<<

. 9
( 18 .)



>>

2 iiiii 2 qqqq V
2 VV
i
iiii qq VV
qqq
ii
iiii √ 2 VVV3
√i √
Q( 3 2 ζ3 ) VV
3 3
Q( 2) Q( 2 ζ3 )
www `` VV
``
www VV
``
www VV
``
www
`3
www
3
www ``` Q(ζ3 )
3
www `` r
2 rr
www `` rrr
` rr
www `
w rrr
Q
√ √2
An alternative strategy would have been to adjoin one of the other roots 3 2 ζ3 or 3 2 ζ3
¬rst. We could also have begun by adjoining ζ3 to form the extension Q(ζ3 )/Q, but none of

the roots of p(X) lie in this ¬eld so the extension Q( 3 2, ζ3 )/Q(ζ3 ) of degree 3 is obtained by
adjoining one and hence all of the roots.

3.3. Monomorphisms between extensions
Definition 3.31. For extensions F/K and L/K, let MonoK (L, F ) denote the set of all
monomorphisms L ’’ F which ¬x the elements of K.
Remark 3.32. We always have AutK (F ) ⊆ MonoK (F, F ) and MonoK (F, F ) is closed under
composition but is not always a group since elements are not necessarily invertible. Of course,
if F/K is ¬nite, then MonoK (F, F ) = AutK (F ) since every injective K-linear transformation is
surjective and so invertible.
We will also use the following notation.
Definition 3.33. Let F/K be an extension and p(X) ∈ K[X]. Set
Roots(p, F ) = {u ∈ F : p(u) = 0},
the set of roots of p(X) in F . This is always a ¬nite set which may of course be empty.
Suppose that p(X) ∈ K[X] is an irreducible polynomial which we might as well assume is
monic, and F/K an extension. Then if t ∈ F is a root of p(X), the evaluation homomorphism
µt : K[X] ’’ F factors through the quotient monomorphism µt : K[X]/(p(X)) ’’ F whose
image is K(t) F . Of course, there is one such monomorphism for each root of p(X) in F . If
we ¬x one such root t0 and identify K[X]/(p(X)) with K(t0 ) via µt0 , then each root of p(X) in
F gives rise to a monomorphism •t = µt —¦ µ’1 : K(t0 ) ’’ F for which •t (t0 ) = t.
t0

•t =et —¦e’1
µ µt
0

et0
µ et
µ
/* F
K(t0 ) o K[X]/(p(X))

=

Notice that if • : K[X]/(p(X)) ’’ F is any homomorphism extending the identity function
on K, then the coset X + (p(X)) must be sent by • to a root of p(X) in F , hence every such
homomorphism arises this way. This discussion is summarized in the following result.
36 3. ALGEBRAIC EXTENSIONS OF FIELDS

Proposition 3.34. Let F/K be a ¬eld extension. Let p(X) ∈ K[X] be an irreducible
polynomial with t0 ∈ F be a root of p(X). Then there is a bijection
Roots(p, F ) ←’ MonoK (K(t0 ), F )
given by t ←’ •t , where •t : K(t0 ) ’’ F has the e¬ect •t (t0 ) = t.

Example 3.35. Show that MonoQ (Q( 2), C) has two elements.

Solution. We have Q( 2) ∼√ = Q[X]/(X 2 ’ 2) where X 2 ’ 2 is irreducible over Q. Hence the

Q-monomorphisms send 2 to ± 2. In fact both possibilities occur, giving monomorphisms

id, ± : Q( 2) ’’ C, where
√ √
±(a + b 2) = a ’ b 2.

