<< Предыдущая стр. 9(из 18 стр.)ОГЛАВЛЕНИЕ Следующая >>
2 iiiii 2 qqqq V
2 VV
i
iiii qq VV
qqq
ii
iiii в€љ 2 VVV3
в€љi в€љ
Q( 3 2 О¶3 ) VV
3 3
Q( 2) Q( 2 О¶3 )
www `` VV
``
www VV
``
www VV
``
www
`3
www
3
www ``` Q(О¶3 )
3
www `` r
2 rr
www `` rrr
` rr
www `
w rrr
Q
в€љ в€љ2
An alternative strategy would have been to adjoin one of the other roots 3 2 О¶3 or 3 2 О¶3
п¬Ѓrst. We could also have begun by adjoining О¶3 to form the extension Q(О¶3 )/Q, but none of
в€љ
the roots of p(X) lie in this п¬Ѓeld so the extension Q( 3 2, О¶3 )/Q(О¶3 ) 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 в€’в†’ F which п¬Ѓx the elements of K.
Remark 3.32. We always have AutK (F ) вЉ† 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 п¬Ѓnite, 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) в€€ K[X]. Set
Roots(p, F ) = {u в€€ F : p(u) = 0},
the set of roots of p(X) in F . This is always a п¬Ѓnite set which may of course be empty.
Suppose that p(X) в€€ K[X] is an irreducible polynomial which we might as well assume is
monic, and F/K an extension. Then if t в€€ F is a root of p(X), the evaluation homomorphism
Оµt : K[X] в€’в†’ F factors through the quotient monomorphism Оµt : K[X]/(p(X)) в€’в†’ F whose
image is K(t) F . Of course, there is one such monomorphism for each root of p(X) in F . If
we п¬Ѓx one such root t0 and identify K[X]/(p(X)) with K(t0 ) via Оµt0 , then each root of p(X) in
F gives rise to a monomorphism П•t = Оµt в—¦ Оµв€’1 : K(t0 ) в€’в†’ F for which П•t (t0 ) = t.
t0

П•t =et в—¦eв€’1
Оµ Оµt
0

et0
Оµ et
Оµ
/* F
K(t0 ) o K[X]/(p(X))
в€ј
=

Notice that if П• : K[X]/(p(X)) в€’в†’ F is any homomorphism extending the identity function
on K, then the coset X + (p(X)) must be sent by П• 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 п¬Ѓeld extension. Let p(X) в€€ K[X] be an irreducible
polynomial with t0 в€€ F be a root of p(X). Then there is a bijection
Roots(p, F ) в†ђв†’ MonoK (K(t0 ), F )
given by t в†ђв†’ П•t , where П•t : K(t0 ) в€’в†’ F has the eп¬Ђect П•t (t0 ) = t.
в€љ
Example 3.35. Show that MonoQ (Q( 2), C) has two elements.
в€љ
Solution. We have Q( 2) в€јв€љ = Q[X]/(X 2 в€’ 2) where X 2 в€’ 2 is irreducible over Q. Hence the
в€љ
Q-monomorphisms send 2 to В± 2. In fact both possibilities occur, giving monomorphisms
в€љ
id, О± : Q( 2) в€’в†’ C, where
в€љ в€љ
О±(a + b 2) = a в€’ b 2.
в€љ
We can replace C by Q( 2) to obtain
в€љ в€љ в€љ в€љ
MonoQ (Q( 2), C) = MonoQ (Q( 2), Q( 2)) = AutQ (Q( 2)).
We will see that this is not always true.
в€љ в€љ в€љ
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.
в€љв€љ
Solution. Here minpolyQ, в€љ2 (X) = X 3 в€’ 2 and there are 3 complex roots 3 2, 3 2 О¶3 ,
3
в€љ2 в€љ в€љ
2 О¶3 . As two of these roots are not real, MonoQ (Q( 2), Q( 3 2)) contains only the identity
3 3
в€љ
since Q( 3 2) R. в€љ в€љ в€љ2
Each of the above roots corresponds to one of the subп¬Ѓelds Q( 3 2), Q( 3 2 О¶3 ) or Q( 3 2 О¶3 )
в€љ
of C and there are 3 monomorphisms О±0 , О±1 , О±2 : Q( 3 2) в€’в†’ C given by
в€љ в€љ2 в€љ в€љ2
3 3 3 3
О±0 (a + b 2 + c( 2) ) = a + b 2 + c( 2) ,
в€љ в€љ в€љ в€љ
О±1 (a + b 2 + c( 2)2 ) = a + b 2 О¶3 + c( 2)2 О¶3 ,
2
3 3 3 3

