Download Solutions to Homework 7 27. (Dummit

Solutions to Homework 7
√
27.√ (Dummit-Foote 13.2 #21)√Let α = a√+ b D. Considering
K as a Q-vector space with basis
√
1, D, we see that α·1 = a+b D and α· D = bD +a D. Therefore the matrix for multiplication
by α is
a bD
.
b a
It is straightforward to verify that α 7→ (the above matrix) is a ring homomorphism, and it is
clearly non-zero. Since K is a field it has no non-zero ideals and thus our map is injective. Since
it is obviously surjective, we are done.
(Dummit-Foote 13.2 #22) Let {αi } be a basis for K1 over F , and let {βj } be a basis for K2
over F . Then {αi ⊗ βj } is a basis for K1 ⊗F K2 over F . Define a map φ : K1 ⊗F K2 → K1 K2 by
φ(αi ⊗ βj ) = αi βj , and extend it by linearity. It is easy to check that φ is an F -algebra homomorphism. The map φ is surjective because the elements αi βj span K1 K2 as an F -vector space.
(⇐) By assumption the F -vector spaces K1 ⊗F K2 and K1 K2 have the same dimension over
F , namely [K1 : F ][K2 : F ]. Thus by linear algebra φ is injective and hence an isomorphism.
Therefore K1 ⊗F K2 is isomorphic to the field K1 K2 .
(⇒) Conversely, if K1 ⊗F K2 is a field then it has no non-zero ideals so ker(φ) = 0. Therefore
φ is an isomorphism and [K1 K2 : F ] = [K1 ⊗F K2 : F ] = [K1 : F ][K2 : F ].
28. (Dummit-Foote 13.4 #5) (⇒) Suppose K is a splitting field. Let f ∈ F [X] be irreducible, and
let α be a root of f in K. Let β be an arbitrary root of f . By Theorems 8 and 27, there is a
commutative diagram
∼ /
K(α)
K(β)
O
O
F (α)
∼
/ F (β)
O
id
/F
O
F
of maps of fields. But α ∈ K implies that [K(β) : K] = [K(α) : K] = 1 and thus β ∈ K as desired.
(⇐) Because K/F is a finite extension, we have K = F (α1 , . . . , αn ) for some αi ∈ K. Let fi be
the minimal polynomial
of αi , then fi splits completely in K by assumption. Therefore K is the
Q
splitting field of i fi .
(Dummit-Foote 13.4 #6) (a) Let K1 be the splitting field of f1 and let K2 be the splitting field
of f2 . Let K be the splitting field of f = f1 f2 . We will show that K = K1 K2 . It is clear that
K ⊆ K1 K2 because all of the roots of f lie in K1 K2 . Conversely, if K splits f then it certainly
splits f1 , hence K1 ⊆ K. Similarly, K2 ⊆ K. Therefore K = K1 K2 .
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(b) We will use the previous exercise. Let f be a polynomial with a root α in K1 ∩ K2 . Then
α ∈ K1 and α ∈ K2 and because each of these is a splitting field, all of the other roots of f must
be in both K1 and K2 . Therefore f splits completely in K1 ∩ K2 , which proves that K1 ∩ K2 is a
splitting field.
√
√
√
30. Let ζ = 2/2 + i 2/2, then ζ is√a primitive√eighth root of
unity. If 8 2 denotes the
real
√
√
8
8
8
8
positive √
eighth root of 2, then
K
=
Q(
2,
ζ)
=
Q(
2,
i).
As
Q(
2)
⊆
R,
we
have
i
∈
6
Q(
2)
so
√
√
8
8
8
Q ( Q( 2) ( K. Since [Q( 2) : Q] = 8 and [K : Q( 2)] = 2, we have [K : Q] = 8 · 2 = 16.
31. Let f ∈ K[X1 , . . . , Xn ] and suppose f is in the intersection of all of the maximal ideals of
K[X1 , . . . , Xn ]. Then 1 + f g is a unit (hence in K × ) for all g ∈ K[X1 , . . . , Xn ]. Setting g = 1 we
have that f must be constant, and as it is contained in every maximal ideal, we must therefore
have f = 0.
32. We have a tower of fields K ⊆ K(a2 ) ⊆ K(a). The degree [K(a) : K] is equal to the degree of
the minimal polynomial of a, which is odd. By multiplicativity of degrees of field extensions, we
see that [K(a) : K(a2 )] must also be odd. However, a satisfies the polynomial X 2 − a2 ∈ K(a2 )[X]
which is degree two. Therefore [K(a) : K(a2 )] ≤ 2, so we have [K(a) : K(a2 )] = 1 and thus
K(a) = K(a2 ).
2