## A Book of Abstract Algebra, Second Edition (1982)

### Chapter 24. RINGS OF POLYNOMIALS

In elementary algebra an important role is played by polynomials in an unknown *x*. These are expressions such as

whose terms are grouped in powers of *x*. The exponents, of course, are positive integers and the coefficients are real or complex numbers.

Polynomials are involved in countless applications—applications of every kind and description. For example, polynomial functions are the easiest functions to compute, and therefore one commonly attempts to approximate arbitrary functions by polynomial functions. A great deal of effort has been expended by mathematicians to find ways of achieving this.

Aside from their uses in science and computation, polynomials come up very naturally in the general study of rings, as the following example will show:

Suppose we wish to enlarge the ring by adding to it the number *π*. It is easy to see that we will have to adjoin to other new numbers besides just *π*; for the enlarged ring (containing *π* as well as all the integers) will also contain such things as − *π*, *π* + 7, 6*π*^{2} − 11, and so on.

As a matter of fact, any ring which contains as a subring and which also contains the number *π* will have to contain every number of the form

*aπ ^{n}* +

*bπ*

^{n}^{”1}+ ⋯ +

*kπ*+

*l*

where *a*, *b*, …, *k*, *l* are integers. In other words, it will contain *all the polynomial expressions in π with integer coefficients*.

But the set of all the polynomial expressions in *π* with integer coefficients is a ring. (It is a subring of because it is obvious that the sum and product of any two polynomials in *π* is again a polynomial in *π*.) This ring contains because every integer *a* is a polynomial with a constant term only, and it also contains *π*.

Thus, if we wish to enlarge the ring by adjoining to it the new number *π*, it turns out that the “next largest” ring after which contains as a subring and includes π, is exactly the ring of all the polynomials in *π* with coefficients in .

As this example shows, aside from their practical applications, polynomials play an important role in the scheme of ring theory because they are precisely what we need when we wish to enlarge a ring by adding new elements to it.

In elementary algebra one considers polynomials whose coefficients are real numbers, or in some cases, complex numbers. As a matter of fact, the properties of polynomials are pretty much independent of the exact nature of their coefficients. All we need to know is that the coefficients are contained in some ring. For convenience, we will assume this ring is a commutative ring with unity.

Let *A* be a commutative ring with unity. Up to now we have used letters to denote elements or sets, but now we will use the letter *x* in a different way. In a polynomial expression such as *ax*^{2} + *bx* + *c*, where *a*, *b*, *c* ∈ *A*, we do not consider *x* to be an element of *A*, but rather *x* is a symbol which we use in an entirely formal way. Later we will allow the substitution of other things for *x*, but at present *x is simply a placeholder*.

Notationally, the terms of a polynomial may be listed in either ascending or descending order. For example, 4*x*^{3} – 3*x*^{2} + *x* + 1 and 1 + *x* – 3*x*^{2} + 4*x*^{3} denote the same polynomial. In elementary algebra descending order is preferred, but for our purposes ascending order is more convenient.

*Let A be a commutative ring with unity*, *and x an arbitrary symbol. Every expression of the form*

*a*_{0} + *a*_{1}*x* + *a*_{2}*x*^{2} + ⋯ + *a _{n}x^{n}*

*is called a polynomial in x with coefficients in A*,

*or more simply*,

*a*.

**polynomial in x over A***The expressions a*,

_{k}x^{k}*for k*∈ {1, …,

*n*},

*are called the*.

**terms**of the polynomialPolynomials in *x* are designated by symbols such as *a*(*x*), *b*(*x*), *q*(*x*), and so on. If *a*(*x*) = *a*_{0} + *a*_{1}*x* + ⋯ + *a _{n}x^{n}* is any polynomial and

*a*is any one of its terms,

_{k}x^{k}*a*is called the

_{k}*coefficient*of

*x*. By the

^{k}*degree*of a polynomial

*a*(

*x*) we mean the

*greatest n such that the coefficient of x*. In other words, if

^{n}is not zero*a*(

*x*) has degree

*n*, this means that

*a*≠ 0 but

_{n}*a*= 0 for every

_{m}*m*>

*n*. The degree of

*a*(

*x*) is symbolized by

deg *a*(*x*)

For example, 1 + 2*x* − 3*x*^{2} + *x*^{3} is a polynomial degree 3.

