Algebra: Section - 3 (Fields) | Mathematics Optional Notes for UPSC PDF Download

 Page 1


Edurev123 
SECTION - III 
FIELDS 
 
7. FIELD EXTENSIONS  
Before we begin our study of fields, we shall present some facts on Polynomial rings. 
The set of polynomials in one Indeterminate with coefficients, in a field ?? ?? forms a ring: 
an integral domain which turns out to be a Principal ideal domain. 
[7:1] DEFINITION  
Let ?? be a field, and ?? a variable (also called indeterminate). We call an expression 
(7.1) ?? (?? )=?? 0
+?? 1
?? +?? +?? ?? ?? ?? ,?? ?? +0 ?? ?? ??? (?? =0,1,…,?? ) a polynomial in ?? with 
coefficients ?? ?? . We call ?? ?? the coefficient to ?? ?? and ?? ?? as the highest coefficient or the 
"leading coefficient"; n(?? ?? ?0) ," is called the "degree" of the polynomial 
 ?? (?? ):?? =?? eg ?? (?? ) . 
(7.2) If ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? 
we define the sum of two polynomials: ?? (?? )=?? (?? )?
 def 
?? =?? and ?? ?? =?? ?? (?? =0,1,…,?? ) 
(7.3) ?? (?? )+?? (?? ):?? ??? (?? 0
+?? 0
) +(?? 1
 
'
+?? 1
)?? +?+(?? ?? +?? ?? )?? ?? +… 
i.e., coefficient of ?? ?? in ?? (?? )+?? (?? ) = coefficient of ?? ?? in ?? (?? )+ coefficient of in 
?? (?? )(?? =0,1,2,… 
 
Example: ?? (?? )=?? 3
-?? ,?? (?? )=1+2?? -?? 3
+?? 5
 
?? (?? )+?? (?? )=?? 5
+?? +1 
Convention: The leading coefficient, by requirement of the very definition, is always ?
0. 
If coefficient of ?? ?? is = 0 in ?? (?? ) , we do not display the ?? ?? -term. Thus e.g. 
?? (?? )=?? 3
-?? =?? 3
+0.?? 2
-1.?? +0 
We define the product of two polynomials ?? (?? ) and ?? (?? ) by, (7.3) 
?? (?? )·?? (?? )=?? 0
+?? 1
?? +?+?? ?? +?? ?? ?? +?? 
wherein ?? ?? =?? 0
?? ?? +?? 1
?? ?? +?? ?? +?? ?? ?? 0
 
we put deg ?? (?? )=0, if ?? (?? )
(?? +1
?? 0
(?? 0
??? ) 
Page 2


Edurev123 
SECTION - III 
FIELDS 
 
7. FIELD EXTENSIONS  
Before we begin our study of fields, we shall present some facts on Polynomial rings. 
The set of polynomials in one Indeterminate with coefficients, in a field ?? ?? forms a ring: 
an integral domain which turns out to be a Principal ideal domain. 
[7:1] DEFINITION  
Let ?? be a field, and ?? a variable (also called indeterminate). We call an expression 
(7.1) ?? (?? )=?? 0
+?? 1
?? +?? +?? ?? ?? ?? ,?? ?? +0 ?? ?? ??? (?? =0,1,…,?? ) a polynomial in ?? with 
coefficients ?? ?? . We call ?? ?? the coefficient to ?? ?? and ?? ?? as the highest coefficient or the 
"leading coefficient"; n(?? ?? ?0) ," is called the "degree" of the polynomial 
 ?? (?? ):?? =?? eg ?? (?? ) . 
(7.2) If ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? 
we define the sum of two polynomials: ?? (?? )=?? (?? )?
 def 
?? =?? and ?? ?? =?? ?? (?? =0,1,…,?? ) 
(7.3) ?? (?? )+?? (?? ):?? ??? (?? 0
+?? 0
) +(?? 1
 
'
+?? 1
)?? +?+(?? ?? +?? ?? )?? ?? +… 
i.e., coefficient of ?? ?? in ?? (?? )+?? (?? ) = coefficient of ?? ?? in ?? (?? )+ coefficient of in 
?? (?? )(?? =0,1,2,… 
 
Example: ?? (?? )=?? 3
-?? ,?? (?? )=1+2?? -?? 3
+?? 5
 
?? (?? )+?? (?? )=?? 5
+?? +1 
Convention: The leading coefficient, by requirement of the very definition, is always ?
0. 
If coefficient of ?? ?? is = 0 in ?? (?? ) , we do not display the ?? ?? -term. Thus e.g. 
?? (?? )=?? 3
-?? =?? 3
+0.?? 2
-1.?? +0 
We define the product of two polynomials ?? (?? ) and ?? (?? ) by, (7.3) 
?? (?? )·?? (?? )=?? 0
+?? 1
?? +?+?? ?? +?? ?? ?? +?? 
wherein ?? ?? =?? 0
?? ?? +?? 1
?? ?? +?? ?? +?? ?? ?? 0
 
we put deg ?? (?? )=0, if ?? (?? )
(?? +1
?? 0
(?? 0
??? ) 
I.e., there is no ?? ?? term for ?? =1,2,.. Thus, fields coefficients are treated as polynomials 
of degree 0. 
We note: (7;4)deg ?? (?? ):?? (?? )=deg ?? (?? )+deg 8(?? ) 
for two polynomials ?? ,?? which are ?0. 
 
