are obviously redundant, so K = Q(α, iα). This
equals Q(α, i). Indeed, both generators α and iα of K are clearly in Q(α, i),
so Q(α, iα) ⊆ Q(α, i). On the other hand, both generators α and i = iα/α
of Q(α, i) are in Q(α, iα) and so Q(α, iα) = Q(α, i).
To prove that [K : Q] = 4, we note that [K : Q] = [Q(α, i) : Q(α)][Q(α) :
Q]. The degree [Q(α) : Q] = 4 because α is a root of an irreducible polynomial x4 − 2 over Q. We claim that [Q(α, i) : Q(α)] = 2. Indeed, this
is obtained by adjoining a root i of a polynomial x2 + 1 of degree 2 over
4
Q(α). So it is sufficient to show that this polynomial is irreducible over
Q(α). If not, since it has degree 2, it would have a root in Q(α). (See Proposition 10 on page 308 of Dummit and Foote.) However the possible i and
−i are not in Q(α) since Q(α) ⊆ R, so this cannot happen. We see that
[Q(α, i) : Q(α)] = 2, so [K : Q] = [Q(α, i) : Q(α)][Q(α) : Q] = 8.
Problem 13.4 #2. Determine the splitting field and its degree over Q for
x4 + 2.
Solution. We will prove that the splitting field for this polynomial x4 + 2 is
the same field K = Q(α, i) as in the previous problem #1, where as before
α = 21/4 .
Let β = εα, where ε = e2πi/8 is a primitive 8-th root of unity. The roots
of the polynomial x4 + 2 = 0 are β, iβ, −β, −iβ. So the splitting field is
Q(β, iβ) = Q(β, i).
But we will show that Q(β, i) = Q(α, i). First note that
1
1
ε = e2πi/4 = √ (1 + i) = 2 (1 + i),
α
2
β = εα =
1
(1 + i).
α
This shows that β ∈ Q(α, i), and obviously i ∈ Q(α, i) so Q(β, i) ⊆ Q(α, i).
On the other hand we may rewrite the last identity
α=
1
(1 + i),
β
and obviously i ∈ Q(β, i), so Q(α, i) = Q(β, i). We have proven that the
splitting fields of the two polynomials x4 − 2 and x4 + 2 are the same.
Problem 13.4 #1. Determine the splitting field and its degree over Q for
x4 + x2 + 1.
Solution. The polynomial x4 + x2 + 1 may be factored: x4 + x2 + 1 =
(x2 + x + 1)(x2 − x + 1). So its roots are the roots of x2 + x + 1 together with
the roots of x2 − x + 1.
√
Let ρ be the primitive cube root of unity ρ = e2πi/3 = − 12 + 23 i.
First consider the polynomial x2 + x + 1 = (x3 − 1)/(x − 1). Its roots are
the primitive cube roots of unity, ρ = e2πi/3 and ρ2 = e−2πi/6 . On the other
hand, the change of variables x → −x turns this polynomial into x2 − x + 1,
so the roots of x2 − x + 1 are −ρ and −ρ2 . All four roots of x4 + x2 + 1 lie
in the field Q(ρ), which is thus the splitting field. Since ρ is a root of an
irreducible quadratic x2 + x + 1, the degree of Q(ρ) over Q is 2.
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Problem 13.4 #5. Let K be a finite extension of F . Prove that K is a
splitting field over F if and only if every irreducible polynomial in F [x] that
has a root in K splits completely in K[x]. (Use Theorems 8 and 27.)
Before proceeding to the solution, a couple of remarks about splitting fields.
First, a remark about the definition of the splitting field K of a polynomial f .
We may assume that f is monic, but we do not assume that it is irreducible.
As we will explain, the definition (Dummit and Foote, page 536) is equivalent
to asserting that K = F (α1 , · · · , αn ), where αi are the roots of f in K, and
that
n
Y
f (x) =
(x − αi ).
(1)
i=1
Indeed, by definition f factors into linear factors in K, so we must have (1).
The definition also assumes that f does not factor into linear factors over
any proper subfield of K. This is equivalent to assuming that K is the field
F (α1 , · · · , αn ) generate by the roots αi . Indeed, letting K 0 = F (α1 , · · · , αn ),
the factorization (1) is valid in K 0 [x] and so K 0 may not be proper, that is,
K = K 0.