The polynomial 0 + 0*x* + 0*x*^{2} + ⋯ all of whose coefficients are equal to zero is called the *zero polynomial*, and is symbolized by 0. It is the only polynomial whose *degree is not defined* (because it has no nonzero coefficient).

If a nonzero polynomial *a*(*x*) = *a*_{0} + *a*_{1}*x* + ⋯ + *a _{n}x^{n}* has degree

*n*, then

*a*is called its

_{n}*leading coefficient*: it is the last nonzero coefficient of

*a*(

*x*). The term

*a*is then called its

_{n}x^{n}*leading term*, while

*a*

_{0}is called its

*constant term*.

If a polynomial *a*(*x*) has degree zero, this means that its constant term *a*_{0} is its only nonzero term: *a*(*x*) is a *constant polynomial*. Beware of confusing a polynomial of degree zero with the zero polynomial.

Two polynomials *a*(*x*) and *b*(*x*) are *equal* if they have the same degree and corresponding coefficients are equal. Thus, if *a*(*x*) = *a*_{0} + ⋯ + *a _{n}x^{n}* is of degree

*n*, and

*b*(

*x*) =

*b*

_{0}+ ⋯ +

*b*is of degree

_{m}x^{m}*m*, then

*a*(

*x*) =

*b*(

*x*) iff

*n*=

*m*and

*a*=

_{k}*b*for each

_{k}*k*from 0 to

*n*.

The familiar sigma notation for sums is useful for polynomials. Thus,

with the understanding that *x*^{0} = 1.

Addition and multiplication of polynomials is familiar from elementary algebra. We will now define these operations formally. Throughout these definitions we let *a*(*x*) and *b*(*x*) stand for the following polynomials:

Here we do *not* assume that *a*(*x*) and *b*(*x*) have the same degree, but allow ourselves to insert zero coefficients if necessary to achieve uniformity of appearance.

*We add polynomials by adding corresponding coefficients*. Thus,

*a*(*x*) + *b*(*x*) = (*a*_{0} + *b*_{0}) + (*a*_{1}, + *b*_{1})*x* + ⋯ + (*a _{n}* +

*b*)

_{n}*x*

^{n}Note that the degree of *a*(*x*) + *b*(*x*) is less than or equal to the *higher* of the two degrees, deg *a*(*x*) and deg *b*(*x*).

Multiplication is more difficult, but quite familiar:

*a*(*x*)*b*(*x*)

= *a*_{0}*b*_{0} + (*a*_{0}*b*_{1} + *b*_{0}*a*_{1})*x* + (*a*_{0}*b*_{2} + *a*_{1} *b*_{1} + *a*_{2} *b*_{0})*x*^{2} + ⋯ + *a _{n} b_{n} x*

^{2n}

In other words, the product of *a*(*x*) and *b*(*x*) is the polynomial

*c*(*x*) = *c*_{0} + *c*_{1}*x* + ⋯ + *c*_{2n} *x*^{2n}

whose *k*th coefficient (for any *k* from 0 to 2*n*) is

This is the sum of all the *a _{i}b_{j}* for which

*i*+

*j*=

*k*. Note that deg [

*a*(

*x*)

*b*(

*x*)] ≤ deg

*a*(

*x*) + deg

*b*(

*x*).

If *A* is any ring, the symbol

*A*[*x*]

designates the set of all the polynomials in *x* whose coefficients are in *A*, with addition and multiplication of polynomials as we have just defined them.

**Theorem 1** *Let A be a commutative ring with unity*. *Then A*[*x*] *is a commutative ring with unity*.

PROOF: To prove this theorem, we must show systematically that *A*[*x*] satisfies all the axioms of a commutative ring with unity. Throughout the proof, let *a*(*x*), *b*(*x*), and *c*(*x*) stand for the following polynomials:

The axioms which involve only addition are easy to check: for example, addition is commutative because

The associative law of addition is proved similarly, and is left as an exercise. The zero polynomial has already been described, and the negative of *a*(*x*) is

– *a*(*x*) = (– *a*_{0}) + (– *a*_{1})*x* + ⋯ + (– *a _{n}*)

*x*

^{n}To prove that multiplication is associative requires some care. Let *b*(*x*)*c*(*x*) = *d*(*x*), where *d*(*x*) = *d*_{0} + *d*_{1}*x* + ⋯ + *d*_{2n}*x*^{2n}. By the definition of polynomial multiplication, the *k*th coefficient of *b*(*x*)*c*(*x*) is

