‫ ﺗﻮﺳﯿﻊ اﻟﺤﻘﻞ‬17 ‫اﻟﻤﺤﺎﺿﺮة‬
‫ وﺳﻨﺘﻄﺮق إﻟﻰ أﻧﻮاع ﻣﮭﻤﺔ‬.‫ﻣﺎدة ھﺬه اﻟﻤﺤﺎﺿﺮة ھﻲ ﻣﻘﺪﻣﺔ ﺿﺮورﯾﺔ ﻟﺪراﺳﺔ ﻧﻈﺮﯾﺔ ﮔﺎﻟﻮا‬
‫ ھﻮ ﺣﻘﻞ ﺟﺰﺋﻲ‬F ‫ إذا ﻛﺎن‬، F ‫ ھﻮ ﺗﻮﺳﯿﻊ ﻟﻠﺤﻘﻞ‬K ‫ ﺑﺸﻜﻞ ﻋﺎم ﻧﻘﻮل أن اﻟﺤﻘﻞ‬. ‫ﻣﻦ ﺗﻮﺳﯿﻊ اﻟﺤﻘﻞ‬
‫ و ﺣﻘﻞ‬، Q ‫ ھﻮ ﺗﻮﺳﯿﻊ ﻟﺤﻘﻞ اﻷﻋﺪاد اﻟﻨﺴﺒﯿﺔ‬R # ‫ ﻣﺜﻼ ﺣﻘﻞ اﻷﻋﺪاد اﻟﺤﻘﯿﻘﯿﺔ‬. K ‫ﻣﻦ اﻟﺤﻘﻞ‬
‫ وﺳﻨﺒﺪأ ﺑﺎﻟﻤﻼﺣﻈﺎت اﻟﺘﺎﻟﯿﺔ ﺣﻮل ھﺬا‬. R # ‫ ھﻮ ﺗﻮﺳﯿﻊ ﻟﺤﻘﻞ اﻷﻋﺪاد اﻟﺤﻘﯿﻘﯿﺔ‬C ‫اﻷﻋﺪاد اﻟﻌﻘﺪﯾﺔ‬
:‫اﻟﻤﻔﮭﻮم‬
E1. By an extension F' of a field F we simply mean any field that contains
F as a subfield.
E2. Assume that F' is an extension field of a field F and let the elements
r 1 , r 2 , …, r n all lie in F'. The subfield of F' generated by the subset F È {
r 1 , r 2 , …, r n } is denoted by F(r 1 , r 2 , …, r n ):
F(r 1 , r 2 , …, r n ) = Ç { K| K is a subfield of F'; F Í K; r i Î K}.
E3. Recall that the symbol F[r] means the set { f(r) | f(x) Î F[x]}, where
r Î F'. F[r] forms an integral domain (being a subring of the field F') . and
therefore has a field of quotients K = Q cl (F[r]) in F'. In fact F(r)=K.
E4. If R is an integral domain, a (classical) field of quotients of R is any
field Q cl (R) such that: 1. R Í Q cl (R) and 2. each element of Q cl (R) is of
the form ab -1 , where a, b Î R and b≠0.
E5. Recall that the substitution homomorphism f r :F[x] →F', for each
element rÎ F', defined by f r f(x)= f(r). The key to classifying simple
extensions is the nature of the kernel of f r . Bear in mined that ker f r
consists of all polynomials in F[x] having r as a root.
Theorem 2.18 Let F' be an extension field of the field F and rÎ F'.
Then either a) F(r) @ Q cl (F[x]) or else
b) F[r] @ F[x]/( f(x)) for some monic irreducible polynomial f(x) Î F[x]
such that f(r)=0; this polynomial is uniquely determined.