The zero-polynomial is the one in which coefficient of ?? ?? =0 for every ?? : 
0=0(?? )=0+0.?? +0.?? 2
…+0.?? ?? =0 
put 1=1(?? )=1+0·?? +?? ·?? 2
+?+0·?? ?? (for any ?? =1) 
Then ?? (?? ).1=?? (?? ) for any polynomial ?? (?? ) with sum and product thus defined the 
polynomials form a ring.  
(7.5) we denote it by ?? [?? ] 
Convention: The indeterminate ?? is shown in the notation. We call ?? [?? ] the polynomial 
ring  in one variable x over the field F. 
Owing to the familiar division process. (“Euclidean Algorithm") we have 
[7.2] Given two polynomials ?? (?? ) and ?? (?? ) ;  
?? ?0,? ?????????????????????? ?? (?? ) ?????? ?? (?? ) ?????? h ?? h???? ?? (?? )=?? (?? )?? (?? )+?? (?? ). 
 
(7.6) where ?? (?? )=0 or deg {?? (?? )]<deg [?? (?? )] is called the quotient and r is called the 
remainder. 
Proof: if deg ?? (?? )<eg ?? (?? ) , we can take ?? (?? )=0, and ?? (?? )=???? ) There is nothing to 
prove. Assume, then  
?????? ?? (?? )>deg ?? (?? ) 
?? (?? )=?? 0
+?? 1
?? +2+?? 1
?? 
?? (?? )=?? 0
+?? 1
?? +?? +?? 1
?? ??  
 
put ?? 1
(?? )=?? (?? )-
?? ?? ?? ?? ;?? -?? ?? ?? (?? ) (we have ?? ?? ?0); 
one gets rid of the ?? term, and deg ?? 1
(?? )=?? -1 
we can prove by induction on ?? ; thus ?? ?? may assume that if ?? 1
(?? )=0, the process 
terminates here, and. 
?? (?? )=
?? ?? 1
?? ?? ?? ?? -?? ·?? (?? )=?? (?? )?? (?? )+?? (?? )
 ?? (?? )-
?? ?? ?? ?? ?? ?? -?? , and ?? (?? )=0
 
Page 3


Edurev123 
SECTION - III 
FIELDS 
 
7. FIELD EXTENSIONS  
Before we begin our study of fields, we shall present some facts on Polynomial rings. 
The set of polynomials in one Indeterminate with coefficients, in a field ?? ?? forms a ring: 
an integral domain which turns out to be a Principal ideal domain. 
[7:1] DEFINITION  
Let ?? be a field, and ?? a variable (also called indeterminate). We call an expression 
(7.1) ?? (?? )=?? 0
+?? 1
?? +?? +?? ?? ?? ?? ,?? ?? +0 ?? ?? ??? (?? =0,1,…,?? ) a polynomial in ?? with 
coefficients ?? ?? . We call ?? ?? the coefficient to ?? ?? and ?? ?? as the highest coefficient or the 
"leading coefficient"; n(?? ?? ?0) ," is called the "degree" of the polynomial 
 ?? (?? ):?? =?? eg ?? (?? ) . 
(7.2) If ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? 
we define the sum of two polynomials: ?? (?? )=?? (?? )?
 def 
?? =?? and ?? ?? =?? ?? (?? =0,1,…,?? ) 
(7.3) ?? (?? )+?? (?? ):?? ??? (?? 0
+?? 0
) +(?? 1
 
'
+?? 1
)?? +?+(?? ?? +?? ?? )?? ?? +… 
i.e., coefficient of ?? ?? in ?? (?? )+?? (?? ) = coefficient of ?? ?? in ?? (?? )+ coefficient of in 
?? (?? )(?? =0,1,2,… 
 
Example: ?? (?? )=?? 3
-?? ,?? (?? )=1+2?? -?? 3
+?? 5
 
?? (?? )+?? (?? )=?? 5
+?? +1 
Convention: The leading coefficient, by requirement of the very definition, is always ?
0. 
If coefficient of ?? ?? is = 0 in ?? (?? ) , we do not display the ?? ?? -term. Thus e.g. 
?? (?? )=?? 3
-?? =?? 3
+0.?? 2
-1.?? +0 
We define the product of two polynomials ?? (?? ) and ?? (?? ) by, (7.3) 
?? (?? )·?? (?? )=?? 0
+?? 1
?? +?+?? ?? +?? ?? ?? +?? 
wherein ?? ?? =?? 0
?? ?? +?? 1
?? ?? +?? ?? +?? ?? ?? 0
 
we put deg ?? (?? )=0, if ?? (?? )
(?? +1
?? 0
(?? 0
??? ) 
I.e., there is no ?? ?? term for ?? =1,2,.. Thus, fields coefficients are treated as polynomials 
of degree 0. 
We note: (7;4)deg ?? (?? ):?? (?? )=deg ?? (?? )+deg 8(?? ) 
for two polynomials ?? ,?? which are ?0. 
 
