Field Extensions and Galois Theory - Field Theory, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


2 Field Extensions
Let K be a ?eld
2
. By a (?eld) extension of K we mean a ?eld containing K as a sub?eld.
Let a ?eld L be an extension of K (we usually express this by saying that L/K [read: L over
K] is an extension). Then L can be considered as a vector space over K. The degree of L over
K, denoted by [L : K], is de?ned as
[L : K] = dim
K
L = the vector space dimension of L over K.
If [L : K] <8, we say that L is a ?nite extension of K or that L is ?nite over K. A sub?eld
K of I C such that [K :
|
Q] <8 is called an algebraic number ?eld or simply a number ?eld.
Lemma 1: Finite over ?nite is ?nite. More precisely, ifL/E andE/K are ?eld extensions,
then
L is finite over K ? L is finite over E and E is finite over K
and, in this case, [L :K] = [L : E][E :K].
Proof: The implication “?” is obvious. The rest follows easily from the observation that
if{u
i
} is an E–basis of L and{v
j
} is a K–basis of E, then{u
i
v
j
} is a K–basis of L. 2
Let L/K bea ?eld extension. An element a? Lis said to be algebraicover K if it satis?es
a nonzero polynomial with coe?cients in K, i.e, ? 0 6= f(X) ? K[X] such that f(a) = 0.
Given a ? L which is algebraic over K, we can ?nd a monic polynomial in K[X] of least
possible degree, satis?ed by a. This is unique and is called the minimal polynomial of a over
K. It is easily seen to be irreducible and we will denote it by Irr(a,K). Note that if f(X)
is any monic irreducible polynomial satis?ed by a, then we must have f(X) =Irr(a,K) and
that it generates the ideal{g(X)? K[X] : g(a) = 0} in K[X].
3
The extension L of K is said
to be algebraic if every element of L is algebraic over K.
Lemma 2: Finite? algebraic. That is, if L/K is a ?nite extension, then it is algebraic.
Proof: For any a? L, there must exist a positive integer n such that {1,a,a
2
,...,a
n
} is
linearly dependent over K, thus showing that a is algebraic over K. 2
Exercise 1: Show, by an example, that the converse of the above lemma is not true, in
general.
We now study extensions for which the converse is true.
2
Fields are usually denoted by K or k since the German word for ?eld is K¨ orper. Much of Modern
Field Theory was created by the German mathematician E. Steinitz; see his paper “Algebraische Theorie der
K¨ orper”, Crelle Journal (1910), pp. 167–308, for an original exposition.
3
It may be instructive to verify the observations made in the last few statements. General Hint: Use the
Division Algorithm in K[X].
Page 2


2 Field Extensions
Let K be a ?eld
2
. By a (?eld) extension of K we mean a ?eld containing K as a sub?eld.
Let a ?eld L be an extension of K (we usually express this by saying that L/K [read: L over
K] is an extension). Then L can be considered as a vector space over K. The degree of L over
K, denoted by [L : K], is de?ned as
[L : K] = dim
K
L = the vector space dimension of L over K.
If [L : K] <8, we say that L is a ?nite extension of K or that L is ?nite over K. A sub?eld
K of I C such that [K :
|
Q] <8 is called an algebraic number ?eld or simply a number ?eld.
Lemma 1: Finite over ?nite is ?nite. More precisely, ifL/E andE/K are ?eld extensions,
then
L is finite over K ? L is finite over E and E is finite over K
and, in this case, [L :K] = [L : E][E :K].
Proof: The implication “?” is obvious. The rest follows easily from the observation that
if{u
i
} is an E–basis of L and{v
j
} is a K–basis of E, then{u
i
v
j
} is a K–basis of L. 2
Let L/K bea ?eld extension. An element a? Lis said to be algebraicover K if it satis?es
a nonzero polynomial with coe?cients in K, i.e, ? 0 6= f(X) ? K[X] such that f(a) = 0.
Given a ? L which is algebraic over K, we can ?nd a monic polynomial in K[X] of least
possible degree, satis?ed by a. This is unique and is called the minimal polynomial of a over
K. It is easily seen to be irreducible and we will denote it by Irr(a,K). Note that if f(X)
is any monic irreducible polynomial satis?ed by a, then we must have f(X) =Irr(a,K) and
that it generates the ideal{g(X)? K[X] : g(a) = 0} in K[X].
3
The extension L of K is said
to be algebraic if every element of L is algebraic over K.
Lemma 2: Finite? algebraic. That is, if L/K is a ?nite extension, then it is algebraic.
Proof: For any a? L, there must exist a positive integer n such that {1,a,a
2
,...,a
n
} is
linearly dependent over K, thus showing that a is algebraic over K. 2
Exercise 1: Show, by an example, that the converse of the above lemma is not true, in
general.
We now study extensions for which the converse is true.
2
Fields are usually denoted by K or k since the German word for ?eld is K¨ orper. Much of Modern
Field Theory was created by the German mathematician E. Steinitz; see his paper “Algebraische Theorie der
K¨ orper”, Crelle Journal (1910), pp. 167–308, for an original exposition.
3
It may be instructive to verify the observations made in the last few statements. General Hint: Use the
Division Algorithm in K[X].