Dummit and Foote (on page 536) define the splitting field of a polynomial,
but then in the definition of a normal extension (on page 537) they talk about
the splitting field of a collection of polynomials. Indeed, if the collection is a
finite set, f1 , · · · , fn , then it is obvious from the definition that the splitting
field of this collection of polynomials is the same as the splitting field of
the single polynomial f1 · · · fn . The more general language of the definition
of a normal extension
√ needed
√ is√only
for extension that are algebraic, but
infinite, such as Q 2, 3, 5, · · · , which is a splitting field for an infinite
collection of polynomials. Since we can’t multiply an infinite number of
polynomials together, the more general notion is needed. In problem 13.4,
the field extension is assumed to be finite, so we will interpret splitting field
to mean the splitting field of a polynomial.
Solution. First suppose that K = F (α1 , · · · , αn ) is the splitting field and
let g ∈ F [x] be another polynomial that has a β root in K. The problem
assumes that g is irreducible over F and asks for us to show that g splits in
K. So let L ⊇ K be a splitting field for
m
Y
g(x) =
(x − βi ).
i=1
6
(2)
Since β is one of the roots, we order them so that β = β1 . We have β1 ∈ K
want to show that the other βi ∈ K. Let L = F (α1 , · · · , αn , β1 , · · · , βm ) be
the splitting field of f g. Since β = β1 and βi are roots of the same irreducible
polynomial g, by Theorem 8 on (page 519) there exists a homomorphism
ϕ : F (β) −→ F (βi ) such that ϕ is the identity map on F , and ϕ(β) = βi .
Now we apply Theorem 27 (page 541) using this ϕ and the polynomial f g
to deduce that ϕ can be extended to a homomorphism L −→ L. Since β ∈
F (α1 , · · · , αn ) we may express β as h(α1 , · · · , αn ), where h ∈ F [x1 , · · · , xn ].
Applying ϕ, we obtain
βi = h(ϕ(α1 ), · · · , ϕ(αn )).
Now we claim that ϕ(α1 ), · · · , ϕ(αn ) are some permutation of α1 , · · · , αn .
Indeed, apply ϕ to (1), and remember that since f ∈ F [x], ϕ does not
change the coefficients of the polynomial f . So ϕ(αi ) is a root of the polynomial f , that is, ϕ(αi ) = αj for some j. Thus βi = h(ϕ(α1 ), · · · , ϕ(αn )) ∈
F (α1 , · · · , αi ) = K. Thus all the βi in the splitting (2) are in K. We have
proved that if K is a splitting field, then every polynomial with coefficients
in F that has one root in K splits completely. The implication in the other
direction is obvious from the definition of a splitting field.
Problem 13.5 #1. Prove that the derivative Dx of a polynomial satisfies
Dx (f (x) + g(x)) = Dx (f (x)) + Dx (g(x))
and
Dx (f (x)g(x)) = Dx (f (x))g(x) + f (x)Dx (g(x))
for any two polynomials f and g.
Solution. Of course we cannot use calculus to prove this, but rather the
definition of the derivative from Page 546 of Dummit and Foote. That is, if
f (x) =
n
X
ak xk ,
then
Dx (f (x)) =
k=0
n
X
k ak−1
k .
k=1
The maps f 7−→ Dx f is obviously linear, so the first identity is trivial. For
the second identity, let us check this first when f and g are monomials, say
f (x) = xn and g(x) = xm . Then
Dx (f g) = (n + m)xn+m−1 = nxn−1 · xm + xn · mxm−1 = Dx (f )g + f Dx (g).
7
If f and g are general, we note that both sides of the equation Dx (f (x)g(x)) =
Dx (f (x))g(x) + f (x)Dx (g(x)) are bilinear in f and g, reducing to the case
where f and g are monomials.
Problem 13.5 #2. Find all irreducible polynomials of degree 1, 2 and 4
over F2 and prove that their product is x16 − x.
Observation. A polynomial over F2 has no root in F2 if and only if it has
an odd number of terms (so 1 is not a root) and constant term 1 (so that 0
is not a root).
Solution. The polynomials x and x + 1 of degree 1 are clearly irreducible.
Remember that a polynomial of degree 2 is irreducible in F [x] if and only
if it does not have a root in F .) By the observation, there is one irreducible
of degree 2: x2 + x + 1.
The irreducibles of degree 4 may be found as follows. First, an irreducible
cannot have a root in F2 , and the observation gives us four such polynomials.
One is ruled out: x4 +x2 +1 = (x2 +x+1)2 has no root in F2 , but is reducible.
So here is the result. The product of these six polynomials is x16 + x.
degree irreducible polynomials
x
1
x+1
2
x2 + x + 1
x4 + x3 + x2 + x + 1
x4 + x3 + 1
4
x4 + x + 1
8