The zero-polynomial is the one in which coefficient of ?? ?? =0 for every ?? : 
0=0(?? )=0+0.?? +0.?? 2
…+0.?? ?? =0 
put 1=1(?? )=1+0·?? +?? ·?? 2
+?+0·?? ?? (for any ?? =1) 
Then ?? (?? ).1=?? (?? ) for any polynomial ?? (?? ) with sum and product thus defined the 
polynomials form a ring.  
(7.5) we denote it by ?? [?? ] 
Convention: The indeterminate ?? is shown in the notation. We call ?? [?? ] the polynomial 
ring  in one variable x over the field F. 
Owing to the familiar division process. (“Euclidean Algorithm") we have 
[7.2] Given two polynomials ?? (?? ) and ?? (?? ) ;  
?? ?0,? ?????????????????????? ?? (?? ) ?????? ?? (?? ) ?????? h ?? h???? ?? (?? )=?? (?? )?? (?? )+?? (?? ). 
 
(7.6) where ?? (?? )=0 or deg {?? (?? )]<deg [?? (?? )] is called the quotient and r is called the 
remainder. 
Proof: if deg ?? (?? )<eg ?? (?? ) , we can take ?? (?? )=0, and ?? (?? )=???? ) There is nothing to 
prove. Assume, then  
?????? ?? (?? )>deg ?? (?? ) 
?? (?? )=?? 0
+?? 1
?? +2+?? 1
?? 
?? (?? )=?? 0
+?? 1
?? +?? +?? 1
?? ??  
 
put ?? 1
(?? )=?? (?? )-
?? ?? ?? ?? ;?? -?? ?? ?? (?? ) (we have ?? ?? ?0); 
one gets rid of the ?? term, and deg ?? 1
(?? )=?? -1 
we can prove by induction on ?? ; thus ?? ?? may assume that if ?? 1
(?? )=0, the process 
terminates here, and. 
?? (?? )=
?? ?? 1
?? ?? ?? ?? -?? ·?? (?? )=?? (?? )?? (?? )+?? (?? )
 ?? (?? )-
?? ?? ?? ?? ?? ?? -?? , and ?? (?? )=0
 
otherwise, suppose the theorem hold for polynomials of degree < n, Thus,  ? 
polynomials ?? 1
(?? ),?? 1
(?? ) such that ?? 1
(?? )=???
1
(?? )?? (x)+?? 1
(?? ) 
wherein either ?? 1
(?? )=0 or deg ?? 1
(?? )<deg ?? (?? ) 
Hence ?? (?? )=?? 1
(?? )+
?? ?? ?? 1
·?? ?? -?? ·?? (?? ) 
=[?? 1
(?? )?? (?? )+?? 1
·?? )]+
?? ?? )
?? ?? ·?? (?? -1)
·?? (?? ) 
=?? (?? )·?? (?? )+?? (:) , 
9:! 
where  ?? (?? )=?? 1
(?? )+
?? ????
?? ?? ·6
?? -?? and ?? (?? )=?? 1
(?? )  
This proves the proposition. 
[7:5] The polynomial ring ?? [?? ], over a field ?? , is a Principal Ideal Domain, 
This follows from [7.2] since ?? [?? ] is a Euclidean ring, and hence in a Principal Ideal 
Domain. 
Consequently, the results regarding prime elements unique factorization  that were 
established for ?? rincipal ideal domains are available for ?? [?? ] in particular: 
[7.4] Given two ?? olynomials ?? (?? ) and ?? (?? ) in ?? [?? ], they have a t.c.d ?? (?? ) : 
?? (?? )=(?? (?? ),?? (?? )) 
which can be expressed in the form  ?? (?? )=?? (?? )·?? (?? )+?? (?? )·?? (?? ) 
In particular, if,?? (?? ) and ?? (?? ) are relatively prime, there exist ?? (?? ),?? (?? ) in ?? [?? ] such that 
a( ?? )?? (?? )+?? (?? )?? (?? )=1 
 
[ 7:5] The prime elements of the principal is ideal domain ?? { x} are those polynomials 
?? (?? ) . ceg ?? (?? )=1 that are divisible only by elements of ?? (i.e. Units of the ring ?? [?? ]) 
and by polynomials (c.p.x), ?? ?0 in ?? . 
 (i.e., associated to ?? (?? ) ). They are characterized 
by the property that they are not divisible by any non-constant ( ??.?? .  deg =1 ) 
polynomial of lower degree. 
(7.7) The prime elements of ?? [?? ] are called 'prime' polynomials" or "irreducible 
polynomials". 
?? every cass if associated prime polynomials, we choose tho one, whose leading 
coefficient is =1; te. Ls call them "normalized". Thus 
Every ?? (?? )??? [?? ] can be uniquely factorised into a product of normalized prime (i.e., 
irreducible) polyn mials. it is not an easy matter to decide whe her a given polynomial is 
irreducible or not. Some sufficient conditions are known these are called "Irreducibility 
Criteria". One of these, easy to prove, and useful in applications, is the following: 
 