De?nition: Given elements a
1
,...,a
n
in an extension L of a ?eld K, we de?ne
K[a
1
,...,a
n
] = the smallest subring of L containing K and a
1
,...,a
n
K(a
1
,...,a
n
) = the smallest sub?eld of L containing K and a
1
,...,a
n
.
Note that K[a
1
,...,a
n
] precisely consists of elements of the form f(a
1
,...,a
n
) where
f(X
1
,...,X
n
)variesoverK[X
1
,...,X
n
](=theringofpolynomialsinthenvariablesX
1
,...,X
n
with coe?cients in K) whereas K(a
1
,...,a
n
) precisely consists of elements of the form
f(a
1
,...,an)
g(a
1
,...,an)
wheref(X
1
,...,X
n
),g(X
1
,...,X
n
)varyoverK[X
1
,...,X
n
]withg(a
1
,...,a
n
)6= 0.
Also note that K(a
1
,...,a
n
) is the quotient ?eld of K[a
1
,...,a
n
] in L.
De?nition: An extension L of K is said to be ?nitely generated over K if there exist
a
1
,...,a
n
in L such that L = K(a
1
,...,a
n
). We say that L is a simple extension of K if
L = K(a) for some a? L.
For simple extensions, the converse to Lemma 2 is true. In fact, we can say much more.
Lemma 3: Let a be an element in an over?eld L of a ?eld K. Then:
K(a)/K is algebraic ? a is algebraic over K ? K[a] =K(a)? [K(a) : K] <8.
Moreover, if a is algebraic over K and f(X) =Irr(a,K), then there exists an isomorphism of
K(a) onto K[X]/(f(X)) which maps a to X, the residue class of X, and the elements of K
to their residue classes.
Proof: Without loss of generality, we can and will assume that a6= 0. The ?rst assertion
trivially implies the second. Now, the map ? : K[X]? L de?ned by f(X)7? f(a) is clearly
a ring homomorphism whose image is K[a]. If a is algebraic over K, then the kernel of ? is
a nonzero prime ideal in K[X] and is hence a maximal ideal (prove!). So K[a]? K[X]/ker?
is a ?eld containing K and a. Therefore K[a] = K(a). Next, if K[a] = K(a), we can write
a
-1
= a
0
+a
1
a+···+a
r
a
r
for some a
0
,...,a
r
? K with a
r
6= 0, which shows that a
r+1
lies
in the K–linear span of 1,a,a
2
,...,a
r
, and consequently so does a
r+j
for any j = 1. And
since 1,a,a
2
,... clearly span K[a] = K(a), it follows that [K(a) : K] = r + 1 < 8. If
[K(a) : K] <8, Lemma 2 shows that K(a) is algebraic over K. Moreover, if a is algebraic
over K and f(X) =Irr(a,K), then, as noted earlier, ker? is generated by f(X), from which
we get the desired isomorphism between K(a) and K[X]/(f(X)). 2
Exercise 2: If a is algebraic over K, then show that [K(a) : K] equals the degree of
Irr(a,K).
Exercise 3: Try to give a more constructive proof of the fact that if a is algebraic over
K, then K[a] = K(a) by showing that for any g(X) ? K[X] with g(a) 6= 0, we can ?nd
h(X)? K[X] such that g(a)
-1
= h(a).
Thefollowinglemmagivesnecessaryandsu?cient conditionsfortheconversetoLemma 2.
Page 3


2 Field Extensions
Let K be a ?eld
2
. By a (?eld) extension of K we mean a ?eld containing K as a sub?eld.
Let a ?eld L be an extension of K (we usually express this by saying that L/K [read: L over
K] is an extension). Then L can be considered as a vector space over K. The degree of L over
K, denoted by [L : K], is de?ned as
[L : K] = dim
K
L = the vector space dimension of L over K.
If [L : K] <8, we say that L is a ?nite extension of K or that L is ?nite over K. A sub?eld
K of I C such that [K :
|
Q] <8 is called an algebraic number ?eld or simply a number ?eld.
Lemma 1: Finite over ?nite is ?nite. More precisely, ifL/E andE/K are ?eld extensions,
then
L is finite over K ? L is finite over E and E is finite over K
and, in this case, [L :K] = [L : E][E :K].
Proof: The implication “?” is obvious. The rest follows easily from the observation that
if{u
i
} is an E–basis of L and{v
j
} is a K–basis of E, then{u
i
v
j
} is a K–basis of L. 2
Let L/K bea ?eld extension. An element a? Lis said to be algebraicover K if it satis?es
a nonzero polynomial with coe?cients in K, i.e, ? 0 6= f(X) ? K[X] such that f(a) = 0.
Given a ? L which is algebraic over K, we can ?nd a monic polynomial in K[X] of least
possible degree, satis?ed by a. This is unique and is called the minimal polynomial of a over
K. It is easily seen to be irreducible and we will denote it by Irr(a,K). Note that if f(X)
is any monic irreducible polynomial satis?ed by a, then we must have f(X) =Irr(a,K) and
that it generates the ideal{g(X)? K[X] : g(a) = 0} in K[X].
3
The extension L of K is said
to be algebraic if every element of L is algebraic over K.
Lemma 2: Finite? algebraic. That is, if L/K is a ?nite extension, then it is algebraic.