Then *a*(*x*)[*b*(*x*)*c*(*x*)] = *a*(*x*)*d*(*x*) = *e*(*x*), where *e*(*x*) = *e*_{0} + *e*_{1}*x* + ⋯ + *e*_{3n}*x*^{3n}. Now, the *l*th coefficient of *a*(*x*)*d*(*x*) is

It is easy to see that the sum on the right consists of all the terms *a _{h} b_{i} c_{j}* such that

*h*+

*i*+

*j*=

*l*Thus,

For each *l* from 0 to 3*n*, *e _{l}* is the

*l*th coefficient of

*a*(

*x*)[

*b*(

*x*)

*c*(

*x*)].

If we repeat this process to find the *l*th coefficient of [*a*(*x*) *b*(*x*)]*c*(*x*), we discover that it, too, is *e _{l}* Thus,

*a*(*x*)[*b*(*x*)*c*(*x*)] = [*a*(*x*)*b*(*x*)]*c*(*x*)

To prove the distributive law, let *a*(*x*)[*b*(*x*) + *c*(*x*)] = *d*{*x*) where *d*(*x*) = *d*_{0} + *d*_{1}*x* + ⋯ + *d*_{2n}*x*^{2n}. By the definitions of polynomial addition and multiplication, the *k*th coefficient *a*(*x*)[*b*(*x*) + *c*(*x*)] is

But Σ_{i}_{+j = k} *a _{i}b_{j}*. is exactly the

*k*th coefficient of

*a*(

*x*)

*b*(

*x*), and Σ

_{i}_{ + j = k}

*a*is the

_{i}c_{j}*k*th coefficient of

*a*(

*x*)

*c*(

*x*), hence

*d*is equal to the

_{k}*k*th coefficient of

*a*(

*x*)

*b*(

*x*) +

*a*(

*x*)

*c*(

*x*). This proves that

*a*(*x*)[*b*(*x*) + *c*(*x*)] = *a*(*x*)*b*(*x*) + *a*(*x*)*c*(*x*)

The commutative law of multiplication is simple to verify and is left to the student. Finally, the unity polynomial is the constant polynomial 1. ■

**Theorem 2** *If A is an integral domain*, *then A*[*x*] *is an integral domain*.

PROOF: If *a*(*x*) and *b*(*x*) are nonzero polynomials, we must show that their product *a*(*x*) *b*(*x*) is not zero. Let *a _{n}* be the leading coefficient of

*a*(

*x*), and

*b*the leading coefficient of

_{m}*b*(

*x*). By definition,

*a*≠ 0, and

_{n}*b*≠ 0.

_{m}*Thus a*≠ 0 because

_{n}b_{m}*A*is an integral domain. It follows that

*a*(

*x*)

*b*(

*x*) has a nonzero coefficient (namely,

*a*), so it is not the zero polynomial. ■

_{n}b_{m}If *A* is an integral domain, we refer to *A*[*x*] as a *domain of polynomials*, because *A*[*x*] is an integral domain. Note that by the preceding proof, if *a _{n}* and

*b*are the leading coefficients of

_{m}*a*(

*x*) and

*b*(

*x*), then

*a*is the leading coefficient of

_{n}b_{m}*a*(

*x*)

*b*(

*x*). Thus, deg

*a*(

*x*)

*b*(

*x*) =

*n*+

*m*:

*In a domain of polynomials A*[

*x*],

*where A is an integral domain*,

deg[*a*(*x*) · *b*(*x*)] = deg *a*(*x*) + deg *b*(*x*)

In the remainder of this chapter we will look at a property of polynomials which is of special interest when all the coefficients lie in a field. Thus, from this point forward, let *F* be a field, and let us consider polynomials belonging to *F*[*x*].

It would be tempting to believe that if *F* is a field then *F*[*x*] also is a field. However, this is not so, for one can easily see that the multiplicative inverse of a polynomial is not generally a polynomial. Nevertheless, by __Theorem 2__, *F*[*x*] *is an integral domain*.

Domains of polynomials over a *field* do, however, have a very special property: any polynomial *a*(*x*) may be divided by any nonzero polynomial *b*(*x*) to yield a quotient *q*(*x*) and a remainder *r*(*x*). The remainder is either 0, or if not, its degree is less than the degree of the divisor *b*(*x*). For example, *x*^{2} may be divided by *x* – 2 to give a quotient of *x* + 2 and a remainder of 4:

This kind of polynomial division is familiar to every student of elementary algebra. It is customarily set up as follows:

The process of polynomial division is formalized in the next theorem.