в€љ в€љ в€љ2 в€љ
О±2 (a + b 2 + c( 2)2 ) = a + b 2 О¶3 + c( 2)2 О¶3 .
3 3 3 3

The images of these mappings are
в€љ в€љ в€љ в€љ в€љ в€љ2
3 3 3 3 3 3
О±0 Q( 2) = Q( 2), О±1 Q( 2) = Q( 2 О¶3 ), О±2 Q( 2) = Q( 2 О¶3 ).
Proposition 3.37. Let F/K and L/K be extensions.
(i) For p(X) в€€ K[X], each monomorphism О± в€€ MonoK (L, F ) restricts to a function
О±p : Roots(p, L) в€’в†’ Roots(p, F ) which is an injection.
(ii) If О± в€€ MonoK (L, L), then О±p : Roots(p, L) в€’в†’ Roots(p, L) is a bijection.
Proof. (i) For u в€€ Roots(p, L) we have
p(О±(u)) = О±(p(u)) = О±(0) = 0,
so О± maps Roots(p, L) into Roots(p, F ). Since О± is an injection its restriction to Roots(p, L) вЉ† L
is also an injection.
(ii) From (i), О±p : Roots(p, L) в€’в†’ Roots(p, L) is an injective function from a п¬Ѓnite set to itself,
hence it is also surjective by the Pigeon Hole Principle. Thus О±p : Roots(p, L) в€’в†’ 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) в€€ K[X]. This gives us a strong hold on the possible automorphisms. In the case of п¬Ѓnite,
or more generally algebraic, extensions it is the key to understanding the automorphism group
and this is a fundamental insight of Galois Theory.
в€љ
Example 3.38. Determine MonoQ (Q( 3 2, О¶3 ), C).
3.4. ALGEBRAIC CLOSURES 37
в€љ
Solution. We have already met the extension Q( 3 2, О¶3 )/Q in Example 3.30 and we will
make use of information from there. We build up the list of monomorphisms в€љ stages.
in
в€љ
First consider monomorphisms that п¬Ѓx 2 and hence п¬Ѓx the subп¬Ѓeld Q( 3 2). These form
3