Proof: For any a? L, there must exist a positive integer n such that {1,a,a
2
,...,a
n
} is
linearly dependent over K, thus showing that a is algebraic over K. 2
Exercise 1: Show, by an example, that the converse of the above lemma is not true, in
general.
We now study extensions for which the converse is true.
2
Fields are usually denoted by K or k since the German word for ?eld is K¨ orper. Much of Modern
Field Theory was created by the German mathematician E. Steinitz; see his paper “Algebraische Theorie der
K¨ orper”, Crelle Journal (1910), pp. 167–308, for an original exposition.
3
It may be instructive to verify the observations made in the last few statements. General Hint: Use the
Division Algorithm in K[X].
De?nition: Given elements a
1
,...,a
n
in an extension L of a ?eld K, we de?ne
K[a
1
,...,a
n
] = the smallest subring of L containing K and a
1
,...,a
n
K(a
1
,...,a
n
) = the smallest sub?eld of L containing K and a
1
,...,a
n
.
Note that K[a
1
,...,a
n
] precisely consists of elements of the form f(a
1
,...,a
n
) where
f(X
1
,...,X
n
)variesoverK[X
1
,...,X
n
](=theringofpolynomialsinthenvariablesX
1
,...,X
n
with coe?cients in K) whereas K(a
1
,...,a
n
) precisely consists of elements of the form
f(a
1
,...,an)
g(a
1
,...,an)
wheref(X
1
,...,X
n
),g(X
1
,...,X
n
)varyoverK[X
1
,...,X
n
]withg(a
1
,...,a
n
)6= 0.
Also note that K(a
1
,...,a
n
) is the quotient ?eld of K[a
1
,...,a
n
] in L.
De?nition: An extension L of K is said to be ?nitely generated over K if there exist
a
1
,...,a
n
in L such that L = K(a
1
,...,a
n
). We say that L is a simple extension of K if
L = K(a) for some a? L.
For simple extensions, the converse to Lemma 2 is true. In fact, we can say much more.
Lemma 3: Let a be an element in an over?eld L of a ?eld K. Then:
K(a)/K is algebraic ? a is algebraic over K ? K[a] =K(a)? [K(a) : K] <8.
Moreover, if a is algebraic over K and f(X) =Irr(a,K), then there exists an isomorphism of
K(a) onto K[X]/(f(X)) which maps a to X, the residue class of X, and the elements of K
to their residue classes.
Proof: Without loss of generality, we can and will assume that a6= 0. The ?rst assertion
trivially implies the second. Now, the map ? : K[X]? L de?ned by f(X)7? f(a) is clearly
a ring homomorphism whose image is K[a]. If a is algebraic over K, then the kernel of ? is
a nonzero prime ideal in K[X] and is hence a maximal ideal (prove!). So K[a]? K[X]/ker?
is a ?eld containing K and a. Therefore K[a] = K(a). Next, if K[a] = K(a), we can write
a
-1
= a
0
+a
1
a+···+a
r
a
r
for some a
0
,...,a
r
? K with a
r
6= 0, which shows that a
r+1
lies
in the K–linear span of 1,a,a
2
,...,a
r
, and consequently so does a
r+j
for any j = 1. And
since 1,a,a
2
,... clearly span K[a] = K(a), it follows that [K(a) : K] = r + 1 < 8. If
[K(a) : K] <8, Lemma 2 shows that K(a) is algebraic over K. Moreover, if a is algebraic
over K and f(X) =Irr(a,K), then, as noted earlier, ker? is generated by f(X), from which
we get the desired isomorphism between K(a) and K[X]/(f(X)). 2
Exercise 2: If a is algebraic over K, then show that [K(a) : K] equals the degree of
Irr(a,K).
Exercise 3: Try to give a more constructive proof of the fact that if a is algebraic over
K, then K[a] = K(a) by showing that for any g(X) ? K[X] with g(a) 6= 0, we can ?nd
h(X)? K[X] such that g(a)
-1
= h(a).
Thefollowinglemmagivesnecessaryandsu?cient conditionsfortheconversetoLemma 2.
Lemma 4: Let L be an extension of a ?eld K. Then:
L is finite over K ? L is algebraic and finitely generated over K.
Proof: If L is ?nite over K, then it is algebraic, and if u
1
,...,u
n
is a K–basis of L, then
clearly L = K(u
1
,...,u
n
). Conversely, if L = K(a
1
,...,a
n
) for some a
1
,...,a
n
? K, then
using Lemmas 1 and 3 and induction on n, it is seen that L is ?nite over K. 2
Let us obtain some useful consequences of the above lemma.
Lemma 5: Algebraic over algebraic is algebraic. More precisely, if L/E and E/K are
?eld extensions, then:
L is algebraic over K ? L is algebraic over E and E is algebraic over K
Proof: The implication “?” is obvious. To prove the other one, take any a ? L. Find
b
0
,b
1
,...,b
n
? E, not all zero, such that b
0
+b
1
a+···+b
n
a
n
= 0. Then a is algebraic over
K(b
0
,b
1
,...,b
n
), and K(b
0
,b
1
,...,b
n
)? E is algebraic over K. Hence, in view of Lemmas 1,
3 and 4, we see that
[K(a) : K] = [K(b
0
,b
1
,...,b
n
,a) :K]
= [K(b
0
,b
1
,...,b
n
,a) :K(b
0
,b
1
,...,b
n
)][K(b
0
,b
1
,...,b
n
) : K]
< 8
which shows that a is algebraic over K. 2
Lemma 6: Let L be an extension of a ?eld K and let
E ={a? L : a is algebraic over K}.