Page 4


Edurev123 
SECTION - III 
FIELDS 
 
7. FIELD EXTENSIONS  
Before we begin our study of fields, we shall present some facts on Polynomial rings. 
The set of polynomials in one Indeterminate with coefficients, in a field ?? ?? forms a ring: 
an integral domain which turns out to be a Principal ideal domain. 
[7:1] DEFINITION  
Let ?? be a field, and ?? a variable (also called indeterminate). We call an expression 
(7.1) ?? (?? )=?? 0
+?? 1
?? +?? +?? ?? ?? ?? ,?? ?? +0 ?? ?? ??? (?? =0,1,…,?? ) a polynomial in ?? with 
coefficients ?? ?? . We call ?? ?? the coefficient to ?? ?? and ?? ?? as the highest coefficient or the 
"leading coefficient"; n(?? ?? ?0) ," is called the "degree" of the polynomial 
 ?? (?? ):?? =?? eg ?? (?? ) . 
(7.2) If ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? 
we define the sum of two polynomials: ?? (?? )=?? (?? )?
 def 
?? =?? and ?? ?? =?? ?? (?? =0,1,…,?? ) 
(7.3) ?? (?? )+?? (?? ):?? ??? (?? 0
+?? 0
) +(?? 1
 
'
+?? 1
)?? +?+(?? ?? +?? ?? )?? ?? +… 
i.e., coefficient of ?? ?? in ?? (?? )+?? (?? ) = coefficient of ?? ?? in ?? (?? )+ coefficient of in 
?? (?? )(?? =0,1,2,… 
 
Example: ?? (?? )=?? 3
-?? ,?? (?? )=1+2?? -?? 3
+?? 5
 
?? (?? )+?? (?? )=?? 5
+?? +1 
Convention: The leading coefficient, by requirement of the very definition, is always ?
0. 
If coefficient of ?? ?? is = 0 in ?? (?? ) , we do not display the ?? ?? -term. Thus e.g. 
?? (?? )=?? 3
-?? =?? 3
+0.?? 2
-1.?? +0 
We define the product of two polynomials ?? (?? ) and ?? (?? ) by, (7.3) 
?? (?? )·?? (?? )=?? 0
+?? 1
?? +?+?? ?? +?? ?? ?? +?? 
wherein ?? ?? =?? 0
?? ?? +?? 1
?? ?? +?? ?? +?? ?? ?? 0
 
we put deg ?? (?? )=0, if ?? (?? )
(?? +1
?? 0
(?? 0
??? ) 
I.e., there is no ?? ?? term for ?? =1,2,.. Thus, fields coefficients are treated as polynomials 
of degree 0. 
We note: (7;4)deg ?? (?? ):?? (?? )=deg ?? (?? )+deg 8(?? ) 
for two polynomials ?? ,?? which are ?0. 
 
The zero-polynomial is the one in which coefficient of ?? ?? =0 for every ?? : 
0=0(?? )=0+0.?? +0.?? 2
…+0.?? ?? =0 
put 1=1(?? )=1+0·?? +?? ·?? 2
+?+0·?? ?? (for any ?? =1) 
Then ?? (?? ).1=?? (?? ) for any polynomial ?? (?? ) with sum and product thus defined the 
polynomials form a ring.  
(7.5) we denote it by ?? [?? ] 
Convention: The indeterminate ?? is shown in the notation. We call ?? [?? ] the polynomial 
ring  in one variable x over the field F. 
Owing to the familiar division process. (“Euclidean Algorithm") we have 
[7.2] Given two polynomials ?? (?? ) and ?? (?? ) ;  
?? ?0,? ?????????????????????? ?? (?? ) ?????? ?? (?? ) ?????? h ?? h???? ?? (?? )=?? (?? )?? (?? )+?? (?? ). 
 
