Polynomial Rings and Irreducibility Criteria - Ring Theory, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Polynomial Rings
If R is a ring, then  , the ring of polynomials in x with coefficients in R, consists of all formal sums 
 , where  for all but finitely many values of i.
If  is a nonzero polynomial, the degree is the largest  such that  . The zero
polynomial has degree  .
 becomes a ring with the usual operations of polynomial addition and multiplication.
If F is a field, the units in  are exactly the nonzero elements of F.
If  is a field,  , and  , there are unique polynomials  such
that
Let R be an integral domain. An element  is irreducible if  , x is not a unit, and if 
implies either y is a unit or z is a unit.
Let R be an integral domain. An element  is prime if  , x is not a unit, and  implies 
 or  .
In an integral domain, primes are irreducible.
Let F be a field. If  are not both zero, then  and  have a greatest common
divisor which is unique up to multiplication by units (elements of F).
Let F be a field, let  , and let  be a greatest common divisor of  and  .
There are polynomials  such that
If R is a ring, the ring of polynomials in x with coefficients in R is denoted  . It consists of all formal
sums
Here  for all but finitely many values of i.
If the idea of "formal sums" worries you, replace a formal sum with the infinite vector whose components are
the coefficients of the sum:
Page 2


Polynomial Rings
If R is a ring, then  , the ring of polynomials in x with coefficients in R, consists of all formal sums 
 , where  for all but finitely many values of i.
If  is a nonzero polynomial, the degree is the largest  such that  . The zero
polynomial has degree  .
 becomes a ring with the usual operations of polynomial addition and multiplication.
If F is a field, the units in  are exactly the nonzero elements of F.
If  is a field,  , and  , there are unique polynomials  such
that
Let R be an integral domain. An element  is irreducible if  , x is not a unit, and if 
implies either y is a unit or z is a unit.
Let R be an integral domain. An element  is prime if  , x is not a unit, and  implies 
 or  .
In an integral domain, primes are irreducible.
Let F be a field. If  are not both zero, then  and  have a greatest common
divisor which is unique up to multiplication by units (elements of F).
Let F be a field, let  , and let  be a greatest common divisor of  and  .
There are polynomials  such that
If R is a ring, the ring of polynomials in x with coefficients in R is denoted  . It consists of all formal
sums
Here  for all but finitely many values of i.
If the idea of "formal sums" worries you, replace a formal sum with the infinite vector whose components are
the coefficients of the sum:
All of the operations which I'll define using formal sums can be defined using vectors. But it's traditional to
represent polynomials as formal sums, so this is what I'll do.
A nonzero polynomial  has degree n if  and  , and n is the largest integer with this
property. The zero polynomial is defined by convention to have degree  . (This is necessary in order to
make the dgree formulas work out.) Alternatively, you can say that the degree of the zero polynomial is
undefined; in that case, you will need to make minor changes to some of the results below.
Polynomials are added componentwise, and multiplied using the "convolution" formula:
These formulas say that you compute sums and products as usual.
Example. ( Polynomial arithmetic) In  ,
Let R be an integral domain. Then If  , write  to denote the degree of f. It's easy to show that the
degree function satisfies the following properties:
The verifications amount to writing out the formal sums, with a little attention paid to the case of the zero
polynomial. These formulas {\it do} work if either f or g is equal to the zero polynomial, provided that  is
understood to behave in the obvious ways (e.g.  for any  ).
Example. ( Degrees of polynomials) Note that in  ,
This shows that equality might not hold in  .
The equality  might not hold if R is not an integral domain. For example, take 
 . Then (since  ),
Page 3


Polynomial Rings
If R is a ring, then  , the ring of polynomials in x with coefficients in R, consists of all formal sums 
 , where  for all but finitely many values of i.
If  is a nonzero polynomial, the degree is the largest  such that  . The zero
polynomial has degree  .
 becomes a ring with the usual operations of polynomial addition and multiplication.
If F is a field, the units in  are exactly the nonzero elements of F.
If  is a field,  , and  , there are unique polynomials  such
that
Let R be an integral domain. An element  is irreducible if  , x is not a unit, and if 
implies either y is a unit or z is a unit.