**Theorem 3: Division algorithm for polynomials** *If a*(*x*) *and b*(*x*) *are polynomials over a field F*, *and b*(*x*) ≠ 0, *there exist polynomials q*(*x*) *and r*(*x*) *over F such that*

*a*(*x*) = *b*(*x*)*q*(*x*) + *r*(*x*)

and

*r*(*x*) = 0 or deg *r*(*x*) < deg *b*(*x*)

PROOF: Let *b*(*x*) remain fixed, and let us show that every polynomial *a*(*x*) satisfies the following condition:

*There exist polynomials q*(*x*) *and r*(*x*) *over F such that a*(*x*) = *b*(*x*) *q*(*x*) + *r*(*x*), *and r*(*x*) = 0 *or* deg *r*(*x*) < deg *b*(*x*).

We will assume there are polynomials *a*(*x*) which do *not* fulfill the condition, and from this assumption we will derive a contradiction. Let *a*(*x*) be a polynomial of *lowest degree* which fails to satisfy the conditions. Note that *a*(*x*) cannot be zero, because we can express 0 as 0 = *b*(*x*) · 0 + 0, whereby *a*(*x*) would satisfy the conditions. Furthermore, deg *a*(*x*) ≥ deg *b*(*x*), for if deg *a*(*x*) < deg *b*(*x*) then we could write *a*(*x*) = *b*(*x*) · 0 + *a*(*x*), so again *a*(*x*) would satisfy the given conditions.

Let *a*(*x*) = *a*_{0} + ⋯ + *a _{n}x^{n}* and

*b*(

*x*) =

*b*

_{0}+ ⋯ +

*b*. Define a new polynomial

_{m}x^{m}This expression is the difference of two polynomials both of degree *n* and both having the same leading term *a _{n}x^{n}*. Because

*a*cancels in the subtraction,

_{n}x^{n}*A*(

*x*) has degree

*less than n*.

Remember that *a*(*x*) is a polynomial of *least degree* which fails to satisfy the given condition; hence *A*(*x*) *does satisfy* it. This means there are polynomials *p*(*x*) and *r*(*x*) such that

*A*(*x*) = *b*(*x*)*p*(*x*) + *r*(*x*)

where *r*(*x*) = 0 or deg *r*(*x*) < deg *b*(*x*). But then

If we let *p*(*x*) + (*a _{n}*/

*b*)

_{m}*x*

^{n}^{ – m}be renamed

*q*(

*x*), then

*a*(

*x*) =

*b*(

*x*)

*q*(

*x*) +

*r*(

*x*), so

*a*(

*x*) fulfills the given condition. This is a contradiction, as required. ■

**EXERCISES**

**A. Elementary Computation in Domains of Polynomials**

REMARK ON NOTATION: In some of the problems which follow, we consider polynomials with coefficients in * _{n}* for various

*n*. To simplify notation, we denote the elements of

*by 1, 2, …,*

_{n}*n*– 1 rather than the more correct .

**# 1** Let

*a*(

*x*) = 2

*x*

^{2}+ 3

*x*+ 1 and

*b*(

*x*) =

*x*

^{3}+ 5

*x*

^{2}+

*x*. Compute

*a*(

*x*) +

*b*(

*x*),

*a*(

*x*) –

*b*(

*x*) and

*a*(

*x*)

*b*(

*x*) in [

*x*],

_{5}[

*x*],

_{6}[

*x*], and

_{7}[

*x*].

**2** Find the quotient and remainder when *x*^{3} + *x*^{2} + *x* + 1 is divided by *x*^{2} + 3*x* + 2 in [*x*] and in _{5}[*x*].

**3** Find the quotient and remainder when *x*^{3} + 2 is divided by 2*x*^{2} + 3*x* + 4 in [*x*], in _{3}[*x*], and in _{5}[*x*].

We call *b*(*x*) a *factor* of *a*(*x*) if *a*(*x*) = *b*(*x*)*q*(*x*) for some *q*(*x*), that is, if the remainder when *a*(*x*) is divided by *b*(*x*) is equal to zero.

**4** Show that the following is true in *A*[*x*] for any ring *A*: For any odd *n*,

(*a*) *x* + 1 is a factor of *x ^{n}* + 1.