the subset в€љ в€љ
3 3
MonoQ( в€љ2) (Q( 2, О¶3 ), C) вЉ† MonoQ (Q( 2, О¶3 ), C).
3
в€љ в€љ
We know that Q( 2, О¶3 ) = Q( 3 2)(О¶3в€љ and that О¶3 is a root of the irreducible cyclotomic
3
)
2 + X + 1 в€€ Q( 3 2)[X]. So there are two monomorphisms id, О± п¬Ѓxing
polynomial О¦3 (X) = X 0
в€љ
3
Q( 2), where О±0 has the eп¬Ђect
в€љ в€љ
3
2 в€’в†’ 3 2
О±0 : .
2
О¶3 в€’в†’ О¶3
в€љ в€љ
Next we consider monomorphisms that send в€љ 2 to 3 2 О¶3 . This time we have 2 distinct ways to
3
в€љ
extend to elements of MonoQ (Q( 3 2, О¶3 ), Q( 3 2, О¶3 )) since again we can send О¶3 to either О¶3 or
2
О¶3 . The possibilities are
в€љ в€љ
в€љ в€љ 3
2 в€’в†’ 3 2 О¶3
3
2 в€’в†’ 3 2 О¶3
О±1 : , О±1 : .
2
О¶3 в€’в†’ О¶3 О¶3 в€’в†’ О¶3
в€љ в€љ2
Finally we consider monomorphisms that send 3 2 to 3 2 О¶3 . There are again two possibilities
в€љ в€љ2
в€љ в€љ2 3
2 в€’в†’ 3 2 О¶3
3 3
2 в€’в†’ 2 О¶3
О±2 : , О±2 : .
2
О¶3 в€’в†’ О¶3 О¶3 в€’в†’ О¶3
These are all 6 of the required monomorphisms. It is also the case here that
в€љ в€љ в€љ в€љ
3 3 3 3
MonoQ (Q( 2, О¶3 ), C) = MonoQ (Q( 2, О¶3 ), Q( 2, О¶3 )) = AutQ (Q( 2, О¶3 )),
в€љ
so these form a group. It is a nice exercise to show that AutQ (Q( 3 2, О¶3 )) в€ј S3 , the symmetric
=в€љ
в€љ
group on 3 objects. It is also worth remarking that | AutQ (Q( 2, О¶3 ))| = [Q( 3 2, О¶3 ) : Q].
3

We end this section with another useful result.
Proposition 3.39. Let L/K be an extension and О± в€€ MonoK (L, L). Then О± restricts to
an automorphism О±alg : Lalg в€’в†’ Lalg .
Proof. Suppose that u в€€ Lalg , say p(u) = 0 for some p(X) в€€ K[X] of positive degree.
Then
p(О±(u)) = О±(p(u)) = О±(0) = 0,
so О± maps Lalg вЉ† L into itself and therefore gives rise to a restriction О±alg : Lalg в€’в†’ Lalg which
is also a monomorphism. We must show that О±alg is a bijection by showing it is surjective.
Let v в€€ Lalg and suppose that q(v) = 0 for some q(X) в€€ K[X] of positive degree. Now
Roots(q, L) = в€… since it contains v, and it is also п¬Ѓnite. Then О±q : Roots(q, L) в€’в†’ Roots(q, L)
is a bijection by Proposition 3.37(ii), hence v = О±q (w) = О±(w) for some w в€€ Roots(q, L) вЉ† Lalg .
This shows that v в€€ im О± and so О±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) в€€ C[X] has a root in C.
Corollary 3.41. Every non-constant polynomial p(X) в€€ C[X] has a factorization
p(X) = c(X в€’ u1 ) В· В· В· (X в€’ ud ),
where c, u1 , . . . , ud в€€ 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 п¬Ѓeld. Is there an algebraically closed п¬Ѓeld F containing K?
By taking F alg we might as well ask that such a п¬Ѓeld be algebraic over K.
38 3. ALGEBRAIC EXTENSIONS OF FIELDS

Definition 3.43. Let K be a п¬Ѓeld. 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 п¬Ѓeld.
(i) There is an algebraic closure of K.
(ii) Let F1 and F2 be algebraical closures of K. Then there is an isomorphism П• : F1 в€’в†’ F2
which п¬Ѓxes the elements of K.

Ke
ee
}} ee
}} ee
}} e
˜}} П•
/ F2
F1

Hence algebraic closures are essentially unique.
в™
Proof. See  for a proof using ZornвЂ™s Lemma (see Axiom 3.48) which is logically
в™Ґв™¦
в™Ј
equivalent to the Axiom of Choice.

Because of the uniqueness we usually п¬Ѓx 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 в€’в†’ E2
п¬Ѓxing the elements of K.
Proposition 3.45. Let K be a п¬Ѓeld.
.
(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-
 << Предыдущая стр. 9(из 18 стр.)ОГЛАВЛЕНИЕ Следующая >>