Let R be an integral domain. An element  is prime if  , x is not a unit, and  implies 
 or  .
In an integral domain, primes are irreducible.
Let F be a field. If  are not both zero, then  and  have a greatest common
divisor which is unique up to multiplication by units (elements of F).
Let F be a field, let  , and let  be a greatest common divisor of  and  .
There are polynomials  such that
If R is a ring, the ring of polynomials in x with coefficients in R is denoted  . It consists of all formal
sums
Here  for all but finitely many values of i.
If the idea of "formal sums" worries you, replace a formal sum with the infinite vector whose components are
the coefficients of the sum:
All of the operations which I'll define using formal sums can be defined using vectors. But it's traditional to
represent polynomials as formal sums, so this is what I'll do.
A nonzero polynomial  has degree n if  and  , and n is the largest integer with this
property. The zero polynomial is defined by convention to have degree  . (This is necessary in order to
make the dgree formulas work out.) Alternatively, you can say that the degree of the zero polynomial is
undefined; in that case, you will need to make minor changes to some of the results below.
Polynomials are added componentwise, and multiplied using the "convolution" formula:
These formulas say that you compute sums and products as usual.
Example. ( Polynomial arithmetic) In  ,
Let R be an integral domain. Then If  , write  to denote the degree of f. It's easy to show that the
degree function satisfies the following properties:
The verifications amount to writing out the formal sums, with a little attention paid to the case of the zero
polynomial. These formulas {\it do} work if either f or g is equal to the zero polynomial, provided that  is
understood to behave in the obvious ways (e.g.  for any  ).
Example. ( Degrees of polynomials) Note that in  ,
This shows that equality might not hold in  .
The equality  might not hold if R is not an integral domain. For example, take 
 . Then (since  ),
Lemma. Let F be a field, and let  be the polynomial ring in one variable over F. The units in  are
exactly the nonzero elements of F.
Proof. It's clear that the nonzero elements of F are invertible in  , since they're already invertible in F.
Conversely, suppose that  is invertible, so  for some  . Then 
 , which is impossible unless f and g both have degree 0. In particular, f is a nonzero
constant, i.e. an element of F.
Theorem. ( Division Algorithm) Let F be a field, and let  . Suppose that  . There exist 
 such that
Proof. The idea is to imitate the proof of the Division Algorithm for  .
Let
The set  is a subset of the nonnegative integers, and therefore must contain a smallest
element by well-ordering. Let  be an element in S of smallest degree, and write
I need to show that  .
If  , then since  ,  .
Suppose then that  . Assume toward a contradiction that  . Write
Assume  , and  .
Consider the polynomial
Its degree is less than n, since the n-th degree terms cancel out.
However,
The latter is an element of S.
I've found an element of S of smaller degree than  , which is a contradiction. It follows that 
 , and this completes the proof.
Page 4


Polynomial Rings
If R is a ring, then  , the ring of polynomials in x with coefficients in R, consists of all formal sums 
 , where  for all but finitely many values of i.
If  is a nonzero polynomial, the degree is the largest  such that  . The zero
polynomial has degree  .
 becomes a ring with the usual operations of polynomial addition and multiplication.
If F is a field, the units in  are exactly the nonzero elements of F.
If  is a field,  , and  , there are unique polynomials  such
that
Let R be an integral domain. An element  is irreducible if  , x is not a unit, and if 
implies either y is a unit or z is a unit.
Let R be an integral domain. An element  is prime if  , x is not a unit, and  implies 
 or  .
In an integral domain, primes are irreducible.
Let F be a field. If  are not both zero, then  and  have a greatest common
divisor which is unique up to multiplication by units (elements of F).
Let F be a field, let  , and let  be a greatest common divisor of  and  .
There are polynomials  such that
If R is a ring, the ring of polynomials in x with coefficients in R is denoted  . It consists of all formal
sums
Here  for all but finitely many values of i.
If the idea of "formal sums" worries you, replace a formal sum with the infinite vector whose components are
the coefficients of the sum:
All of the operations which I'll define using formal sums can be defined using vectors. But it's traditional to
represent polynomials as formal sums, so this is what I'll do.