We can replace C by Q( 2) to obtain
√ √ √ √
MonoQ (Q( 2), C) = MonoQ (Q( 2), Q( 2)) = AutQ (Q( 2)).
We will see that this is not always true.
√ √ √
Example 3.36. Show that MonoQ (Q( 3 2), C) has 3 elements but MonoQ (Q( 3 2), Q( 3 2))
only contains the identity function.
√√
Solution. Here minpolyQ, √2 (X) = X 3 ’ 2 and there are 3 complex roots 3 2, 3 2 ζ3 ,
3
√2 √ √
2 ζ3 . As two of these roots are not real, MonoQ (Q( 2), Q( 3 2)) contains only the identity
3 3

since Q( 3 2) R. √ √ √2
Each of the above roots corresponds to one of the sub¬elds Q( 3 2), Q( 3 2 ζ3 ) or Q( 3 2 ζ3 )

of C and there are 3 monomorphisms ±0 , ±1 , ±2 : Q( 3 2) ’’ C given by
√ √2 √ √2
3 3 3 3
±0 (a + b 2 + c( 2) ) = a + b 2 + c( 2) ,
√ √ √ √
±1 (a + b 2 + c( 2)2 ) = a + b 2 ζ3 + c( 2)2 ζ3 ,
2
3 3 3 3

√ √ √2 √
±2 (a + b 2 + c( 2)2 ) = a + b 2 ζ3 + c( 2)2 ζ3 .
3 3 3 3



The images of these mappings are
√ √ √ √ √ √2
3 3 3 3 3 3
±0 Q( 2) = Q( 2), ±1 Q( 2) = Q( 2 ζ3 ), ±2 Q( 2) = Q( 2 ζ3 ).
Proposition 3.37. Let F/K and L/K be extensions.
(i) For p(X) ∈ K[X], each monomorphism ± ∈ MonoK (L, F ) restricts to a function
±p : Roots(p, L) ’’ Roots(p, F ) which is an injection.
(ii) If ± ∈ MonoK (L, L), then ±p : Roots(p, L) ’’ Roots(p, L) is a bijection.
Proof. (i) For u ∈ Roots(p, L) we have
p(±(u)) = ±(p(u)) = ±(0) = 0,
so ± maps Roots(p, L) into Roots(p, F ). Since ± is an injection its restriction to Roots(p, L) ⊆ L
is also an injection.
(ii) From (i), ±p : Roots(p, L) ’’ Roots(p, L) is an injective function from a ¬nite set to itself,
hence it is also surjective by the Pigeon Hole Principle. Thus ±p : Roots(p, L) ’’ Roots(p, L)
is a bijection.

Part (ii) says that any automorphism of L/K permutes the set of roots in L of a polynomial
p(X) ∈ K[X]. This gives us a strong hold on the possible automorphisms. In the case of ¬nite,
or more generally algebraic, extensions it is the key to understanding the automorphism group
and this is a fundamental insight of Galois Theory.

Example 3.38. Determine MonoQ (Q( 3 2, ζ3 ), C).
3.4. ALGEBRAIC CLOSURES 37

Solution. We have already met the extension Q( 3 2, ζ3 )/Q in Example 3.30 and we will
make use of information from there. We build up the list of monomorphisms √ stages.
in

First consider monomorphisms that ¬x 2 and hence ¬x the sub¬eld Q( 3 2). These form
3


the subset √ √
3 3
MonoQ( √2) (Q( 2, ζ3 ), C) ⊆ MonoQ (Q( 2, ζ3 ), C).
3
√ √
We know that Q( 2, ζ3 ) = Q( 3 2)(ζ3√ and that ζ3 is a root of the irreducible cyclotomic
3
)
2 + X + 1 ∈ Q( 3 2)[X]. So there are two monomorphisms id, ± ¬xing
polynomial ¦3 (X) = X 0

3
Q( 2), where ±0 has the e¬ect
√ √
3
2 ’’ 3 2
±0 : .
2
ζ3 ’’ ζ3
√ √
Next we consider monomorphisms that send √ 2 to 3 2 ζ3 . This time we have 2 distinct ways to
3