Then E is a sub?eld of L containing K.
Proof: Clearly K ? E? L. Given any a,ß? E, by Lemma 3, we see that
[K(a,ß) : K] = [K(a,ß) : K(a)][K(a) : K] <8
and therefore every element of K(a,ß) is algebraic over K. So a+ß,a-ß,aß ? E and if
ß6= 0, then
a
ß
? E, and hence E is a sub?eld of L. 2
Exercise 4: Givenelementsa,ß,algebraicovera?eldK,canyouexplicitly?ndpolynomials
in K[X] satis?ed by a + ß, aß? Find, for instance, a polynomial, preferably irreducible,
satis?ed by
v
2+
v
3.
5
Page 4


2 Field Extensions
Let K be a ?eld
2
. By a (?eld) extension of K we mean a ?eld containing K as a sub?eld.
Let a ?eld L be an extension of K (we usually express this by saying that L/K [read: L over
K] is an extension). Then L can be considered as a vector space over K. The degree of L over
K, denoted by [L : K], is de?ned as
[L : K] = dim
K
L = the vector space dimension of L over K.
If [L : K] <8, we say that L is a ?nite extension of K or that L is ?nite over K. A sub?eld
K of I C such that [K :
|
Q] <8 is called an algebraic number ?eld or simply a number ?eld.
Lemma 1: Finite over ?nite is ?nite. More precisely, ifL/E andE/K are ?eld extensions,
then
L is finite over K ? L is finite over E and E is finite over K
and, in this case, [L :K] = [L : E][E :K].
Proof: The implication “?” is obvious. The rest follows easily from the observation that
if{u
i
} is an E–basis of L and{v
j
} is a K–basis of E, then{u
i
v
j
} is a K–basis of L. 2
Let L/K bea ?eld extension. An element a? Lis said to be algebraicover K if it satis?es
a nonzero polynomial with coe?cients in K, i.e, ? 0 6= f(X) ? K[X] such that f(a) = 0.
Given a ? L which is algebraic over K, we can ?nd a monic polynomial in K[X] of least
possible degree, satis?ed by a. This is unique and is called the minimal polynomial of a over
K. It is easily seen to be irreducible and we will denote it by Irr(a,K). Note that if f(X)
is any monic irreducible polynomial satis?ed by a, then we must have f(X) =Irr(a,K) and
that it generates the ideal{g(X)? K[X] : g(a) = 0} in K[X].
3
The extension L of K is said
to be algebraic if every element of L is algebraic over K.
Lemma 2: Finite? algebraic. That is, if L/K is a ?nite extension, then it is algebraic.
Proof: For any a? L, there must exist a positive integer n such that {1,a,a
2
,...,a
n
} is
linearly dependent over K, thus showing that a is algebraic over K. 2
Exercise 1: Show, by an example, that the converse of the above lemma is not true, in
general.
We now study extensions for which the converse is true.
2
Fields are usually denoted by K or k since the German word for ?eld is K¨ orper. Much of Modern
Field Theory was created by the German mathematician E. Steinitz; see his paper “Algebraische Theorie der
K¨ orper”, Crelle Journal (1910), pp. 167–308, for an original exposition.
3
It may be instructive to verify the observations made in the last few statements. General Hint: Use the
Division Algorithm in K[X].
De?nition: Given elements a
1
,...,a
n
in an extension L of a ?eld K, we de?ne
K[a
1
,...,a
n
] = the smallest subring of L containing K and a
1
,...,a
n
K(a
1
,...,a
n
) = the smallest sub?eld of L containing K and a
1
,...,a
n
.
Note that K[a
1
,...,a
n
] precisely consists of elements of the form f(a
1
,...,a
n
) where
f(X
1
,...,X
n
)variesoverK[X
1
,...,X
n
](=theringofpolynomialsinthenvariablesX
1
,...,X
n
with coe?cients in K) whereas K(a
1
,...,a
n
) precisely consists of elements of the form
f(a
1
,...,an)
g(a
1
,...,an)
wheref(X
1
,...,X
n
),g(X
1
,...,X
n
)varyoverK[X
1
,...,X
n
]withg(a
1
,...,a
n
)6= 0.
Also note that K(a
1
,...,a
n
) is the quotient ?eld of K[a
1
,...,a
n
] in L.
De?nition: An extension L of K is said to be ?nitely generated over K if there exist
a
1
,...,a
n
in L such that L = K(a
1
,...,a
n
). We say that L is a simple extension of K if
L = K(a) for some a? L.
For simple extensions, the converse to Lemma 2 is true. In fact, we can say much more.
Lemma 3: Let a be an element in an over?eld L of a ?eld K. Then:
K(a)/K is algebraic ? a is algebraic over K ? K[a] =K(a)? [K(a) : K] <8.
Moreover, if a is algebraic over K and f(X) =Irr(a,K), then there exists an isomorphism of
K(a) onto K[X]/(f(X)) which maps a to X, the residue class of X, and the elements of K
to their residue classes.