(7.6) where ?? (?? )=0 or deg {?? (?? )]<deg [?? (?? )] is called the quotient and r is called the 
remainder. 
Proof: if deg ?? (?? )<eg ?? (?? ) , we can take ?? (?? )=0, and ?? (?? )=???? ) There is nothing to 
prove. Assume, then  
?????? ?? (?? )>deg ?? (?? ) 
?? (?? )=?? 0
+?? 1
?? +2+?? 1
?? 
?? (?? )=?? 0
+?? 1
?? +?? +?? 1
?? ??  
 
put ?? 1
(?? )=?? (?? )-
?? ?? ?? ?? ;?? -?? ?? ?? (?? ) (we have ?? ?? ?0); 
one gets rid of the ?? term, and deg ?? 1
(?? )=?? -1 
we can prove by induction on ?? ; thus ?? ?? may assume that if ?? 1
(?? )=0, the process 
terminates here, and. 
?? (?? )=
?? ?? 1
?? ?? ?? ?? -?? ·?? (?? )=?? (?? )?? (?? )+?? (?? )
 ?? (?? )-
?? ?? ?? ?? ?? ?? -?? , and ?? (?? )=0
 
otherwise, suppose the theorem hold for polynomials of degree < n, Thus,  ? 
polynomials ?? 1
(?? ),?? 1
(?? ) such that ?? 1
(?? )=???
1
(?? )?? (x)+?? 1
(?? ) 
wherein either ?? 1
(?? )=0 or deg ?? 1
(?? )<deg ?? (?? ) 
Hence ?? (?? )=?? 1
(?? )+
?? ?? ?? 1
·?? ?? -?? ·?? (?? ) 
=[?? 1
(?? )?? (?? )+?? 1
·?? )]+
?? ?? )
?? ?? ·?? (?? -1)
·?? (?? ) 
=?? (?? )·?? (?? )+?? (:) , 
9:! 
where  ?? (?? )=?? 1
(?? )+
?? ????
?? ?? ·6
?? -?? and ?? (?? )=?? 1
(?? )  
This proves the proposition. 
[7:5] The polynomial ring ?? [?? ], over a field ?? , is a Principal Ideal Domain, 
This follows from [7.2] since ?? [?? ] is a Euclidean ring, and hence in a Principal Ideal 
Domain. 
Consequently, the results regarding prime elements unique factorization  that were 
established for ?? rincipal ideal domains are available for ?? [?? ] in particular: 
[7.4] Given two ?? olynomials ?? (?? ) and ?? (?? ) in ?? [?? ], they have a t.c.d ?? (?? ) : 
?? (?? )=(?? (?? ),?? (?? )) 
which can be expressed in the form  ?? (?? )=?? (?? )·?? (?? )+?? (?? )·?? (?? ) 
In particular, if,?? (?? ) and ?? (?? ) are relatively prime, there exist ?? (?? ),?? (?? ) in ?? [?? ] such that 
a( ?? )?? (?? )+?? (?? )?? (?? )=1 
 
[ 7:5] The prime elements of the principal is ideal domain ?? { x} are those polynomials 
?? (?? ) . ceg ?? (?? )=1 that are divisible only by elements of ?? (i.e. Units of the ring ?? [?? ]) 
and by polynomials (c.p.x), ?? ?0 in ?? . 
 (i.e., associated to ?? (?? ) ). They are characterized 
by the property that they are not divisible by any non-constant ( ??.?? .  deg =1 ) 
polynomial of lower degree. 
(7.7) The prime elements of ?? [?? ] are called 'prime' polynomials" or "irreducible 
polynomials". 
?? every cass if associated prime polynomials, we choose tho one, whose leading 
coefficient is =1; te. Ls call them "normalized". Thus 
Every ?? (?? )??? [?? ] can be uniquely factorised into a product of normalized prime (i.e., 
irreducible) polyn mials. it is not an easy matter to decide whe her a given polynomial is 
irreducible or not. Some sufficient conditions are known these are called "Irreducibility 
Criteria". One of these, easy to prove, and useful in applications, is the following: 
 
[7:6] EISENSTEIN'S IRREDUCIBILITY CRITERION 
We consider polynomials with coefficients in Z, the ring of rational integers. 
Given: a polynomial,  ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? 
Suppose ? a prime number ?? such that 
1
0
 ?? ?? ?0(mod?? )
2
°
 ?? 1
=0(mod?? ),  for ?? =0,1,…,?? -1
3
0
 ?? 0
?0(mod?? 2
)
 
Claim: ?? (?? ) is irreducible over ?? (the field of rational numbers) (see [7:8]) 
Proof: Suppose, on the contrary, that ?? (?? ) is reducible: 
? polynomials A(x).B(x) with coefficients in Z, such that: 
?? (?? )=?? (?? )?? (?? )·{
?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? (?? =1,?? =1),(?? 1
=0,?? 5
?0)
 
then comparing coefficients. 
(7.8) 
?? 0
=?? 0
?? 0
,?? 1
=?? 0
?? 1
+?? ?? ?? 0
….
?? ?? =?? 0
?? ?? =?? 1
?? ?? -1
+?+?? ?? ?? 0
.
(?? 0
,…,?? ?? and ?? 0
,…,??ˆ
5
 in ?? )
 
Since 
?? 0
=?? 0
?? 0
,?? 0
=0(mod?? ),?? 0
=0(mod?? 2
) 
?? must divice either ?? 0
 or ?? 0
 but not both of them, it is a matter of notation to assume. 
(7.9)?? 0
=0(mod?? ),?? 0
?0(mod?? ) 
 