A nonzero polynomial  has degree n if  and  , and n is the largest integer with this
property. The zero polynomial is defined by convention to have degree  . (This is necessary in order to
make the dgree formulas work out.) Alternatively, you can say that the degree of the zero polynomial is
undefined; in that case, you will need to make minor changes to some of the results below.
Polynomials are added componentwise, and multiplied using the "convolution" formula:
These formulas say that you compute sums and products as usual.
Example. ( Polynomial arithmetic) In  ,
Let R be an integral domain. Then If  , write  to denote the degree of f. It's easy to show that the
degree function satisfies the following properties:
The verifications amount to writing out the formal sums, with a little attention paid to the case of the zero
polynomial. These formulas {\it do} work if either f or g is equal to the zero polynomial, provided that  is
understood to behave in the obvious ways (e.g.  for any  ).
Example. ( Degrees of polynomials) Note that in  ,
This shows that equality might not hold in  .
The equality  might not hold if R is not an integral domain. For example, take 
 . Then (since  ),
Lemma. Let F be a field, and let  be the polynomial ring in one variable over F. The units in  are
exactly the nonzero elements of F.
Proof. It's clear that the nonzero elements of F are invertible in  , since they're already invertible in F.
Conversely, suppose that  is invertible, so  for some  . Then 
 , which is impossible unless f and g both have degree 0. In particular, f is a nonzero
constant, i.e. an element of F.
Theorem. ( Division Algorithm) Let F be a field, and let  . Suppose that  . There exist 
 such that
Proof. The idea is to imitate the proof of the Division Algorithm for  .
Let
The set  is a subset of the nonnegative integers, and therefore must contain a smallest
element by well-ordering. Let  be an element in S of smallest degree, and write
I need to show that  .
If  , then since  ,  .
Suppose then that  . Assume toward a contradiction that  . Write
Assume  , and  .
Consider the polynomial
Its degree is less than n, since the n-th degree terms cancel out.
However,
The latter is an element of S.
I've found an element of S of smaller degree than  , which is a contradiction. It follows that 
 , and this completes the proof.
Example. ( Polynomial division) Division of polynomials should be familiar to you --- at least over  and 
.
In this example, I'll divide  by  in  . Remember as you follow the division that 
 ,  , and  --- I'm doing arithmetic mod 5.
If you prefer, you can do long division without writing the powers of x --- i.e. just writing down the coefficients.
Here's how it looks:
Either way, the quotient is  and the remainder is  :
Definition. Let R be a commutative ring and let  . An element  is a root of  if 
.
Note that polynomials are actually formal sums, not functions. However, it is obvious how to plug a number into
a polynomial. Specifically, let
For  , define
Observe that a polynomial can be nonzero as a polynomial even if it equals 0 for every input! For example,
take  is a nonzero polynomial. However, plugging in the two elements of the coefficient
ring  gives
Page 5


Polynomial Rings
If R is a ring, then  , the ring of polynomials in x with coefficients in R, consists of all formal sums 
 , where  for all but finitely many values of i.
If  is a nonzero polynomial, the degree is the largest  such that  . The zero
polynomial has degree  .
 becomes a ring with the usual operations of polynomial addition and multiplication.
If F is a field, the units in  are exactly the nonzero elements of F.
If  is a field,  , and  , there are unique polynomials  such
that
Let R be an integral domain. An element  is irreducible if  , x is not a unit, and if 
implies either y is a unit or z is a unit.
Let R be an integral domain. An element  is prime if  , x is not a unit, and  implies 
 or  .
In an integral domain, primes are irreducible.
Let F be a field. If  are not both zero, then  and  have a greatest common
divisor which is unique up to multiplication by units (elements of F).
Let F be a field, let  , and let  be a greatest common divisor of  and  .
There are polynomials  such that
If R is a ring, the ring of polynomials in x with coefficients in R is denoted  . It consists of all formal
sums
Here  for all but finitely many values of i.
If the idea of "formal sums" worries you, replace a formal sum with the infinite vector whose components are
the coefficients of the sum:
All of the operations which I'll define using formal sums can be defined using vectors. But it's traditional to
represent polynomials as formal sums, so this is what I'll do.