Proof: Without loss of generality, we can and will assume that a6= 0. The ?rst assertion
trivially implies the second. Now, the map ? : K[X]? L de?ned by f(X)7? f(a) is clearly
a ring homomorphism whose image is K[a]. If a is algebraic over K, then the kernel of ? is
a nonzero prime ideal in K[X] and is hence a maximal ideal (prove!). So K[a]? K[X]/ker?
is a ?eld containing K and a. Therefore K[a] = K(a). Next, if K[a] = K(a), we can write
a
-1
= a
0
+a
1
a+···+a
r
a
r
for some a
0
,...,a
r
? K with a
r
6= 0, which shows that a
r+1
lies
in the K–linear span of 1,a,a
2
,...,a
r
, and consequently so does a
r+j
for any j = 1. And
since 1,a,a
2
,... clearly span K[a] = K(a), it follows that [K(a) : K] = r + 1 < 8. If
[K(a) : K] <8, Lemma 2 shows that K(a) is algebraic over K. Moreover, if a is algebraic
over K and f(X) =Irr(a,K), then, as noted earlier, ker? is generated by f(X), from which
we get the desired isomorphism between K(a) and K[X]/(f(X)). 2
Exercise 2: If a is algebraic over K, then show that [K(a) : K] equals the degree of
Irr(a,K).
Exercise 3: Try to give a more constructive proof of the fact that if a is algebraic over
K, then K[a] = K(a) by showing that for any g(X) ? K[X] with g(a) 6= 0, we can ?nd
h(X)? K[X] such that g(a)
-1
= h(a).
Thefollowinglemmagivesnecessaryandsu?cient conditionsfortheconversetoLemma 2.
Lemma 4: Let L be an extension of a ?eld K. Then:
L is finite over K ? L is algebraic and finitely generated over K.
Proof: If L is ?nite over K, then it is algebraic, and if u
1
,...,u
n
is a K–basis of L, then
clearly L = K(u
1
,...,u
n
). Conversely, if L = K(a
1
,...,a
n
) for some a
1
,...,a
n
? K, then
using Lemmas 1 and 3 and induction on n, it is seen that L is ?nite over K. 2
Let us obtain some useful consequences of the above lemma.
Lemma 5: Algebraic over algebraic is algebraic. More precisely, if L/E and E/K are
?eld extensions, then:
L is algebraic over K ? L is algebraic over E and E is algebraic over K
Proof: The implication “?” is obvious. To prove the other one, take any a ? L. Find
b
0
,b
1
,...,b
n
? E, not all zero, such that b
0
+b
1
a+···+b
n
a
n
= 0. Then a is algebraic over
K(b
0
,b
1
,...,b
n
), and K(b
0
,b
1
,...,b
n
)? E is algebraic over K. Hence, in view of Lemmas 1,
3 and 4, we see that
[K(a) : K] = [K(b
0
,b
1
,...,b
n
,a) :K]
= [K(b
0
,b
1
,...,b
n
,a) :K(b
0
,b
1
,...,b
n
)][K(b
0
,b
1
,...,b
n
) : K]
< 8
which shows that a is algebraic over K. 2
Lemma 6: Let L be an extension of a ?eld K and let
E ={a? L : a is algebraic over K}.
Then E is a sub?eld of L containing K.
Proof: Clearly K ? E? L. Given any a,ß? E, by Lemma 3, we see that
[K(a,ß) : K] = [K(a,ß) : K(a)][K(a) : K] <8
and therefore every element of K(a,ß) is algebraic over K. So a+ß,a-ß,aß ? E and if
ß6= 0, then
a
ß
? E, and hence E is a sub?eld of L. 2
Exercise 4: Givenelementsa,ß,algebraicovera?eldK,canyouexplicitly?ndpolynomials
in K[X] satis?ed by a + ß, aß? Find, for instance, a polynomial, preferably irreducible,
satis?ed by
v
2+
v
3.
5
3 Splitting Fields and Normal Extensions
Galois Theory, at least in its original version, has to do with roots of polynomial equations.
This motivates much of what is done in this section.
Let K be a ?eld. By a root of a polynomial f(X) ? K[X] we mean an element a in an
over?eld of K such that f(a) = 0. It is easy to see that a nonzero polynomial in K[X] of
degree n has at most n roots (Verify!). The following lemma, usually attributed to Kronecker,
shows, by a method not unlike witchcraft, that roots can always be found.
Lemma 7: Let f(X)? K[X] be a nonconstant polynomial of degree n. Then there exists
an extension E of K such that [E : K]= n and f(X) has a root in E.
Proof: Let g(X) be a monic irreducible factor of f(X). Then (g(X)), the ideal generated
byg(X)inK[X],isamaximalidealandhenceE =K[X]/(g(X))isa?eld. Lets : K[X]? E
be the canonical homomorphism which maps an element in K[X] to its residue class modulo
(g(X)). Note that s|
K
is injective and hence K may be regarded as a sub?eld of E. Let
a = s(X). Then g(a) = g(s(X)) = s(g(X)) = 0, and hence f(a) = 0. From Lemma 3 and
Exercise 2, it follows that [E : K] = degg(X)= n. 2
Remark: The above proof, though common in many texts, is slightly imprecise. To be
pedantic, an actual extension E of K as in the statement of Lemma 6 can be constructed by
putting
E = (s(K[X])\s(K))?K
where s is as in the above proof, and by de?ning ?eld operations on E in an obvious manner.