(*b*) *x* + 1 is a factor of *x ^{n}* +

*x*

^{n}^{–1}+ ⋯ +

*x*+ 1.

**5** Prove the following: In _{3}[*x*], *x* + 2 is a factor of *x ^{m}* + 2, for all

*m*. In

*[*

_{n}*x*],

*x*+ (

*x*– 1) is a factor of

*x*+ (

^{m}*n*− 1), for all

*m*and

*n*.

**6** Prove that there is no *integer m* such that 3*x*^{2} + 4*x* + *m* is a factor of 6*x*^{4} + 50 in [*x*].

**7** For what values of *n* is *x*^{2} + 1 a factor of *x*^{5} + 5*x* + 6 in * _{n}*[

*x*]?

**B. Problems Involving Concepts and Definitions**

**1** Is *x*^{8} + 1 = *x*^{3} + 1 in _{5}[*x*]? Explain your answer.

**2** Is there any ring *A* such that in *A*[*x*], some polynomial of degree 2 is equal to a polynomial of degree 4? Explain.

**# 3** Write all the quadratic polynomials in

_{5}[

*x*]. How many are there? How many cubic polynomials are there in

_{5}[

*x*]? More generally, how many polynomials of degree

*m*are there in

*[*

_{n}*x*]?

**4** Let *A* be an integral domain; prove the following:

If (*x* + 1)^{2} = *x*^{2} + 1 in *A*[*x*], then *A* must have characteristic 2.

If (*x* + 1)^{4} = *x*_{4} + 1 in *A*[*x*], then *A* must have characteristic 2.

If (*x* + 1)^{6} = *x*^{6} + 2*x*^{3} + 1 in *A*[*x*], then *A* must have characteristic 3.

**5** Find an example of each of the following in _{8}[*x*]: a divisor of zero, an invertible element. (Find nonconstant examples.)

**6** Explain why *x* cannot be invertible in any *A*[*x*], hence no domain of polynomials can ever be a field.

**7** There are rings such as *P*_{3} in which every element ≠0,1 is a divisor of zero. Explain why this cannot happen in any ring of polynomials *A*[*x*], even when *A* is *not* an integral domain.

**8** Show that in every *A*[*x*], there are elements ≠0,1 which are not idempotent, and elements ≠0,1 which are not nilpotent.

**C. Rings A[x] Where A Is Not an Integral Domain**

**1** Prove: If *A* is not an integral domain, neither is *A*[*x*].

**2** Give examples of divisors of zero, of degrees 0, 1, and 2, in _{4}[*x*].

**3** In _{10}[*x*], (2*x* + 2)(2*x* + 2) = (2*x* + 2)(5*x*^{3} + 2*x* + 2), yet (2*x* + 2) cannot be canceled in this equation. Explain why this is possible in _{10}[*x*], but not in _{5}[*x*].

**4** Give examples in _{4}[*x*], in _{6}[*x*], and in *Z*_{9}[*x*] of polynomials *a*(*x*) and *b*(*x*) such that deg *a*(*x*)*b*(*x*) < deg *a*(*x*) + deg *b*(*x*).

**5** If *A* is an integral domain, we have seen that in *A*[*x*],

deg *a*(*x*)*b*(*x*) = deg *a*(*x*) + deg *b*(*x*)

Show that if *A* is *not* an integral domain, we can always find polynomials *a*(*x*) and *b*(*x*) such that deg *a*(*x*)*b*(*x*) < deg *a*(*x*) + deg *b*(*x*).

**6** Show that if *A* is an integral domain, the only invertible elements in *A*[*x*] are the constant polynomials with inverses in *A*. Then show that in _{4}[*x*] there are invertible polynomials of all degrees.

**# 7** Give all the ways of factoring

*x*

^{2}into polynomials of degree 1 in

_{9}[

*x*]; in

_{5}[

*x*]. Explain the difference in behavior.

**8** Find all the square roots of *x*^{2} + *x* + 4 in _{5}[*x*]. Show that in _{8}[*x*], there are infinitely many square roots of 1.

**D. Domains A[x] Where A Has Finite Characteristic**

In each of the following, let *A* be an integral domain:

**1** Prove that if *A* has characteristic *p*, then *A*[*x*] has characteristic *p*.

**2** Use part 1 to give an example of an infinite integral domain with finite characteristic.

**3** Prove: If *A* has characteristic 3, then *x* + 2 is a factor of *x ^{m}* + 2 for all

*m*. More generally, if

*A*has characteristic

*p*, then

*x*+ (

*p*– 1)isa factor of

*x*+ (

^{m}*p*− 1) for all

*m*.