Now ?? 0
,…..,?? , 
 cannot all be divisible by ?? . since that would imply, (see [7.8]), that 
?? ?? (?? =0,1,…,?? ) would be divisible by ?? ; but, by hypothesis 1
°
,?? ?? ?0(mod,?? ?? . 
Hence ? , among ?? 0
,…,?? ?? some NOT divisible by p : 
Choose the very first one in the sequence ?? 0
,?? 1
,…,?? ?? which is ?0 (mod.p); let it be ?? ?? , 
thus: 
(7.10)?? =0(mod,?? ),…,?? ?? -1
=0(mod?? ) but ?? ?? =0(mod?? ) 
since ?? =?? +?? ,?? =1,?? =1, we have ?? =?? =?? -?? =?? -1,?? =?? 
(7.11) Now we have, ?? ?? =(?? 0
?? ?? +?+?? ?? -1
?? 1
)
1
+?? ?? ?? 0
 
By (7.10) ,   ???? (?? 0
?? ?? +?+?? ?? -1
?? 1
)+?? |?? ?? .  
By (7.9) ,   p |?? 0
. so ?? |?? ?? ?? 0
 
Whence follows that  p |?? ?? 
Page 5


Edurev123 
SECTION - III 
FIELDS 
 
7. FIELD EXTENSIONS  
Before we begin our study of fields, we shall present some facts on Polynomial rings. 
The set of polynomials in one Indeterminate with coefficients, in a field ?? ?? forms a ring: 
an integral domain which turns out to be a Principal ideal domain. 
[7:1] DEFINITION  
Let ?? be a field, and ?? a variable (also called indeterminate). We call an expression 
(7.1) ?? (?? )=?? 0
+?? 1
?? +?? +?? ?? ?? ?? ,?? ?? +0 ?? ?? ??? (?? =0,1,…,?? ) a polynomial in ?? with 
coefficients ?? ?? . We call ?? ?? the coefficient to ?? ?? and ?? ?? as the highest coefficient or the 
"leading coefficient"; n(?? ?? ?0) ," is called the "degree" of the polynomial 
 ?? (?? ):?? =?? eg ?? (?? ) . 
(7.2) If ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? 
we define the sum of two polynomials: ?? (?? )=?? (?? )?
 def 
?? =?? and ?? ?? =?? ?? (?? =0,1,…,?? ) 
(7.3) ?? (?? )+?? (?? ):?? ??? (?? 0
+?? 0
) +(?? 1
 
'
+?? 1
)?? +?+(?? ?? +?? ?? )?? ?? +… 
i.e., coefficient of ?? ?? in ?? (?? )+?? (?? ) = coefficient of ?? ?? in ?? (?? )+ coefficient of in 
?? (?? )(?? =0,1,2,… 
 
Example: ?? (?? )=?? 3
-?? ,?? (?? )=1+2?? -?? 3
+?? 5
 
?? (?? )+?? (?? )=?? 5
+?? +1 
Convention: The leading coefficient, by requirement of the very definition, is always ?
0. 
If coefficient of ?? ?? is = 0 in ?? (?? ) , we do not display the ?? ?? -term. Thus e.g. 
?? (?? )=?? 3
-?? =?? 3
+0.?? 2
-1.?? +0 
We define the product of two polynomials ?? (?? ) and ?? (?? ) by, (7.3) 
?? (?? )·?? (?? )=?? 0
+?? 1
?? +?+?? ?? +?? ?? ?? +?? 
wherein ?? ?? =?? 0
?? ?? +?? 1
?? ?? +?? ?? +?? ?? ?? 0
 
we put deg ?? (?? )=0, if ?? (?? )
(?? +1
?? 0
(?? 0
??? ) 
I.e., there is no ?? ?? term for ?? =1,2,.. Thus, fields coefficients are treated as polynomials 
of degree 0. 
We note: (7;4)deg ?? (?? ):?? (?? )=deg ?? (?? )+deg 8(?? ) 
for two polynomials ?? ,?? which are ?0. 
 
The zero-polynomial is the one in which coefficient of ?? ?? =0 for every ?? : 
0=0(?? )=0+0.?? +0.?? 2
…+0.?? ?? =0 
put 1=1(?? )=1+0·?? +?? ·?? 2
+?+0·?? ?? (for any ?? =1) 
Then ?? (?? ).1=?? (?? ) for any polynomial ?? (?? ) with sum and product thus defined the 
polynomials form a ring.  
(7.5) we denote it by ?? [?? ] 
Convention: The indeterminate ?? is shown in the notation. We call ?? [?? ] the polynomial 
ring  in one variable x over the field F. 
Owing to the familiar division process. (“Euclidean Algorithm") we have 
[7.2] Given two polynomials ?? (?? ) and ?? (?? ) ;  
?? ?0,? ?????????????????????? ?? (?? ) ?????? ?? (?? ) ?????? h ?? h???? ?? (?? )=?? (?? )?? (?? )+?? (?? ). 
 