Note that we then have E? s(K[X]).
To study the roots of a polynomial f(X) ? K[X], it seems natural to be in a nice set
containing all the roots of f(X) and which, in some sense, is the smallest such. This is
a?orded by the following.
De?nition: Let f(X)? K[X] be a nonconstant polynomial. By a splitting ?eld of f(X)
over K we mean an extension L of K such that f(X) splits into linear factors in L and L is
generated over K by the roots of f(X) in L, i.e.,
(i) f(X) =c(X-a
1
)...(X-a
n
) for some c? K and a
1
,...,a
n
? L.
(ii) L =K(a
1
,...,a
n
).
Lemma 8: Given any nonconstant polynomial f(X)? K[X] of degree n, there exists a
splitting ?eld L of f(X) over K such that [L : K]= n!.
Proof: Induct on n. If n = 1, then L = K does the job. For n > 1, by Lemma
7, we can ?nd an extension E of K such that [E : K] = n and f(X) = (X - a)g(X)
for some a ? E and g(X) ? E[X]. Since degg(X) = n- 1 = 1, a splitting ?eld, say
L, of g(X) over E exists. Clearly, L is also a splitting ?eld of f(X) over K; moreover,
[L : K] = [L :E][E : K]= (n-1)!n =n!. 2
Page 5


2 Field Extensions
Let K be a ?eld
2
. By a (?eld) extension of K we mean a ?eld containing K as a sub?eld.
Let a ?eld L be an extension of K (we usually express this by saying that L/K [read: L over
K] is an extension). Then L can be considered as a vector space over K. The degree of L over
K, denoted by [L : K], is de?ned as
[L : K] = dim
K
L = the vector space dimension of L over K.
If [L : K] <8, we say that L is a ?nite extension of K or that L is ?nite over K. A sub?eld
K of I C such that [K :
|
Q] <8 is called an algebraic number ?eld or simply a number ?eld.
Lemma 1: Finite over ?nite is ?nite. More precisely, ifL/E andE/K are ?eld extensions,
then
L is finite over K ? L is finite over E and E is finite over K
and, in this case, [L :K] = [L : E][E :K].
Proof: The implication “?” is obvious. The rest follows easily from the observation that
if{u
i
} is an E–basis of L and{v
j
} is a K–basis of E, then{u
i
v
j
} is a K–basis of L. 2
Let L/K bea ?eld extension. An element a? Lis said to be algebraicover K if it satis?es
a nonzero polynomial with coe?cients in K, i.e, ? 0 6= f(X) ? K[X] such that f(a) = 0.
Given a ? L which is algebraic over K, we can ?nd a monic polynomial in K[X] of least
possible degree, satis?ed by a. This is unique and is called the minimal polynomial of a over
K. It is easily seen to be irreducible and we will denote it by Irr(a,K). Note that if f(X)
is any monic irreducible polynomial satis?ed by a, then we must have f(X) =Irr(a,K) and
that it generates the ideal{g(X)? K[X] : g(a) = 0} in K[X].
3
The extension L of K is said
to be algebraic if every element of L is algebraic over K.
Lemma 2: Finite? algebraic. That is, if L/K is a ?nite extension, then it is algebraic.
Proof: For any a? L, there must exist a positive integer n such that {1,a,a
2
,...,a
n
} is
linearly dependent over K, thus showing that a is algebraic over K. 2
Exercise 1: Show, by an example, that the converse of the above lemma is not true, in
general.
We now study extensions for which the converse is true.
2
Fields are usually denoted by K or k since the German word for ?eld is K¨ orper. Much of Modern
Field Theory was created by the German mathematician E. Steinitz; see his paper “Algebraische Theorie der
K¨ orper”, Crelle Journal (1910), pp. 167–308, for an original exposition.
3
It may be instructive to verify the observations made in the last few statements. General Hint: Use the
Division Algorithm in K[X].
De?nition: Given elements a
1
,...,a
n
in an extension L of a ?eld K, we de?ne
K[a
1
,...,a
n
] = the smallest subring of L containing K and a
1
,...,a
n
K(a
1
,...,a
n
) = the smallest sub?eld of L containing K and a
1
,...,a
n
.
Note that K[a
1
,...,a
n
] precisely consists of elements of the form f(a
1
,...,a
n
) where
f(X
1
,...,X
n
)variesoverK[X
1
,...,X
n
](=theringofpolynomialsinthenvariablesX
1
,...,X
n
with coe?cients in K) whereas K(a
1
,...,a
n
) precisely consists of elements of the form
f(a
1
,...,an)
g(a
1
,...,an)
wheref(X
1
,...,X
n
),g(X
1
,...,X
n
)varyoverK[X
1
,...,X
n
]withg(a
1
,...,a
n
)6= 0.
Also note that K(a
1
,...,a
n
) is the quotient ?eld of K[a
1
,...,a
n
] in L.
De?nition: An extension L of K is said to be ?nitely generated over K if there exist
a
1
,...,a
n
in L such that L = K(a
1
,...,a
n
). We say that L is a simple extension of K if
L = K(a) for some a? L.
For simple extensions, the converse to Lemma 2 is true. In fact, we can say much more.