**4** Prove that if *A* has characteristic *p*, then in *A*[*x*], (*x* + *c*)* ^{p}* =

*x*+

^{p}*c*. (You may use essentially the same argument as in the proof of

^{p}__Theorem 3__,

__Chapter 20__.)

**5** Explain why the following “proof of part 4 is not valid: (*x* + *c*)* ^{p}* =

*x*+

^{p}*c*in

^{p}*A*[

*x*] because (

*a*+

*c*)

*=*

^{p}*a*+

^{p}*c*for all

^{p}*a*,

*c*∈

*A*. (Note the following example: in

_{2},

*a*

^{2}+ 1 =

*a*

^{4}+ 1 for every

*a*, yet

*x*

^{2}+ 1 ≠

*x*

^{4}+ 1 in

_{2}[

*x*].)

**# 6** Use the same argument as in part 4 to prove that if

*A*has characteristic

*p*, then [

*a*(

*x*) +

*b*(

*x*)]

*=*

^{p}*a*(

*x*)

*+*

^{p}*b*(

*x*)

*for any*

^{p}*a*(

*x*),

*b*(

*x*) ∈

*A*[

*x*]. Use this to prove:

**E. Subrings and Ideals in A[x]**

**1** Show that if *B* is a subring of *A*, then *B*[*x*] is a subring of *A*[*x*].

**2** If *B* is an *ideal* of *A*, *B*[*x*] is an ideal of *A*[*x*].

**3** Let *S* be the set of all the polynomials *a*(*x*) in *A*[*x*] for which every coefficient *a _{i}* for

*odd i*is equal to zero. Show that

*S*is a subring of

*A*[

*x*]. Why is the same not true when “odd” is replaced by “even”?

**4** Let *J* consist of all the elements in *A*[*x*] whose constant coefficient is equal to zero. Prove that *J* is an ideal of *A*[*x*].

**# 5** Let

*J*consist of all the polynomials

*a*

_{0}+

*a*

_{1}

*x*+ ⋯ +

*a*in

_{n}x^{n}*A*[

*x*] such that

*a*

_{0}+

*a*

_{1}+ ⋯ +

*a*= 0. Prove that

_{n}*J*is an ideal of

*A*[

*x*].

**6** Prove that the ideals in both parts 4 and 5 are *prime* ideals. (Assume *A* is an integral domain.)

**F. Homomorphisms of Domains of Polynomials**

Let *A* be an integral domain.

**1** Let *h* : *A*[*x*]→ *A* map every polynomial to its constant coefficient; that is,

*h*(*a*_{0} + *a*_{1}*x* + ⋯ + *a _{n}x^{n}*) =

*a*

_{0}

Prove that *h* is a homomorphism from *A*[*x*] *onto A*, and describe its kernel.

**2** Explain why the kernel of *h* in part 1 consists of all the products *xa*(*x*), for all *a*(*x*) ∈ *A*[*x*]. Why is this the same as the principal ideal ⟨x⟩ in *A*[*x*]?

**3** Using parts 1 and 2, explain why *A*[*x*]/⟨x⟩ ≅ *A*.

**4** Let *g* : *A*[*x*] → *A* send every polynomial to the sum of its coefficients. Prove that *g* is a surjective homomorphism, and describe its kernel.

**5** If *c* ∈ *A*, let *h* : *A*[*x*] → *A*[*x*] be defined by *h*(*a*(*x*)) = *a*(*cx*), that is,

*h*(*a*_{0} + *a*_{1}*x* + ⋯ + *a _{n}x^{n}*) =

*a*

_{0}+

*a*

_{1}

*cx*+

*a*

_{2}

*c*

^{2}

*x*

^{2}+ ⋯ +

*a*

_{n}c^{n}x^{n}Prove that *h* is a homomorphism and describe its kernel.

**6** If *h* is the homomorphism of part 5, prove that *h* is an automorphism (isomorphism from *A*[*x*] to itself) iff *c* is invertible.