(7.6) where ?? (?? )=0 or deg {?? (?? )]<deg [?? (?? )] is called the quotient and r is called the 
remainder. 
Proof: if deg ?? (?? )<eg ?? (?? ) , we can take ?? (?? )=0, and ?? (?? )=???? ) There is nothing to 
prove. Assume, then  
?????? ?? (?? )>deg ?? (?? ) 
?? (?? )=?? 0
+?? 1
?? +2+?? 1
?? 
?? (?? )=?? 0
+?? 1
?? +?? +?? 1
?? ??  
 
put ?? 1
(?? )=?? (?? )-
?? ?? ?? ?? ;?? -?? ?? ?? (?? ) (we have ?? ?? ?0); 
one gets rid of the ?? term, and deg ?? 1
(?? )=?? -1 
we can prove by induction on ?? ; thus ?? ?? may assume that if ?? 1
(?? )=0, the process 
terminates here, and. 
?? (?? )=
?? ?? 1
?? ?? ?? ?? -?? ·?? (?? )=?? (?? )?? (?? )+?? (?? )
 ?? (?? )-
?? ?? ?? ?? ?? ?? -?? , and ?? (?? )=0
 
otherwise, suppose the theorem hold for polynomials of degree < n, Thus,  ? 
polynomials ?? 1
(?? ),?? 1
(?? ) such that ?? 1
(?? )=???
1
(?? )?? (x)+?? 1
(?? ) 
wherein either ?? 1
(?? )=0 or deg ?? 1
(?? )<deg ?? (?? ) 
Hence ?? (?? )=?? 1
(?? )+
?? ?? ?? 1
·?? ?? -?? ·?? (?? ) 
=[?? 1
(?? )?? (?? )+?? 1
·?? )]+
?? ?? )
?? ?? ·?? (?? -1)
·?? (?? ) 
=?? (?? )·?? (?? )+?? (:) , 
9:! 
where  ?? (?? )=?? 1
(?? )+
?? ????
?? ?? ·6
?? -?? and ?? (?? )=?? 1
(?? )  
This proves the proposition. 
[7:5] The polynomial ring ?? [?? ], over a field ?? , is a Principal Ideal Domain, 
This follows from [7.2] since ?? [?? ] is a Euclidean ring, and hence in a Principal Ideal 
Domain. 
Consequently, the results regarding prime elements unique factorization  that were 
established for ?? rincipal ideal domains are available for ?? [?? ] in particular: 
[7.4] Given two ?? olynomials ?? (?? ) and ?? (?? ) in ?? [?? ], they have a t.c.d ?? (?? ) : 
?? (?? )=(?? (?? ),?? (?? )) 
which can be expressed in the form  ?? (?? )=?? (?? )·?? (?? )+?? (?? )·?? (?? ) 
In particular, if,?? (?? ) and ?? (?? ) are relatively prime, there exist ?? (?? ),?? (?? ) in ?? [?? ] such that 
a( ?? )?? (?? )+?? (?? )?? (?? )=1 
 
[ 7:5] The prime elements of the principal is ideal domain ?? { x} are those polynomials 
?? (?? ) . ceg ?? (?? )=1 that are divisible only by elements of ?? (i.e. Units of the ring ?? [?? ]) 
and by polynomials (c.p.x), ?? ?0 in ?? . 
 (i.e., associated to ?? (?? ) ). They are characterized 
by the property that they are not divisible by any non-constant ( ??.?? .  deg =1 ) 
polynomial of lower degree. 
(7.7) The prime elements of ?? [?? ] are called 'prime' polynomials" or "irreducible 
polynomials". 
?? every cass if associated prime polynomials, we choose tho one, whose leading 
coefficient is =1; te. Ls call them "normalized". Thus 
Every ?? (?? )??? [?? ] can be uniquely factorised into a product of normalized prime (i.e., 
irreducible) polyn mials. it is not an easy matter to decide whe her a given polynomial is 
irreducible or not. Some sufficient conditions are known these are called "Irreducibility 
Criteria". One of these, easy to prove, and useful in applications, is the following: 
 
[7:6] EISENSTEIN'S IRREDUCIBILITY CRITERION 
We consider polynomials with coefficients in Z, the ring of rational integers. 
Given: a polynomial,  ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? 
Suppose ? a prime number ?? such that 
1
0
 ?? ?? ?0(mod?? )
2
°
 ?? 1
=0(mod?? ),  for ?? =0,1,…,?? -1
3
0
 ?? 0
?0(mod?? 2
)
 
Claim: ?? (?? ) is irreducible over ?? (the field of rational numbers) (see [7:8]) 
Proof: Suppose, on the contrary, that ?? (?? ) is reducible: 
? polynomials A(x).B(x) with coefficients in Z, such that: 
?? (?? )=?? (?? )?? (?? )·{
?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? (?? =1,?? =1),(?? 1
=0,?? 5
?0)
 
then comparing coefficients. 
(7.8) 
?? 0
=?? 0
?? 0
,?? 1
=?? 0
?? 1
+?? ?? ?? 0
….
?? ?? =?? 0
?? ?? =?? 1
?? ?? -1
+?+?? ?? ?? 0
.
(?? 0
,…,?? ?? and ?? 0
,…,??ˆ
5
 in ?? )
 