Lemma 3: Let a be an element in an over?eld L of a ?eld K. Then:
K(a)/K is algebraic ? a is algebraic over K ? K[a] =K(a)? [K(a) : K] <8.
Moreover, if a is algebraic over K and f(X) =Irr(a,K), then there exists an isomorphism of
K(a) onto K[X]/(f(X)) which maps a to X, the residue class of X, and the elements of K
to their residue classes.
Proof: Without loss of generality, we can and will assume that a6= 0. The ?rst assertion
trivially implies the second. Now, the map ? : K[X]? L de?ned by f(X)7? f(a) is clearly
a ring homomorphism whose image is K[a]. If a is algebraic over K, then the kernel of ? is
a nonzero prime ideal in K[X] and is hence a maximal ideal (prove!). So K[a]? K[X]/ker?
is a ?eld containing K and a. Therefore K[a] = K(a). Next, if K[a] = K(a), we can write
a
-1
= a
0
+a
1
a+···+a
r
a
r
for some a
0
,...,a
r
? K with a
r
6= 0, which shows that a
r+1
lies
in the K–linear span of 1,a,a
2
,...,a
r
, and consequently so does a
r+j
for any j = 1. And
since 1,a,a
2
,... clearly span K[a] = K(a), it follows that [K(a) : K] = r + 1 < 8. If
[K(a) : K] <8, Lemma 2 shows that K(a) is algebraic over K. Moreover, if a is algebraic
over K and f(X) =Irr(a,K), then, as noted earlier, ker? is generated by f(X), from which
we get the desired isomorphism between K(a) and K[X]/(f(X)). 2
Exercise 2: If a is algebraic over K, then show that [K(a) : K] equals the degree of
Irr(a,K).
Exercise 3: Try to give a more constructive proof of the fact that if a is algebraic over
K, then K[a] = K(a) by showing that for any g(X) ? K[X] with g(a) 6= 0, we can ?nd
h(X)? K[X] such that g(a)
-1
= h(a).
Thefollowinglemmagivesnecessaryandsu?cient conditionsfortheconversetoLemma 2.
Lemma 4: Let L be an extension of a ?eld K. Then:
L is finite over K ? L is algebraic and finitely generated over K.
Proof: If L is ?nite over K, then it is algebraic, and if u
1
,...,u
n
is a K–basis of L, then
clearly L = K(u
1
,...,u
n
). Conversely, if L = K(a
1
,...,a
n
) for some a
1
,...,a
n
? K, then
using Lemmas 1 and 3 and induction on n, it is seen that L is ?nite over K. 2
Let us obtain some useful consequences of the above lemma.
Lemma 5: Algebraic over algebraic is algebraic. More precisely, if L/E and E/K are
?eld extensions, then:
L is algebraic over K ? L is algebraic over E and E is algebraic over K
Proof: The implication “?” is obvious. To prove the other one, take any a ? L. Find
b
0
,b
1
,...,b
n
? E, not all zero, such that b
0
+b
1
a+···+b
n
a
n
= 0. Then a is algebraic over
K(b
0
,b
1
,...,b
n
), and K(b
0
,b
1
,...,b
n
)? E is algebraic over K. Hence, in view of Lemmas 1,
3 and 4, we see that
[K(a) : K] = [K(b
0
,b
1
,...,b
n
,a) :K]
= [K(b
0
,b
1
,...,b
n
,a) :K(b
0
,b
1
,...,b
n
)][K(b
0
,b
1
,...,b
n
) : K]
< 8
which shows that a is algebraic over K. 2
Lemma 6: Let L be an extension of a ?eld K and let
E ={a? L : a is algebraic over K}.
Then E is a sub?eld of L containing K.
Proof: Clearly K ? E? L. Given any a,ß? E, by Lemma 3, we see that
[K(a,ß) : K] = [K(a,ß) : K(a)][K(a) : K] <8
and therefore every element of K(a,ß) is algebraic over K. So a+ß,a-ß,aß ? E and if
ß6= 0, then
a
ß
? E, and hence E is a sub?eld of L. 2
Exercise 4: Givenelementsa,ß,algebraicovera?eldK,canyouexplicitly?ndpolynomials
in K[X] satis?ed by a + ß, aß? Find, for instance, a polynomial, preferably irreducible,
satis?ed by
v
2+
v
3.
5
3 Splitting Fields and Normal Extensions
Galois Theory, at least in its original version, has to do with roots of polynomial equations.
This motivates much of what is done in this section.
Let K be a ?eld. By a root of a polynomial f(X) ? K[X] we mean an element a in an
over?eld of K such that f(a) = 0. It is easy to see that a nonzero polynomial in K[X] of
degree n has at most n roots (Verify!). The following lemma, usually attributed to Kronecker,
shows, by a method not unlike witchcraft, that roots can always be found.
Lemma 7: Let f(X)? K[X] be a nonconstant polynomial of degree n. Then there exists
an extension E of K such that [E : K]= n and f(X) has a root in E.
