COMPUTATIONS IN QUOTIENT RINGS

(iv)p(x)|[{q(x) − s(x)} · p(x) + (r(x) − t (x))].

(v)p(x)|[r(x) − t (x)].

(vi) r(x) = t (x).

Hence every coset of F [x] by (p(x)) contains the zero polynomial or a polynomial of degree less than that of p(x).

Theorem 10.8. If F is a field, let P be the ideal (p(x)) in F [x] generated by the polynomial p(x) of degree n > 0. The different elements of F [x]/(p(x)) are precisely those of the form

remainder when f (x) is divided by p(x). Then, by Lemma 10.7, P + f (x) = P + r(x), which is of the required form.

Suppose that P + r(x) = P + t (x), where r(x) and t (x) are zero or have

The tables for the quotient ring are given in Table 10.2. The addition table is straightforward to calculate. Multiplication is computed as follows:

(P + x)2 = P + x2 = P + (x2 + x + 1) + (x + 1) = P + x + 1

There are often easier ways of finding the remainder of f (x) when divided by p(x) than by applying the division algorithm directly. If deg p(x) = n and P = (p(x)), the problem of finding the remainder reduces to the problem of finding a polynomial r(x) of degree less than n such that f (x) ≡ r(x) modP . This can often be solved by manipulating congruences, using the fact that p(x) ≡ 0 modP .

Consider Example 10.10, in which P is the ideal generated by x2 − 2. Then x2 − 2 ≡ 0 modP and x2 ≡ 2 modP . Hence, in any congruence modulo P , we can always replace x2 by 2. For example,

15x2 + 2x − 24 ≡ 15(2) + 2x − 24 modP

n − 1 instead of as [0], [1], . . . , [n − 1] or as (n) + 0, (n) + 1, . . . , (n) + n − 1. In a similar way, when there is no confusion, we henceforth write the elements

of F [x]/(p(x)) simply as a0 + a1x + · · · + an−1xn−1 instead of (p(x)) + a0 + a1x + · · · + an−1xn−1.

MORPHISM THEOREM

Proposition 10.11. If f : R → S is a ring morphism, then Kerf is an ideal of R.

Proof. Since any ring morphism is a group morphism, it follows from Proposition 4.23 that Kerf is a subgroup of (R, +). If x Kerf and r R, then

210 10 QUOTIENT RINGS

f (xr) = f (x)f (r) = 0 · f (r) = 0 and xr Kerf . Similarly, rx Kerf , so Kerf is an ideal of R.

Furthermore, any ideal I of a ring R is the kernel of a morphism, for example, the ring morphism π : R → R/I defined by π(r) = I + r.

The image of a morphism f : R → S can easily be verified to be a subring of S.

Theorem 10.12. Morphism Theorem for Rings. If f : R → S is a ring morphism, then R/Kerf is isomorphic to Imf .

This result is also known as the first isomorphism theorem for rings; the second and third isomorphism theorems are given in Exercises 10.19 and 10.20.

Proof. Let K = Kerf . It follows from the morphism theorem for groups (Theorem 4.23), that ψ: R/K → Imf , defined by ψ(K + r) = f (r), is a group isomorphism. Hence we need only prove that ψ is a ring morphism. We have

ψ{(K + r)(K + s)} = ψ{K + rs} = f (rs) = f (r)f (s)

QUOTIENT POLYNOMIAL RINGS THAT ARE FIELDS

We now determine when a quotient of a polynomial ring is a field. This result allows us to construct many new fields.

Theorem 10.15. Let a be an element of the euclidean ring R. The quotient ring R/(a) is a field if and only if a is irreducible in R.

Proof. Suppose that a is an irreducible element of R and let (a) + b be a nonzero element of R/(a). Then b is not a multiple of a, and since a is irreducible, gcd(a, b) = 1. By Theorem 9.9, there exist s, t R such that

sa + tb = 1.

Now sa (a), so [(a) + t] · [(a) + b] = (a) + 1, the identity of R/(a). Hence

(a) + t is the inverse of (a) + b in R/(a) and R/(a) is a field.