Since 
?? 0
=?? 0
?? 0
,?? 0
=0(mod?? ),?? 0
=0(mod?? 2
) 
?? must divice either ?? 0
 or ?? 0
 but not both of them, it is a matter of notation to assume. 
(7.9)?? 0
=0(mod?? ),?? 0
?0(mod?? ) 
 
Now ?? 0
,…..,?? , 
 cannot all be divisible by ?? . since that would imply, (see [7.8]), that 
?? ?? (?? =0,1,…,?? ) would be divisible by ?? ; but, by hypothesis 1
°
,?? ?? ?0(mod,?? ?? . 
Hence ? , among ?? 0
,…,?? ?? some NOT divisible by p : 
Choose the very first one in the sequence ?? 0
,?? 1
,…,?? ?? which is ?0 (mod.p); let it be ?? ?? , 
thus: 
(7.10)?? =0(mod,?? ),…,?? ?? -1
=0(mod?? ) but ?? ?? =0(mod?? ) 
since ?? =?? +?? ,?? =1,?? =1, we have ?? =?? =?? -?? =?? -1,?? =?? 
(7.11) Now we have, ?? ?? =(?? 0
?? ?? +?+?? ?? -1
?? 1
)
1
+?? ?? ?? 0
 
By (7.10) ,   ???? (?? 0
?? ?? +?+?? ?? -1
?? 1
)+?? |?? ?? .  
By (7.9) ,   p |?? 0
. so ?? |?? ?? ?? 0
 
Whence follows that  p |?? ?? 
for a value of ?? <?? (See 7.11)). But this contradicts the condition 2
°
 of the hypothesis. 
The proof is complete. 
 
[7:7] GAUSS'S LEMMA 
Given: Two polynomials 
?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? ?? (?? )=?? 0
+?? 1
?? +?+?? ?? ?? ?? 
such that g.c.d. (?? 0
,?? 1
,…,?? ?? )=1 
gcc ?? (?? 0
,?? 1
,…,?? ?? )=1 
Claim: The g.c.d. of the coefficients in the product,  ?? (?? )?? (?? ) is also =1 
 
DEFINITION: A polynomial with, INTEGER coefficients is said to be "primitive" if the 
g.c.d, of its coefficients is =1. 
Proof of GAUSS'S LEMMA: Suppose, on the contrary a prime number ?? that divides all 
the coefficient of ?? (?? ) , ?? (?? ) since ?? (?? ) is primitive, ?? cannot divide all the ?? 0
,?? 1
,…,?? ?? . 
Let ?? ?? be the first coefficlent in the sequence ?? 0
,?? 1
,…,?? ?? that ?? does not divide, thus, 
(7,12) p|?? 0
,…,,?? |?? ?? but p|?? ?? . 
similarly, in the case of the primitive polynomial ?? (?? ). Let ?? be the first coefficient in the 
sequence ?? 0
,?? 1
,…,?? ?? that ?? does not divide, thus:  ?? |?? 0
,…,?? |?? ?? -1
 but ???? ?? ?? 
put ?? (?? )·?? (?? )=?? 0
+?? 1
?? +?? 2
?? 2
+??+?? ?? +?? ?? 1+?? +? 
Then: The coefficient of ?? ?? +?? in ?? (?? )·?? (?? ) is 
=?? ?? +?? =?? ?? ?? ?? +(?? ?? +1
?? ?? -1
+?? ?? +2?? ?? -2
+?+?? ?? +?? 0
)+(?? ?? -1
?? ?? +1
+?? ?? -2
?? ?? +2
+?+?? 0
?? ?? +1
) 
Or   (7,14)?? ?? +?? =?? ?? ?? ?? +?? +?? 1
 
where
?? =(?? ?? +1
?? ?? -1
+?? ?? +2
?? ?? -2
+?+?? ?? +?? ?? 0
)
?? =(?? ?? -1
?? ?? +1
+?? ?? -2
?? ?? +2
+?+?? 0
?? ?? +1
)
 
By (7.13)?? |?? 0
,…?? |?? ?? -1
, hence ?? |?? . By (7.12),  ?? |?? 0
,?? 1
,…,?? ?? =1, hence ?? /?? .  
Also, according to our supposition;   ?? |?? ?? +?? From (7.14),  it now follows that ?? |?? ?? ?? ?? 
 
From(7.12)and (7.13), ?? |?? ?? , ?? |?? . Since ?? is a prime this impossible. Our supposition is 
wrong. C(x) must be primitive.  
[7:8] If the primitive polynomial ?? (?? ) can be expressed as a product of two polynomials 
over ?? (the field of rational numbers), then it can be expressed as a product of two