Proof: Let g(X) be a monic irreducible factor of f(X). Then (g(X)), the ideal generated
byg(X)inK[X],isamaximalidealandhenceE =K[X]/(g(X))isa?eld. Lets : K[X]? E
be the canonical homomorphism which maps an element in K[X] to its residue class modulo
(g(X)). Note that s|
K
is injective and hence K may be regarded as a sub?eld of E. Let
a = s(X). Then g(a) = g(s(X)) = s(g(X)) = 0, and hence f(a) = 0. From Lemma 3 and
Exercise 2, it follows that [E : K] = degg(X)= n. 2
Remark: The above proof, though common in many texts, is slightly imprecise. To be
pedantic, an actual extension E of K as in the statement of Lemma 6 can be constructed by
putting
E = (s(K[X])\s(K))?K
where s is as in the above proof, and by de?ning ?eld operations on E in an obvious manner.
Note that we then have E? s(K[X]).
To study the roots of a polynomial f(X) ? K[X], it seems natural to be in a nice set
containing all the roots of f(X) and which, in some sense, is the smallest such. This is
a?orded by the following.
De?nition: Let f(X)? K[X] be a nonconstant polynomial. By a splitting ?eld of f(X)
over K we mean an extension L of K such that f(X) splits into linear factors in L and L is
generated over K by the roots of f(X) in L, i.e.,
(i) f(X) =c(X-a
1
)...(X-a
n
) for some c? K and a
1
,...,a
n
? L.
(ii) L =K(a
1
,...,a
n
).
Lemma 8: Given any nonconstant polynomial f(X)? K[X] of degree n, there exists a
splitting ?eld L of f(X) over K such that [L : K]= n!.
Proof: Induct on n. If n = 1, then L = K does the job. For n > 1, by Lemma
7, we can ?nd an extension E of K such that [E : K] = n and f(X) = (X - a)g(X)
for some a ? E and g(X) ? E[X]. Since degg(X) = n- 1 = 1, a splitting ?eld, say
L, of g(X) over E exists. Clearly, L is also a splitting ?eld of f(X) over K; moreover,
[L : K] = [L :E][E : K]= (n-1)!n =n!. 2
Notation: Given any ?elds K and K
'
, a homomorphism s : K ? K
'
, and a polynomial
f(X) ? K[X], by f
s
(X) we denote the corresponding polynomial in K
'
[X], i.e., if f(X) =
P
a
i
X
i
then f
s
(X) =
P
s(a
i
)X
i
. Note that f(X) 7? f
s
(X) gives a homomorphism of
K[X]? K
'
[X] which is an isomorphism if s is an isomorphism.
The following lemma will help us prove that a splitting ?eld is unique up to isomorphism.
Lemma 9: LetK andK
'
be ?elds ands :K ? K
'
be an isomorphism. Let g(X)? K[X]
be an irreducible polynomial and let a and a
'
be roots of g(X) and g
s
(X) in some extensions
of K and K
'
respectively. Then there exists an isomorphism ? : K(a) ? K
'
(a
'
) such that
?|
K
= s and ?(a) = a
'
.
Proof: Clearly s gives an isomorphism of K[X] onto K
'
[X], which, in turn, induces an
isomorphism of K[X]/(g(X)) onto K
'
[X]/(g
s
(X)). By Lemma 3, we get an isomorphism of
K(a) onto the former and of K
'
(a
'
) onto the latter. By suitably composing these maps, we
obtain an isomorphism ? : K(a)? K
'
(a
'
) such that ?|
K
=s and ?(a) = a
'
. 2
Note: A ?eld has no proper ideals. This means that a homomorphism of a ?eld (into
a ring) is either injective or maps everything to 0. If L is an extension of K, by a K–
homomorphism of L we mean a homomorphism s : L ? L
'
, where L
'
is some extension of
K, which is identity on K, i.e., s(c) = c ?c ? K. Observe that a K–homomorphism is
always injective.
4
Also observe that, when L/K is ?nite, a K–homomorphism s : L? L is
necessarily an automorphism (= isomorphism onto itself) of L [because s(L) is a subspace of
L and the vector space dimension over K of L and s(L) is the same].
Before proving the uniqueness of splitting ?elds, let us deduce an important consequence
of the above lemma.
Corollary: Let a be algebraic over K and f(X) = Irr(a,K). Let L be any extension of
K containing a splitting ?eld of f(X). Then the number of K–homomorphisms of K(a) to L
is equal to the number of distinct roots of f(X); in particular, this number is = [K(a) : K]
with equality holding if and only if all roots of f(X) are distinct.
Proof: Let a
1
,...,a
r
? L be all possible distinct roots of f(X). By Lemma 9, there
exist K–isomorphisms ?
i
: K(a) ? K(a
i
) such that ?
i
(a) = a
i
(1 = i = r). Moreover, if
s : K(a)? L is any K–homomorphism, then f
s
(X) = f(X), and hence s(a) = a
i
for some
i, which shows that s = ?
i
. The inequality r= [K(a) :K] follows from Exercise 2. 2
Lemma10: LetK andK
'
be?eldsands :K ? K
'
beanisomorphism. Letf(X)? K[X]
be any nonconstant polynomial and let L and L
'
be splitting ?elds of f(X) and f
s
(X) over
K and K
'
respectively. Then there exists an isomorphism t : L ? L
'
such that t|
K
= s.
Moreover, the number of such isomorphisms is= [L : K].
Proof: Let n = degf(X) = degf
s
(X)= 1. We proceed by induction on n. If n = 1, we
4
Indeed, 1?K and s(1) = 16=0.