‫( وﻟﻜﻦ‬-‫ﻛﺘﺎب ﺑﺮﺗﻦ‬-‫ﺳﻮف ﻟﻦ ﻧﺒﺮھﻦ اﻟﻤﺒﺮھﻨﺔ اﻋﻼه ) وﯾﻤﻜﻦ ﻟﻠﻄﺎﻟﺐ اﻟﻤﮭﺘﻢ أن ﯾﻌﻮد ﻟﻠﻤﺼﺪر‬
‫ أو‬ker f r ={0}: ‫ﻧﺸﯿﺮ إﻟﻰ أن اﻟﺤﺎﻟﺘﯿﻦ اﻟﻤﺬﻛﻮرة ﻓﻲ اﻟﻤﺒﺮھﻨﺔ ﺗﻌﻮد إﻟﻰ اﻻﺣﺘﻤﺎﻟﯿﻦ‬
‫ إﻟﻰ ﺻﻨﻔﯿﻦ ﻛﻤﺎ ﻓﻲ اﻟﺘﻌﺮﯾﻒ‬F' ‫ وھﺬان اﻻﺣﺘﻤﺎﻻن ﺳﺘﻔﻀﻲ إﻟﻰ ﺗﺼﻨﯿﻒ ﻋﻨﺎﺻﺮ‬، ker f r ≠{0}
:‫اﻟﺘﺎﻟﻲ‬
Definition 2.8 Let F' be an extension field of a field F. An element rÎ F'
is said to be algebraic over F if ker f r ‹{0}; otherwise, r termed
transcendental over F.
‫ ﯾﺴﻤﻰ ﻋﻨﺼﺮ ﺟﺒﺮي إذا ﻛﺎن ﺟﺬرا ﻟﻤﺘﻌﺪدة ﺣﺪود ﻏﯿﺮ‬r ‫ﻣﻦ اﻟﺘﻌﺮﯾﻒ أﻋﻼه ﯾﻤﻜﻦ اﻟﻘﻮل أن‬
f(x) ‫ ﻟﻜﻞ‬f(r)≠0 ‫ ﯾﻌﻨﻲ ﻋﻨﺪﻣﺎ ﯾﻜﻮن‬،‫ ﻣﺘﻌﺎﻟﻲ أو ﻣﺘﺴﺎﻣﻲ‬r ‫ ﻓﯿﻤﺎ ﻋﺪا ذﻟﻚ ﻓﺎن‬F[x] ‫ﺻﻔﺮﯾﺔ ﻓﻲ‬
Q ‫ ھﻮ ﺟﺒﺮي ﻋﻠﻰ‬2 + 3 ‫ ﻓﺎن‬F=Q ‫ و‬F' =R # ‫ ﻣﺜﺎل ﻋﻠﻰ ذﻟﻚ إذا ﻓﺮﺿﻨﺎ أن‬.F[x] ‫ﻓﻲ‬
. Q ‫ ﻛﻼھﻤﺎ ﻣﺘﻌﺎﻟﻲ ﻋﻠﻰ‬2 2 ‫ و‬p ‫ ﺑﯿﻨﻤﺎ‬x 4 -10x 2 + 1 Î Q[ x ] ‫ﻷﻧﮫ ﺟﺬر ﻟﻤﺘﻌﺪدة اﻟﺤﺪود‬
‫ وﻟﻜﻦ ﻧﺤﺪد ﻣﺎ ﻧﻌﻨﻲ ﺑﺎﻟﺪوال‬،‫اﻟﻤﺒﺮھﻨﺘﺎن اﻵﺗﯿﺔ ﺗﻌﻄﯿﻨﺎ ﺗﻔﺎﺻﯿﻞ أﻛﺜﺮ ﺣﻮل ﺗﻮﺳﯿﻊ اﻟﺤﻘﻞ‬
. F ‫اﻟﻨﺴﺒﯿﺔ ﻋﻠﻰ اﻟﺤﻘﻞ‬
Note: The field of quotients of the polynomial domain F[x] is called the
field of rational functions over F. Thus a rational function (in an
indeterminant x) over F can be written as a quotient f(x)/g(x), where f(x)
and g(x)≠0 are both polynomials in F[x].
Theorem 2.19 (Simple Transcendental Field Extensions) If r Î F' Ê F is
transcendental over F, then F(r) is isomorphic to the field of rational
functions over F. In fact, there is an isomorphism a r of F(r) into
Q cl (F[x]) such that a r (r) =x and a r (a)= a for every a Î F.
Note: This theorem completely determines the structure of simple
transcendental over F; they are all isomorphic to the field of rational
functions over F and hence to each other. Thus, for instance,
Q( p ) @ Q( 2 2 ).
Theorem 2.20 (Simple Algebraic Field Extensions) If rÎ F' Ê F is
algebraic over F, then a unique monic irreducible polynomial f(x) Î F[x]
exists such that f(r) = 0. Furthermore, if g(x) is a polynomial in F[x] for
which g(r) = 0, necessarily f(x)| g(x) .
Notes: 1. The unique polynomial f(x) of Theorem 2.20 is referred to as
the minimum polynomial of (or belonging to) r over F.