extend to elements of MonoQ (Q( 3 2, ζ3 ), Q( 3 2, ζ3 )) since again we can send ζ3 to either ζ3 or
2
ζ3 . The possibilities are
√ √
√ √ 3
2 ’’ 3 2 ζ3
3
2 ’’ 3 2 ζ3
±1 : , ±1 : .
2
ζ3 ’’ ζ3 ζ3 ’’ ζ3
√ √2
Finally we consider monomorphisms that send 3 2 to 3 2 ζ3 . There are again two possibilities
√ √2
√ √2 3
2 ’’ 3 2 ζ3
3 3
2 ’’ 2 ζ3
±2 : , ±2 : .
2
ζ3 ’’ ζ3 ζ3 ’’ ζ3
These are all 6 of the required monomorphisms. It is also the case here that
√ √ √ √
3 3 3 3
MonoQ (Q( 2, ζ3 ), C) = MonoQ (Q( 2, ζ3 ), Q( 2, ζ3 )) = AutQ (Q( 2, ζ3 )),

so these form a group. It is a nice exercise to show that AutQ (Q( 3 2, ζ3 )) ∼ S3 , the symmetric
=√

group on 3 objects. It is also worth remarking that | AutQ (Q( 2, ζ3 ))| = [Q( 3 2, ζ3 ) : Q].
3



We end this section with another useful result.
Proposition 3.39. Let L/K be an extension and ± ∈ MonoK (L, L). Then ± restricts to
an automorphism ±alg : Lalg ’’ Lalg .
Proof. Suppose that u ∈ Lalg , say p(u) = 0 for some p(X) ∈ K[X] of positive degree.
Then
p(±(u)) = ±(p(u)) = ±(0) = 0,
so ± maps Lalg ⊆ L into itself and therefore gives rise to a restriction ±alg : Lalg ’’ Lalg which
is also a monomorphism. We must show that ±alg is a bijection by showing it is surjective.
Let v ∈ Lalg and suppose that q(v) = 0 for some q(X) ∈ K[X] of positive degree. Now
Roots(q, L) = … since it contains v, and it is also ¬nite. Then ±q : Roots(q, L) ’’ Roots(q, L)
is a bijection by Proposition 3.37(ii), hence v = ±q (w) = ±(w) for some w ∈ Roots(q, L) ⊆ Lalg .
This shows that v ∈ im ± and so ±alg is surjective.

3.4. Algebraic closures
An important property of the complex numbers is that C is algebraically closed.
Theorem 3.40 (Fundamental Theorem of Algebra for C). Every non-constant polynomial
p(X) ∈ C[X] has a root in C.
Corollary 3.41. Every non-constant polynomial p(X) ∈ C[X] has a factorization
p(X) = c(X ’ u1 ) · · · (X ’ ud ),
where c, u1 , . . . , ud ∈ C and this is unique apart from the order of the roots uj .
It is natural to pose the following question.
Question 3.42. Let K be a ¬eld. Is there an algebraically closed ¬eld F containing K?
By taking F alg we might as well ask that such a ¬eld be algebraic over K.
38 3. ALGEBRAIC EXTENSIONS OF FIELDS

Definition 3.43. Let K be a ¬eld. An extension F/K is called an algebraic closure of K
if F is algebraic over K and algebraically closed.
Theorem 3.44. Let K be a ¬eld.
(i) There is an algebraic closure of K.
(ii) Let F1 and F2 be algebraical closures of K. Then there is an isomorphism • : F1 ’’ F2
which ¬xes the elements of K.

Ke
ee
}} ee
}} ee
}} e
˜}} •
/ F2
F1

Hence algebraic closures are essentially unique.

Proof. See [3] for a proof using Zorn™s Lemma (see Axiom 3.48) which is logically
™™¦

equivalent to the Axiom of Choice.

Because of the uniqueness we usually ¬x some choice of algebraic closure of K and write K
or K alg cl , referring to it as the algebraic closure of K. We are already familiar with the example
C = C. There are some immediate consequences of Theorem 3.44. We will temporarily write
.
E1 = E2 to indicate that for extensions E1 /K and E2 /K there is an isomorphism E1 ’’ E2
¬xing the elements of K.
Proposition 3.45. Let K be a ¬eld.
.
(i) If L/K is an algebraic extension, then L = K.
.
(ii) If L/K is an extension, then so is L/K and (L)alg = K.

Proof. (i) By Proposition 3.16, every element of L is algebraic over K. Since L is alge-

<<

. 9
( 18 .)



>>