The ring is called the ring of polynomials over .
If has an identity, and we write polynomials as . If is a ring without identity, embed to a ring with identity. Identify with its image under the embedding map so that is a subring of . Then is a subring of . Thus every polynomial can be written uniquely as where , . If , the elements are called the coefficients of . The element is called the constant term. Elements of which have the form are called constant polynomials. If and , then is called the leading coefficient of . If has an identity and has leading coefficient , then is said to be a monic polynomial.
Let be a ring with identity. The element of is called an indeterminate. If is another ring with identity, the indeterminate is not the same element as . We can also define polynomials in more than one indeterminate.
The ring is called the ring of polynomials in determinates over . is considered a subring of . Let be a positive integer and for , let where is the th coordinate of . Every element of may be written as .
If is a ring with identity, the elements are called indeterminates. The elements are called coefficients of the polynomial . A polynomial of the form is called a monomial. The notation and terminology is extended to polynomial rings where has no identity. Embed the ring to a ring with identity and consider as a subring of . If is any ring, for any subset , the monomorphism exists.
Let be a homomorphism of rings, , . Let . Let
Proof. If , then . The map given by is a well-defined map such that and . It is easy to verify that is a homomorphism. Suppose that is a homomorphism with , . Then by direct computation. Define a category whose objects are tuples where is a commutative ring with identity, , a homomorphism with . A morphism in from to is a homomorphism such that , and . Verify that is an equivalence in iff is an equivalence of rings. If is the inclusion map, is a universal object in . Thus is completely determined up to isomorphism. ◻
Proof. The proof that is a homomorphism does not rely on containing an identity. ◻
The map is called the evaluation homomorphism. The corollary may be false when and are not commutative.
Proof. Given a homomorphism of commutative rings with identity, elements , there exists a homomorphism such that . Applying the theorem with in place of gives such that and for all . Due to the uniqueness up to isomorphism, . The other isomorphism follows similarly. ◻
The ring is called the ring of formal power series over the ring . Its elements are called power series.
Proof. (i). If such that , it follows immediately that . Whence is a unit. Suppose is a unit in . If such that , then Conversely, if a solution exists for this system of equations in , then satisfies . Take . Similarly, . Thus the system of equations is solvable. A similar argument shows the existence of a left inverse for in . (ii) is an immediate consequence of (i). ◻
The degree of a nonzero monomial is the nonnegative integer . If is a nonzero polynomial, the degree of is the maximum of the degrees of the monomials making up . The degree of is denoted . A polynomial which is a sum of monomials, each with the same degree , is said to be homogeneous of degree . The degree of in is the degree of considered as a polynomial in one indeterminate over the ring . We define the degree of the zero polynomial to be .
Proof. If , let and . If , with , a unit in . Proceed by induction on . If and . Assume the existence is true for polynomials with degree less than . The polynomial has degree and leading coefficient . Hence is a polynomial of degree less than . There exist polynomials such that and . Thus if . For uniqueness, suppose , and . implies . Since the leading coefficient of is a unit, Since , the above inequality is true only if and . ◻
Proof. Since is an integral domain, is an integral domain. Define by . By the division algorithm, is a Euclidean domain. Since each unit in must have degree , the units of are precisely the nonzero constant polynomials. ◻
Proof. . If , then for . Whence . Thus substituting gives . If , . ◻
Proof. Let be the distinct roots of in . whence . Since and is an integral domain, . Thus . An inductive argument shows for distinct roots , divides . But . Thus . ◻
Proof. implies that , . If , then and . ◻
Let be an integral domain and . If and is a root of , then there is a greatest integer such that where and . The integer is called the multiplicity of the root of . If has multiplicity , is said to be a simple root. If has multiplicity greater than , is called a multiple root.
The polynomial is called the formal derivative of .
Proof. (i).
where
.
If
, then
. Conversely, if
, then
.If
, then
. Thus
means
which is a contradiction. Therefore
.
(ii). Since
is a Euclidean domain,
such that
. If
is a multiple root of
, then
, a contradiction. Thus
is a simple root.
(iii). If
is irreducible and
, then
and
are relatively prime since
. Therefore,
has no multiple roots in
. Conversely, suppose
has no multiple roots in
and
is a root of
in
. If
, then
is a multiple root of
, a contradiction. Hence
. ◻
Let be an integral domain, the following facts hold:
The units of are precisely the constant polynomials that are units in .
If and is irreducible in , is irreducible in .
Every first degree polynomial whose leading coefficient is a unit in is irreducible in . In particular, every first degree polynomial over a field is irreducible.
Suppose is a subring of integral domain and . Then may be irreducible in but not in and vice versa.
For the last point, note that is irreducible in but not in . is irreducible in but not in .
Let be a unique factorization domain and a nonzero polynomial in . A greatest common divisor of is called a content of and is denoted . Write whenever and are associates in . Then is an equivalence relation on . Since is an integral domain, and is a unit. If and then . If and is a unit in , then is said to be primitive. For any polynomial , with primitive.
Proof. Let , primitive. It suffices to prove that is primitive. Let and . with . If is not primitive, there exists an irreducible element such that . Since is a unit, whence there is a least integer such that for and . Similarly, there is a least integer such that for and . Since divides and implying or , a contradiction. ◻
Proof. If and are associates in , for some a unit. so where . Thus . Since are units in , . Thus a unit such that . . Thus whence and are associates in . The converse is obvious. ◻
Proof. Suppose is irreducible in and with , . and with and . Let and for each let . If , then with and primitive. and . Similarly, with , primitive and . thus implying . a unit such that . so . Then is reducible in , a contradiction. Thus is irreducible in . Conversely, if is irreducible in , with , then one of them, say is a constant. Thus . Since is primitive, is a unit in and hence in . Thus is irreducible in . ◻
Proof. We only need to prove is UFD since . If has positive degree, with primitive and positive degree. Since is a UFD, is a unit or with each irreducible in and hence in . Let be the quotient field of . Since is a UFD containing , with each an irreducible polynomial in . For each , with , , , primitive. Each is irreducible in whence each is irreducible in . If , , . Thus . Since are primitive, and are associates in . Thus , with a unit in . Thus if is a nonunit, . Remove if is a unit.
For uniqueness, suppose is a nonprimitive polynomial in of positive degree. Any factorization of as a product of irreducible elements may be written as with each irreducible in , each irreducible and hence primitive in of positive degree. Suppose where irreducible in , irreducible in of positive degree. Then and are associates in . Unique factorization in implies and after reindexing, is an associate of . is associate to in . Since is a UFD, and each is associate of in after reindexing. They are associates in by the lemma. ◻
Proof. , primitive in , . Since is a unit in , it suffices to show is irreducible in . We only need to prove is irreducible in . Suppose with Since , has the same divisibility conditions to , the coefficients of , as it does to . Since , either or , say . Since , . Some coefficient of is not divisible by otherwise would not be primitive. Let be the least integer such that for and . Then . Since and , whence or , a contradiction. Thus must be irreducible in . ◻