M502 Midterm
1. Let α, β ∈ C each have degree 3 over Q. Determine the possibilities for
|Q(α, β) : Q|.
√
√
√
3
3
2. Show Q is square
root
closed
in
Q[
2],
that
is,
if
d
∈
Q[
2] for some
√
d ∈ Q, then d ∈ Q.
3. Write down an explicit generator for the cyclic group Gal(Q[ζ7 ]/Q)
4. Give an example of an inseparable field extension.
5. Express the cyclotomic polynomial Φ10 (x) as a polynomial with integral
coefficients.
6. Suppose K/Q is a normal extension of degree 25. Is every element of
K expressible by radicals? Why or why not?
7. Let L/F be a field extension and α ∈ L. If dimF F [α] < ∞, show that
F [α] is a field.
8. Define a normal field extension L/K. Define an algebraic field extension
L/K. Define a separable field extension L/K.
9. Show that if K/F is a finite separable extension, then there are a finite
number of fields L so that F ⊂ L ⊂ K.
10. Let L/K be a Galois extension with Galois group G and degree n.
Show that G is isomorphic to a transitive subgroup of Sn . (You may
assume the theorem of the primitive element.)
1
Solutions
1. Let α, β ∈ C each have degree 3 over Q. Determine the possibilities for
|Q(α, β) : Q|.
|Q(α, β) : Q| = |Q(α) : Q||Q(α, β) : Q(α)| = 3|Q(α, β) : Q(α)| =
3(deg mβ,Q[α] (x)) = 3, 6, or 9, since mβ,Q[α] (x) is a divisor of mβ,Q (x)
which
has
degree
3.
Examples
are
given
by
√
√
√
√
√
3
3
3
3
3
(α, β) = ([ 2], −[ 2]), ([ 2], ζ3 ), ([ 2], −[ 3])
√
√
√
3
3
2. Show Q is square
root
closed
in
Q[
2],
that
is,
if
d
∈
Q[
2] for some
√
d ∈ Q, then d ∈ Q.
√
√
√
√
√
3
3 = |Q[ 3 2]
√: Q| = |Q[ 2] : Q[ d]||Q[ √d]| : Q| and |Q[ d] : Q| = 1, 2.
Hence |Q[ d]| : Q| = 1 which implies d ∈ Q
3. Write down an explicit generator for the cyclic group Gal(Q[ζ7 ]/Q)
ζ7 7→ ζ73 (since 3 generates the cyclic group Z/7× ).
4. Give an example of an inseparable field extension.
√
F2 ( t)/F2 (t)
5. Express the cyclotomic polynomial Φ10 (x) as a polynomial with integral
coefficients.
Φ10 (x) =
x10 −1
(x5 −1)(x+1)
= x4 − x3 + x2 − x + 1
6. Suppose K/Q is a normal extension of degree 25. Is every element of
K expressible by radicals? Why or why not?
Yes. The Galois group is solvable since it is a p-group. Then use
Galois’ Theorem
7. Let L/F be a field extension and α ∈ L. If dimF F [α] < ∞, show that
F [α] is a field.
We just need to show that nonzero elements have multiplicative inverses. Let β ∈ F [α] − 0. Then ·β : F [α] → F [α] is an injective linear
transformation. Since dimF F [α] equals the rank plus the nullity, ·β is
onto, so 1 is in the image.
2
8. Define a normal field extension L/K. Define an algebraic field extension
L/K. Define a separable field extension L/K.
L/K is normal if L/K is algebraic and every homomorphism σ : K →
K which is the identity on K satisfies σL = L. L/K is algebraic if
every element α ∈ L is a root of a monic polynomial with coefficients
in K. L/K is separable if L/K is algebraic and the minimal polynomial
of every element of L has no multiple roots in its algebraic closure.
9. Show that if K/F is a finite separable extension, then there are a finite
number of fields L so that F ⊂ L ⊂ K.
There is a Galois extension E/F so that K ⊂ E, for example, if K =
F [α1 , . . . , αk ] one could let E be the field over K generated by all the
conjugates of the αi . Since there are a finite number of subgroups of
Gal(E/F ) the Fundamental Theorem of Galois Theorem shows that
there are a finite number of intermediate fields F ⊂ L ⊂ E. The result
follows.
10. Let L/K be a Galois extension with Galois group G and degree n.
Show that G is isomorphic to a transitive subgroup of Sn . (You may
assume the theorem of the primitive element.)
Let L = K[θ1 ] and let m(x) be the minimal polynomial of θ1 . Since
K[θ1 ] ∼
= K[x]/hm(x)i, the degree of m(x) is n. Let m(x) = (x −
θ1 ) · · · (x − θn ) ∈ K[x] with the θi distinct elements of L. Note that for
σ ∈ G, 0 = σ(0) = σ(m(θi )) = m(σ(θi )), hence σ(θi ) = θj some j.
Hence there is a function Φ : G → Sn satisfying σ(θi ) = θΦ(σ)i which
is easily seen to be a homomorphism. Φ is injective since L = K[θ1 ].
Φ(G) is transitive since for any i, j, K[θi ] ∼
= K[x]/hm(x)i ∼
= K[θj ], so
there is a σ ∈ G so that σ(θi ) = θj .
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