2. As the name suggests, the minimum polynomial of r is the monic
polynomial in F[x] of minimal degree having r as a root.
3. The degree of an algebraic element r Î F' Ê F is just the degree of its
minimum polynomial. This degree is 1 if and only if r Î F.
4. Whenever r is algebraic over F, the integral domain F[r] becomes a
field , so F(r) = F[r]. That is, every element of F(r) is of the form f(r),
where f(x) is a polynomial in F[x] .
Example 2.19 If n≠1 is any square-free integer, the element n Î R # Ê Q
is a root of the quadratic polynomial x 2 - nÎ Q[x] and hence is algebraic
over Q. From the last note we know that Q[ n ] is a field, and so Q[ n ]
=Q( n ); in other words, every member of Q( n ) is of the form f( n ),
where f(x) is a polynomial in Q[x] :
Q( n )= { a 0 +a 1 n + a 2 ( n ) 2 + ... + a k ( n ) k | a i Î Q; k ³ 0 }.
But ( n ) 2 =n, ( n ) 3 =n n , …, so Q( n ) can be described more simply
as
Q( n )={ a+ b n | a, b Î Q}. Note also that an arbitrary element
a + b n Î Q( n ) satisfies the polynomial x 2 -2ax + (a 2 - b 2 n) Î Q[x].
Theorem 2.21 Let F' be an extension of F and rÎ F' be algebraic over F
of degree n. Then every element of F(r) can be uniquely represented in
the form c 0 + c1 r + c 2 r 2 + ... + c n -1 r n -1 where the coefficients c i Î F.
Corollary Let rÎ F' Ê F be algebraic over F of degree n. Then each
element of the simple extension F(r) is also algebraic over F.
Example 2.20 Consider the element r = 2 + i Î C Ê Q, C as usual being
the complex number field. Then r 2 = 1 + 2 2 i, so (r 2 - 1) 2 = -8 or
equivalently, r 4 -2r 2 + 9 = 0 . Thus r is a root of the polynomial
f(x) = x 4 - 2x 2 + 9 Î Q[x] and hence is an algebraic element over Q.
Now f(x) has the irreducible factorization over C,
f(x) = (x- 2 + i)(x- 2 - i)(x + 2 + i)(x + 2 - i),
which indicates that f(x) has no linear or quadratic factors in Q[x].
Therefore f(x) is irreducible as a member of Q[x] and serves as the
minimum polynomial of r over Q; in particular, the element r has degree
4. By Theorem 2.21 the simple extension Q(r) can be described as
Q(r)= { c 0 + c1 r + c 2 r 2 + c 3 r 3 | c i Î Q }.
Since r = 2 + i, r 2 = 1 + 2 2 i, and r 3 = - 2 + 5i, this simplifies to
Q( 2 + i) = { (a+ b 2 ) + (c +d 2 i)| a, b, c, d Î Q}.
At the same time, r is a root of the polynomial x 2 - 2 2x + 3 Î R # [x],
with x 2 - 2 2 x + 3 irreducible over R # ; thus r is of degree 2 over R # .
Example 2.21 For a second illustration we turn to the extension field
Q( 2, 3 ). The elements 2 and 3 are clearly algebraic over Q, being
roots of the polynomials x 2 - 2, x 3 - 3 Î Q[x], respectively. Our contention
is that the field Q( 2, 3 ) is actually a simple algebraic extension of Q; in
fact Q( 2, 3 ) = Q( 2 + 3 ), with 2 + 3 algebraic over Q. Since the
element 2 + 3 belongs to the field Q( 2, 3 ), we certainly must have
Q( 2 + 3 ) Í Q( 2, 3 ). As regards the reverse inclusion, a simple
computation shows that 2 2 = ( 2 + 3 ) 3 - 9( 2 + 3 ) is a member of
Q( 2 + 3 ), and therefore so is 2 . But then 3 = ( 2 + 3 ) - 2 also lie
in Q( 2 + 3 ). This leads to the inclusion Q( 2, 3 ) Í Q( 2 + 3 ) and the
asserted equality.
To see that r = 2 + 3 is an algebraic element over Q, note that
r 2 = 5 + 2 6 , (r 2 - 5) 2 = 24, and hence the polynomial f(x) = x 4 -10x 2 + 1
has r as a root. One may verify that f(x) is irreducible in Q[x], making it
the minimum polynomial of r.
