Quotient Rings of Polynomial Rings

If (R, +, ·) is any ring and (S, +) is any subgroup of the abelian group (R, +), then the quotient group (R/S, +) has already been defined. However, R/S does not have a ring structure induced on it by R unless S is a special kind of subgroup called an ideal.
A nonempty subset I of a ring R is called an ideal of R if the following conditions are satisfied for all x, y ∈ I and r ∈ R:

1. x − y ∈ I.

2. x · r and r · x ∈ I.

Condition (i) implies that (I, +) is a subgroup of (R, +). In any ring R, R itself is an ideal, and {0} is an ideal.

Let F be a field, and suppose p(x) ∈ F[x]. ⟨p(x)⟩ is the set of all multiples (by polynomials) of p(x), F[x] the (principal) ideal generated by p(x), that is the set ⟨p(x)⟩ = {q(x) p(x)| q(x) in F[x]}. When you form the quotient ring, it is as if you’ve set ⟨p(x)⟩ multiples of p(x) equal to 0.
If a(x) ∈ F[x], then a(x) + ⟨p(x)⟩ is the coset of ⟨p(x)⟩ represented by a(x).
Define a(x) = b(x) (mod p(x)) (a(x) is congruent to b(x) mod p(x)) to mean that

In words, this means that a(x) and b(x) are congruent mod p(x) if they differ by a multiple of p(x). In equation form, this says a(x) − b(x) = k(x) · p(x) for some k(x) ∈ F[x], or a(x) = b(x) + k(x) · p(x) for some k(x) ∈ F[x].

Lemma 8.2.1. Let R be a commutative ring, and suppose a(x), b(x), p(x) ∈ R[x]. Then a(x) = b(x) (mod p(x)) if and only if a(x) + ⟨p(x) = b(x) + ⟨p(x).

Proof. Suppose a(x) = b(x) (mod p(x)). Then a(x) = b(x) + k(x) · p(x) for some k(x) ∈ R[x]. Hence,

Conversely, suppose a(x) + ⟨p(x)⟩ = b(x) + ⟨p(x)⟩. Then

Hence,

This means that a(x) = b(x) (mod p(x)).  □

Depending on the situation, it can be written a(x) = b(x) (mod p(x)) or a(x) + ⟨p(x)⟩ = b(x) + ⟨p(x)⟩.

Remark. In analogy to what we studied with ℤ/nℤ the set of residue classes modulo an integer n (the construction here is based on the polynomial ring K[x] rather than the ring ℤ), we can identify a congruence class by those polynomials which admit the same remainder r(x) when divided by p(x), as the next lemma shows.

Lemma 8.2.2. f(x) ≡ g(x) mod(p(x)) if and only if f(x) and g(x) have the same remainder when divided by p(x).

Proof. Let f(x) = q(x) · p(x) + r(x) and g(x) = s(x) · p(x) + t(x), where r(x) and t(x) are zero or have degrees less than that of p(x). Now the lemma follows because the following statements are equivalent:

1. f(x) ≡ g(x) mod(p(x)).

2. f(x) − g(x) ∈ (p(x)).

3. p(x)|f(x) − g(x).

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

5. p(x)|[r(x) − t(x)].

6. r(x) = t(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 ⟨p(x)⟩ + a₀ + a₁x + ··· + a_n−1 xⁿ⁻¹ where a₀, a₁, ..., a_n−1 ∈ F.

Proof. Let f(x) + P be any element of F[x]/⟨p(x)⟩ and let r(x) be the remainder when f(x) is divided by p(x). Then, by Lemma 8.2.2, f(x) + P = r(x) + P, which is of the required form. Suppose that r(x) + P = t(x) + P, where r(x) and t (x) are zero or have degree less than n. Then

Example 8.2.3. Write down the tables for ℤ₂[x]/(x² + x + 1).

Solution. Let P = (x² + x + 1), so that

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

The multiplicative inverse of x is x + 1, since x(x + 1) = x² + x = (x + 1) + x = 1.  ■

Nota bene. We have usually written the elements of ℤ_n = ℤ/(n) simply as 0, 1, ..., n − 1, or [0], [1], ..., [n − 1] instead of as 0 + (n), 1 + (n), ..., n − 1 + (n).

Quotient Polynomial rings that are fields

We shall see in next theorem that F[x]/⟨p(x)⟩ is a field when p(x) is irreducible.  ■

Theorem 8.2.4. 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

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 4.8.3, δ(s) < δ(st) = δ(a) and δ(t) < δ(st) = δ(a) (where δ is the Euclidean valuation). 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,

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

Example 8.2.4. The polynomial x² − 1 is clearly not irreducible in ℝ[x]. The quotient ℝ[x]/⟨x² − 1⟩ is not a field. In fact,

Exercises

1. Take p(x) = x − 2 in Q[x]. Show that 2x² + 3x + 5 = x² + 4x + 7 (mod x − 2).

2. Find a rational number r such that x³ − 4x² + x + 12 = r   (mod x − 2).

Solutions

1. Two polynomials are congruent mod x − 2 if they differ by a multiple of x − 2. The difference between the two polynomials is x² − x −2 = (x − 2)(x+1), hence they are congruent modulo x − 2. ■

2. If the two polynomials differ by a multiple of x − 2, then 2 must be a root of their difference, then r = 6. ■