Now suppose that a is not irreducible in R so that there exist elements s and t, which are not invertible, with st = a. By Lemma 9.17, δ(s) < δ(st) = δ(a) and δ(t) < δ(st) = δ(a). Hence s is not divisible by a, and s / (a). Similarly, t / (a), and neither (a) + s nor (a) + t is the zero element of R/(a). However,

[(a) + s] · [(a) + t] = (a) + st = (a), the zero element of R/(a).

Therefore, the ring R/(a) has zero divisors and cannot possibly be a field.

For example, in the quotient ring Q[x]/P , where P = (x2 − 1), the elements P + x + 1 and P + x − 1 are zero divisors because

Another particular case of Theorem 10.15 is the following important theorem.

Theorem 10.17. The ring F [x]/(p(x)) is a field if and only if p(x) is irreducible over the field F . Furthermore, the ring F [x]/(p(x)) always contains a subring isomorphic to the field F .

Proof. The first part of the theorem is just Theorem 10.15. Let F = {(p(x)) + r|r F }. This can be verified to be a subring of F [x]/(p(x)), which is isomorphic to the field F by the isomorphism that takes r F to (p(x))+ r F [x]/(p(x)).

Example 10.18. Show that Z2[x]/(x2 + x + 1) is a field with four elements.

Solution. We showed in Example 9.34 that x2 + x + 1 is irreducible over Z2 and in Example 10.9 that the quotient ring has four elements. Hence the quotient ring is a field containing four elements. Its tables are given in Table 10.2.

Example 10.19. Write down the multiplication table for the field Z3[x]/(x2 + 1).

Solution. If x = 0, 1, or 2 in Z3, then x2 + 1 = 1, 2, or 2; thus, by the factor theorem, x2 + 1 has no linear factors. Hence x2 + 1 is irreducible over Z3 and, by Theorem 10.17, the quotient ring Z3[x]/(x2 + 1) is a field. By Theorem 10.8, the elements of this field can be written as

Z3[x]/(x2 + 1) = {a0 + a1x|a0, a1 Z3}.

Hence the field contains nine elements. Its multiplication table is given in Table 10.3. This can be calculated by multiplying the polynomials in Z3[x] and replacing x2 by −1 or 2, since x2 ≡ −1 ≡ 2mod (x2 + 1).

Example 10.20. Show that Q[x]/(x3 − 5) = {a0 + a1x + a2x2|ai Q} is a field and find the inverse of the element x + 1.

Solution. By the rational roots theorem (Theorem 9.25), (x3 − 5) has no linear factors and hence is irreducible over Q. Therefore, by Theorem 10.17,

We can find such polynomials s(x) and t (x) by the euclidean algorithm. We have (see below)

x3 − 5 = (x2 − x + 1)(x + 1) − 6,

so

6 ≡ (x2 − x + 1)(x + 1) mod(x3 − 5)

and

1 ≡ 16 (x2 − x + 1)(x + 1) mod(x3 − 5).

TABLE 10.3. Multiplication in Z3[x ]/(x 2 + 1)

Example 10.21. Show that Z3[x]/(x3 + 2x + 1) is a field with 27 elements and find the inverse of the element x2.

Solution. If x = 0, 1, or 2 in Z3, then x3 + 2x + 1 = 1; hence x3 + 2x + 1 has no linear factors and is irreducible. Therefore,

Z3[x]/(x3 + 2x + 1) = {a0 + a1x + a2x2|ai Z3}

is a field that has 33 = 27 elements.

As in Example 10.20, to find the inverse of x2, we apply the euclidean algorithm to x3 + 2x + 1 and x2 in Z3[x].

We have x3 + 2x + 1 = x(x2) + (2x + 1) and x2 = (2x + 2)(2x + 1) + 1. Hence

1= x2 − (2x + 2){(x3 + 2x + 1) − x · x2}

= x2(2x2 + 2x + 1) − (2x + 2)(x3 + 2x + 1),

so

1 ≡ x2(2x2 + 2x + 1) mod(x3 + 2x + 1) and the inverse of x2 in Z3[x]/(x3 + 2x + 1) is 2x2 + 2x + 1.

x+0 x+2