**G. Homomorphisms of Polynomial Domains Induced by a Homomorphism of the Ring of Coefficients**

Let *A* and *B* be rings and let *h* : *A* → *B* be a homomorphism with kernel *K*. Define : *A*[*x*] → *B*[*x*] by

(*a*_{0} + *a*_{1}*x* + ⋯ + *a _{n}x^{n}*) =

*h*(

*a*

_{0}) +

*h*(

*a*

_{1})

*x*+ ⋯ +

*h*(

*a*)

_{n}*x*

^{n}(We say that is *induced by h*.)

**1** Prove that is a homomorphism from *A*[*x*] to *B*[*x*].

**2** Describe the kernel of .

**# 3** Prove that is surjective iff

*h*is surjective.

**4** Prove that is injective iff *h* is injective.

**5** Prove that if *a*(*x*) is a factor of *b*(*x*), then (*a*(*x*)) is a factor of (*b*{*x*)).

**6** If *h* : → * _{n}* is the natural homomorphism, let : [

*x*] →

*[*

_{n}*x*] be the homomorphism induced by

*h*. Prove that (

*a*(

*x*)) = 0 iff

*n*divides every coefficient of

*a*(

*x*).

**7** Let be as in part 6, and let *n* be a prime. Prove that if *a*(*x*)*b*(*x*) ∈ ker , then either *a*(*x*) or *b*(*x*) is in ker . (HINT: Use __Exercise F2__ of __Chapter 19__.)

**H. Polynomials in Several Variables**

*A*[*x*_{1}, *x*_{2}] denotes the ring of all the polynomials in *two letters x*_{1} and *x*_{2} with coefficients in *A*. For example, *x*^{2} – 2*xy* + *y*^{2} + *x* − 5 is a quadratic polynomial in [*x*, *y*]. More generally, *A*[*x*_{1}, …, *x _{n}*] is the ring of the polynomials in

*n letters x*

_{1}, …,

*x*with coefficients in

_{n}*A*. Formally it is defined as follows: Let

*A*[

*x*

_{1}] be denoted by

*A*

_{1}; then

*A*

_{1}[

*x*

_{2}] is

*A*[

*x*

_{1},

*x*

_{2}]. Continuing in this fashion, we may adjoin one new letter

*x*at a time, to get

_{i}*A*[

*x*

_{1}, …,

*x*].

_{n}**1** Prove that if *A* is an integral domain, then *A*[*x*_{1}, …, *x _{n}*] is an integral domain.

**2** Give a reasonable definition of the *degree* of any polynomial *p*(*x, y*) in *A*[*x*, *y*] and then list all the polynomials of degree ≤ 3 in *Z*_{3}[*x*, *y*].

Let us denote an arbitrary polynomial *p*(*x*, *y*) in *A*[*x*, *y*] by Σ *a _{ij}x^{i}y^{j}* where Σ ranges over

*some*pairs

*i*,

*j*of nonnegative integers.

**3** Imitating the definitions of sum and product of polynomials in *A*[*x*], give a definition of sum and product of polynomials in *A*[*x*, *y*].

**4** Prove that deg *a*(*x*, *y*)*b*(*x*, *y*) = deg *a*(*x*, *y*) + deg *b*(*x*, *y*) if *A* is an integral domain.

**I. Fields of Polynomial Quotients**

Let *A* be an integral domain. By the closing part of __Chapter 20__, every integral domain can be extended to a “field of quotients.” Thus, *A*[*x*] can be extended to a field of polynomial quotients, which is denoted by *A*(*x*). Note that *A*(*x*) consists of all the fractions *a*(*x*)/*b*(*x*) for *a*(*x*) and *b*(*x*) ≠ 0 in *A*[*x*], and these fractions are added, subtracted, multiplied, and divided in the customary way.

**1** Show that *A*(*x*) has the same characteristic as *A*.

**2** Using part 1, explain why there is an infinite field of characteristic *p*, for every prime *p*.

**3** If *A* and *B* are integral domains and *h* : *A* → *B* is an isomorphism, prove that *h* determines an isomorphism : *A*(*x*) → *B*(*x*).

**J. Division Algorithm: Uniqueness of Quotient and Remainder**

In the division algorithm, prove that *q*(*x*) and *r*(*x*) are uniquely determined. [HINT: Suppose *a*(*x*) = *b*(*x*)*q*_{1}(*x*) + *r*_{1}(*x*) = *b*(*x*)*q*_{2}(*x*) + *r*_{2}(*x*), and subtract these two expressions, which are both equal to *a*(*x*).]