GALOIS THEORY EXERCISES ASSIGNMENT 2:
SOLUTIONS
1. Suppose L : K is a field extension with K ⊆ L and τ : L → L is
a K-homomorphism. Suppose f ∈ K[t] with deg f ≥ 1 and α ∈ L.
Show that if f (α) = 0 then f (τ (α)) = 0. Then show that if τ is an
automorphism of L, we have that f (α) = 0 if and only if f (τ (α)) = 0.
Solution: (Proposition 3.1) Suppose that f (α) = 0. Write f =
c0 + c1 t + · · · + cn tn (where n = deg f ). [Since f ∈ K[t], we know all
the ci lie in K.] Using that τ is a K-homomorphism, we have
0 = τ (f (α)) = c0 + c1 τ (α) + · · · + cn τ (α)n = f (τ (α)).
If τ is an automorphism of L, then τ −1 : L → L exists and is a Khomomorphism, so the argument above gives us that if f (τ (α)) = 0
then
0 = τ −1 (f (τ (α))) = f (τ −1 (τ (α))) = f (α).
2. Let L : K be a field extension. Show that Gal(L : K) is a subgroup
of Aut(L).
Solution: First suppose that K ⊆ L. We know the identity map
on L is in Aut(L) and it leaves K pointwise fixed, so the identity
map on L is in Gal(L : K). Now take σ, τ ∈ Gal(L : K). Thus
σ, τ ∈ Aut(L), so σ ◦ τ, σ −1 ∈ Aut(L). Also, for all α ∈ K, σ(α) = α
and τ (α) = α leave K pointwise fixed, so
σ ◦ τ (α) = σ(τ (α)) = σ(α) = α.
Also, σ −1 (α) = α for all α ∈ K [as σ −1 (α) = β when σ(β) = α,
α, β ∈ L]. Hence σ ◦ τ, σ −1 ∈ Gal(L : K). Thus Gal(L : K) is a
subgroup of Aut(L).
Now suppose that L : K is a field extension relative to an embedding ϕ : K → L. Then in the above argument, for α ∈ K
we have σ(ϕ(α)) = ϕ(α), τ (ϕ(α)) = ϕ(α) so σ ◦ τ (ϕ(α)) = ϕ(α)
and σ −1 (ϕ(α)) = ϕ(α). Thus the identity map, σ ◦ τ and σ −1 are
K-homomorphisms. Thus Gal(L : K) is a subgroup of Aut(L).
3. Let M be a field. Show that the following are equivalent:
(i) M is algebraically closed.
(ii) Every nonconstant polynomial f ∈ M [t] factors in M [t] as a
product of linear factors.
(iii) Every irreducible polynomial in M [t] has degree 1.
(iv) The only algebraic extension of M is M itself.
[(Lemma 4.1) There are many ways to do this; here we do not present
all the possibilities. We first show that (i) implies (ii), (ii) implies
(iii), (iii) implies (iv), (iv) implies (i). Then we show the reverse.]
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GALOIS THEORY EXERCISES ASSIGNMENT 2: SOLUTIONS
Solution 1:
Suppose (i) holds. Take f ∈ M [t]rM . Thus f has a root α1 ∈ M .
With n = deg f , define gi inductively as follows. Take g1 ∈ M [t]
so that f = (t − α1 )g1 . For 1 < i ≤ n, take gi ∈ M [t] so that
gi−1 = (t − αi )gi . Since deg gi = n − i, gi−1 is nonconstant for
1 < i ≤ n, and hence has a root αi ∈ M . [Note that gn ∈ M × is the
leading coefficient of f .] Thus
f = gn
n
Y
(t − αi ).
i=1
So (i) implies (ii).
Suppose (ii) holds, and suppose f ∈ M [t] is irreducible. So f
is nonzero and nonconstant [as the nonzero constants in M [t] are
units]. As f factors as a product of deg f linear factors, we must
have deg f = 1. So (ii) implies (iii).
Suppose (iii) holds, and suppose α lies in some algebraic extension
field N : M . Assume M ⊆ N . So α is algebraic over M , and
hence there is some mα (M ) ∈ M [t]. Since mα (M ) is necessarily
irreducible, it has degree 1. Since it is also monic, we have t − α =
mα (M ) ∈ M [t], so α ∈ M . Thus N = M . So (iii) implies (iv).
Suppose (iv) holds. Say f ∈ M [t] r M . Let N be a field extension
of M containing a root α of f (and assume M ⊆ N ). Then M (α) : M
is an algebraic extension, so by (iv), M (α) = M . Hence α ∈ M .
Thus (iv) implies (i).
This shows the equivalence of (i), (ii), (iii), (iv).
Solution 2:
Suppose (iv) holds. Take f ∈ M [t] so that f is irreducible. Since
M is algebraically closed, f has a root α ∈ M . Thus t − α ∈ M [t]
is a factor of f , and so f = (t − α)h for some h ∈ M [t]. Since f is
irreducible, h must be a unit, meaning h ∈ M × . So deg f = 1. So
(iv) implies (iii).
Suppose (iii) holds. Take nonconstant f ∈ M [t]. We know f
factors as a product of irreducible elements of M [t], so f factors as
a product of linear factors. So (iii) implies (ii).
Suppose (ii) holds. Suppose f ∈ M [t] is nonconstant. Then f
factors as a product of linear factors. Let βt − γ be one of these
linear factors; so β 6= 0. Thus β −1 γ is a root of f . So (ii) implies (i).
Suppose (i) holds. Let N be an algebraic extension of M (and
assume M ⊆ N ). Let α ∈ N . Since N : M is an algebraic extension,
mα (M ) exists, and by (i), mα (M ) must have a root β ∈ M . So
in M [t], t − β is a factor of mα (M ). Since mα (M ) is monic and
irreducible, we must have mα (M ) = t − β. Hence α is a root of
t − β, which means that α = β. Thus α = β ∈ M . As this holds for
all α ∈ N , we have N ⊆ M . Since M ⊆ N , we have M = N . Hence
(i) implies (iv).
This shows the equivalence of (i), (ii), (iii), (iv).
GALOIS THEORY EXERCISES ASSIGNMENT 2: SOLUTIONS
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4. (Propostion 4.7) Suppose L, M are fields so that L is algebraically
closed, and ψ : L → M is a homomorphism. Show that ψ(L) is
algebraically closed.
Solution: Suppose f 0 ∈ ψ(L)[t] is irreducible. Thus f 0 = ψ(f ) for
some f ∈ L[t] and deg f 0 = deg f . For the sake of contradiction,
suppose deg f 0 > 1. Thus deg f > 1. Since L is algebraically closed,
irreducible polynomials in L[t] have degree 1; so f is reducible, and
hence f = gh where g, h ∈ L[t] with deg g, deg h ≥ 1. Consequently
f 0 = g 0 h0 where g 0 = ψ(g), h0 = ψ(h), and deg g 0 , deg h0 ≥ 1. But
this contradicts the assumption that f 0 is irreducible in ψ(L)[t]. So
we must have that deg f 0 = 1. Hence ψ(L0 ) is algebraically closed.