A nonzero polynomial  has degree n if  and  , and n is the largest integer with this
property. The zero polynomial is defined by convention to have degree  . (This is necessary in order to
make the dgree formulas work out.) Alternatively, you can say that the degree of the zero polynomial is
undefined; in that case, you will need to make minor changes to some of the results below.
Polynomials are added componentwise, and multiplied using the "convolution" formula:
These formulas say that you compute sums and products as usual.
Example. ( Polynomial arithmetic) In  ,
Let R be an integral domain. Then If  , write  to denote the degree of f. It's easy to show that the
degree function satisfies the following properties:
The verifications amount to writing out the formal sums, with a little attention paid to the case of the zero
polynomial. These formulas {\it do} work if either f or g is equal to the zero polynomial, provided that  is
understood to behave in the obvious ways (e.g.  for any  ).
Example. ( Degrees of polynomials) Note that in  ,
This shows that equality might not hold in  .
The equality  might not hold if R is not an integral domain. For example, take 
 . Then (since  ),
Lemma. Let F be a field, and let  be the polynomial ring in one variable over F. The units in  are
exactly the nonzero elements of F.
Proof. It's clear that the nonzero elements of F are invertible in  , since they're already invertible in F.
Conversely, suppose that  is invertible, so  for some  . Then 
 , which is impossible unless f and g both have degree 0. In particular, f is a nonzero
constant, i.e. an element of F.
Theorem. ( Division Algorithm) Let F be a field, and let  . Suppose that  . There exist 
 such that
Proof. The idea is to imitate the proof of the Division Algorithm for  .
Let
The set  is a subset of the nonnegative integers, and therefore must contain a smallest
element by well-ordering. Let  be an element in S of smallest degree, and write
I need to show that  .
If  , then since  ,  .
Suppose then that  . Assume toward a contradiction that  . Write
Assume  , and  .
Consider the polynomial
Its degree is less than n, since the n-th degree terms cancel out.
However,
The latter is an element of S.
I've found an element of S of smaller degree than  , which is a contradiction. It follows that 
 , and this completes the proof.
Example. ( Polynomial division) Division of polynomials should be familiar to you --- at least over  and 
.
In this example, I'll divide  by  in  . Remember as you follow the division that 
 ,  , and  --- I'm doing arithmetic mod 5.
If you prefer, you can do long division without writing the powers of x --- i.e. just writing down the coefficients.
Here's how it looks:
Either way, the quotient is  and the remainder is  :
Definition. Let R be a commutative ring and let  . An element  is a root of  if 
.
Note that polynomials are actually formal sums, not functions. However, it is obvious how to plug a number into
a polynomial. Specifically, let
For  , define
Observe that a polynomial can be nonzero as a polynomial even if it equals 0 for every input! For example,
take  is a nonzero polynomial. However, plugging in the two elements of the coefficient
ring  gives
Corollary. Let F be a field, and let  , where  .
(a) ( The Root Theorem) c is a root of  in F if and only if  .
(b)  has at most n roots in F.
Proof. (a) Suppose  . Write
Then  or  .
In the first case, r is a nonzero constant. However, this implies that
This contradiction shows that  , and  .
Conversely, if  is a factor of  , then  for some  . Hence,
and c is a root of f.
(b) If  are the distinct roots of f in F, then
Taking degrees on both sides gives  .
Example. ( Applying the Root Theorem) Consider polynomials in  .
If  , it's obvious that  is a root. Therefore,  is a factor of  .
Likewise,  must be a factor of  for any  , since  is a root of  .
Example. ( Applying the Root Theorem) Prove that  is divisible by  in  .
Plugging in  gives
Since  is a root,  is a factor, by the Root Theorem.
Example. ( A polynomial with more roots than its degree) The quadratic polynomial 
 has roots  ,  ,  ,  . The previous result does not apply, because 
 is not a field.
On the other hand,  has at most 3 roots over  , since  is a field. (In fact, this polynomial has no
roots in  , as you can verify by plugging in 0, 1, 2, 3, and 4.