Theorem 2.22 ( Kronecker) If f(x) is an irreducible polynomial in F[x],
then there is an extension field of F in which f(x) has a root.
Corollary If the polynomial f(x) Î F[x] is of positive degree, then there
exists an extension field of F containing a root of f(x) .
Example 2.22 Consider Z 2 , the field of integers modulo 2, and the
polynomial f(x)= x 3 +x +1 Î Z 2 [x]. Since neither of the elements 0 and 1
a root of x 3 +x +1, f(x) must be irreducible in Z 2 [x].
Theorem 2.22 thus guarantees the existence of an extension of Z 2 ,
specifically, the field Z 2 [x]/( f(x)), in which the given polynomial has a
root. If we denote the root by l , then we have
Z 2 [x]/( f(x)) = { a + b l + c l 2 | a, b, c Î Z 2 }
= { 0, 1, l , 1+ l , l 2 , 1+ l 2 , l + l 2 , 1+ l + l 2 }, where, of
course, l 3 + l +1 =0.
It is worth noting that x 3 +x +1 factors completely into linear factors in
Z 2 [x]/( f(x)) and has the three roots l , l 2 and l + l 2 :
x 3 +x +1= (x- l )(x- l 2 )(x- ( l + l 2 )).
Example 2.23 The quadratic polynomial x 2 +1 is irreducible in R # [x]. If
x 2 +1 were reducible, it would be of the form
x 2 +1 =(ax +b )(cx + d) = ac x 2 + (ad +bc)x +bd , where a, b, c, dÎ R # .
It follows at once that a
ac = bd =1 and ad + bc= 0. Therefore bc = -( ad), and
1= (ac)(bd) =(ad)(bc) = -(ad) 2 , or rather (ad) 2 = -1, which is impossible.
In this instance the extension field R # [x]/( f(x)) is described by
R # [x]/( x 2 +1) = { a +b l | a, b Î R # ; l 2 +1= 0}.
In fact R # [x]/( x 2 +1) is isomorphic to the field C of complex numbers
under the mapping F : R # [x]/( x 2 +1) → C given by F ( a +b l ) = a + bi.
Theorem 2.23 If f(x) Î F[x] is a polynomial of positive degree, then an
extension field F' of F exists in which f(x) factors completely into linear
polynomials.
Corollary Let f(x) Î F[x] , deg f(x) =n>0. Then an extension of F exists
in which f(x) has n (not necessarily distinct)roots.
Example 2.24 Consider the polynomial f(x)= x 4 - 2 in Z 5 [x]. Since f(x)
has no roots in Z 5 , it can have no linear factors in Z 5 [x]. As regards a
possible quadratic factorization, suppose that
x 4 - 2= (x 2 + ax + b)( x 2 + cx +d), where a, b, c, dÎ Z 5 .
Then, on equating corresponding coefficients,
a + c= 0, ac + b + d = 0, ad + bc = 0, bd = -2 .
Substitution of c = -a into ad + bc = 0 gives a(d – b) =0. If a≠ 0, then
d= b, which leads to b 2 = -2, an impossibility in Z 5 . On the other hand, if
a = 0, then the system of equations reduced to b + d =0, bd = -2.
These in their turn yield b 2 = 2, which again is impossible. This means
that f(x) cannot be factored into either linear or quadratic polynomials and
is therefore irreducible over Z 5 .
Appealing to Theorem 2.22, we see that the quotient ring
Z 5 [x]/( x 4 - 2) = { a +b l + c l 2 + d l 3 | a, b, c, d Î Z 5 ; l 4 = 2}
is a field containing 5 4 = 625 elements. Notice that because
2 4 º 3 4 º 4 4 º 1(mod 5), l 4 =(2 l ) 4 = (3 l ) 4 = (4 l ) 4 = 2 ; and so l , 2 l ,
3 l , 4 l are all roots of f(x) in this extension field of Z 5 . Here we obtain a
complete factorization into linear factors
f(x) = (x - l )(x - 2 l )( x - 3 l )( x - 4 l ) or, by conveniently writing l as
4
2 , f(x) = (x - 4 2 )(x - 2 4 2 )( x - 3 4 2 )( x - 4 4 2 ).
Example 2.25 A second example is afforded by considering the
polynomial f(x) = (x 2 - 2)( x 2 - 3) over the field Q of rational numbers.
From a previous example, x 2 - 2 ( and by similar reasoning , x 2 - 3) is
already known to be irreducible in Q[x]. So we begin by extending Q to
the field F 1 , where F 1 = Q[x]/(x 2 - 2) = { a +b l |a, b Î Q; l 2 - 2= 0};
and obtain the factorization
f(x) =(x - l )(x + l )( x 2 - 3) = (x - 2 )(x + 2 )( x 2 - 3) .
( As l 2 =2, one customarily identifies l with 2 .)
However, f(x) does not split completely, since the polynomial x 2 - 3
remains irreducible in F 1 [x]. Suppose to the contrary that x 2 - 3 has a root
in F 1 , say c + d 2 , with c, d Î Q. Substituting, we find that
( c 2 + 2d 2 - 3) + 2cd 2 = 0, or, what amounts to the same thing,
c 2 + 2d 2 - 3 = 0 , cd = 0.
The later equation implies that either c = 0 or d = 0. But neither c nor d
can be zero, since this would mean that d 2 =3/2 or c 2 = 3, which is clearly
impossible. Accordingly, x 2 - 3 does not split in F 1 [x].
To factor f(x) into linear factors, it becomes necessary to extend the
coefficient field further. We therefore construct a second extension F 2 ,
where F 2 = F 1 [x]/ (x 2 - 3) = { a + bm | a, b Î F1 ; m 2 - 3 = 0 }.
The elements of F 2 can be expressed alternatively in the form
(a + b 2 ) + (c + d 2 ) 3 = a + b 2 + c 3 +d 6 .
Where, of course, the coefficients a, b, c, d all lie in Q. It follows without
difficulty that the original polynomial now factors in F 2 [x] as
f(x) =(x - l )(x + l )(x - m )(x + m ) = (x - 2 )(x + 2 )(x - 3 )(x + 3 ).
Notes: 1. Let a field F be given and consider a nonconstant polynomial
f(x) Î F[x]. An extension field F' of F is said to be a splitting field for f(x)
over F provided that f(x) can be factored completely into linear factors in
F'[x] but not so factored over any proper subfield of F' containing F.
2. A splitting field is the smallest extension field F' in which the
prescribed polynomial factors linearly:
f(x) = a(x - r 1 )(x - r 2 ) … (x - r n ), r i Î F' .
3. The existence of such a splitting field can be guaranteed (with the aid
of Theorem 2.22. Also the uniqueness (up to isomorphism) can be
proved, so we can speak about the splitting field of a given polynomial.
.‫وأﻻن ﻧﻌﺮف ﺻﻨﻒ آﺧﺮ ﻣﻦ أﺻﻨﺎف ﺗﻮﺳﯿﻊ اﻟﺤﻘﻞ‬
Definition 2.9 An extension K of the field F is said to be a finite
extension of degree n if K is a finite dimensional vector space of
dimension n over F. We shall use the symbol [ K:F] to represent the
degree of K over F.
Notes: 1. As a consequence of this definition, we can recast Theorem
2.21 as follows: If K = F[r] is a simple algebraic extension of F, where
the minimum polynomial of r over F has degree n, then [F(r) : F] = n.
2. While any simple algebraic extension is finite, the converse is not true
without additional hypotheses.
3. In fact all finite extensions of a field of characteristic zero ca be
realized as simple extensions.
.‫ﺳﻨﺤﺘﺎج إﻟﻰ ﺻﻨﻒ أﺧﯿﺮ ﻣﻦ أﺻﻨﺎف ﺗﻮﺳﯿﻊ اﻟﺤﻘﻞ ﻗﺒﻞ أن ﻧﺪﺧﻞ ﻓﻲ ﻧﻈﺮﯾﺔ ﮔﺎﻟﻮا ﻟﻠﺤﻘﻮل‬
Definition 2.10 An extension K of the field F is said to be normal over F
if every irreducible polynomial f(x) Î F[x] that has one root in K splits
completely in K.
Note: Not all finite extension fields are normal. For instance, Q( 3 2 ) is
not a normal extension of Q, where 3 2 is the real cube root of 2. The
irreducible polynomial x 3 - 2Î Q[x] has one root in Q( 3 2 ), namely 3 2 ,
but does not split completely in Q( 3 2 ) . if w = (-1 + i 3 ) / 2 , then the other
two roots of x 3 - 2 in a splitting field are w 3 2 and w 2 3 2 ( since w 3 =1);
Q( 3 2 ), being a subfield of the real field R # , contains neither of these
complex roots.