Quotient Rings and Finite Fields

The commutative ring R/I constructed in Theorem 3.110 is called the quotient ring14 of R modulo I (briefly, R mod I). 14Presumably, quotient rings are so called in analogy with quotient groups. Quotient Rings and Finite Fields We saw in Example 2.68 that the additive abelian group Z/(m) is identical to Im. They have the same elements: the coset a+(m) and the congruence class [a] are the same subset of Z; they have the same addition: a + (m) + b + (m) = a + b + (m) = [a + b] = [a] + [b]. We can now prove a converse to Proposition 3.50.

If k is a field, the intersection of all the subfields of k is called the prime field of k. Every subfield of C contains Q, and so the prime field of C and of R is Q. The prime field of a finite field is just the integers mod p, as we show next. Notation. From now on, we will denote Ip by Fp when we are regarding it as a field. Borrowing terminology from group theory, call the intersection of all the subfields of a field containing a subset X the subfield generated by X; it is the smallest subfield containing X in the sense that if F is any subfield containing X, then F contains the subfield generated by X. The prime field is the subfield generated by 1, and the prime field of Fp(x) is Fp.

A field k has characteristic 0 if its prime field is isomorphic to Q; a field k has characteristic p if its prime field is isomorphic to Fp for some prime p. The fields Q, R, C have characteristic 0, as does any subfield of them; every finite field has characteristic p for some prime p, as does Fp(x), the ring of all rational functions over Fp.

If K is a field containing k as a subfield, then K is called a (field) extension of k, and we write “K/k is a field extension.”15 An extension field K of a field k is a finite extension of k if K is a finite-dimensional vector space over k. The dimension of K, denoted by [K : k], is called the degree of K/k.

Let K/k be a field extension. An element α ∈K is algebraic over k if there is some nonzero polynomial f (x) ∈k[x] having α as a root; otherwise, α is transcendental over k. An extension K/k is algebraic if every α ∈K is algebraic over k. When a real number is called transcendental, it usually means that it is transcendental over Q.

If K/k is an extension and α ∈K, then k(α) is the intersection of all those subfields of K that contain k and α; we call k(α) the subfield of K obtained by adjoining α to k. More generally, if A is a (possibly infinite) subset of K, define k(A) to be the intersection of all the subfields of K that contain k ∪A; we call k(A) the subfield of K obtained by adjoining A to k. In particular, if A = {z1, . . . , zn} is a finite subset, then we may denote k(A) by k(z1, . . . , zn). It is clear that k(A) is the smallest subfield of K containing k and A; that is, if B is any subfield of K containing k and A, then k(A) ⊆B. We now show that the field k[x]/(p(x)), where p(x) ∈k[x] is irreducible, is intimately related to adjunction.

If K/k is a field extension and α ∈K is algebraic over k, then the unique monic irreducible polynomial p(x) ∈k[x] having α as a root is called the minimal polynomial of α over k, and it is denoted by irr(α, k) = p(x). The minimal polynomial irr(α, k) does depend on k. For example, irr(i, R) = x2 + 1, while irr(i, C) = x −i. The following formula is quite useful, especially when proving a theorem by induction on degrees.

Let k be a subfield of a field K, and let f (x) ∈k[x]. We say that f (x) splits over K if f (x) = a(x −z1) · · · (x −zn), where z1, . . . , zn are in K and a ∈k is nonzero. If f (x) ∈k[x] is a polynomial, then a field extension E/k is called a splitting field of f (x) over k if f (x) splits over E, but f (x) does not split over any proper subfield of E. For example, consider f (x) = x2 + 1 ∈Q[x]. The roots of f (x) are ±i, and so f (x) splits over C; that is, f (x) = (x −i)(x + i) is a product of linear polynomials in C[x]. However, C is not a splitting field over Q, for C is not the smallest field containing Q and all the roots of f (x). The splitting field of f (x) ∈k[x] depends on k as well as on f (x): Here, the splitting field over Q is Q(i); the splitting field over R is R(i) = C. In Example 3.122, we proved that E = Q (√ 2 + √ ) is a splitting field of f (x) = x4 −10x2 + 1, as well as a splitting field of g(x) = (x2 −2)(x2 −3).

If I is an ideal in a commutative ring R, then the additive abelian group R/I can be made into a commutative ring in such a way that the natural map π : R →R/I is a surjective ring homomorphism.

If I is an ideal in a commutative ring R, then there are a commutative ring A and a ring homomorphism π : R →A with I = ker π.

If k is a field, then its prime field is isomorphic to Q or to Fp for some prime p.

If k is a field and I = (p(x)), where p(x) is a nonzero polynomial in k[x], then the following are equivalent: p(x) is irreducible; k[x]/I is a field; k[x]/I is a domain.

Let k be a field, let p(x) ∈k[x] be a monic irreducible polynomial of degree d, let K = k[x]/I, where I = (p(x)), and let β = x + I ∈K. (i) K is a field and k′ = {a + I : a ∈k} is a subfield of K isomorphic to k. Therefore, if k′ is identified with k, then k is a subfield of K. (ii) β is a root of p(x) in K. (iii) If g(x) ∈k[x] and β is a root of g(x), then p(x) | g(x) in k[x]. (iv) p(x) is the unique monic irreducible polynomial in k[x] having β as a root. (v) The list 1, β, β2, . . . , βd−1 is a basis of K as a vector space over k, and so dimk(K) = d.

(i) If K/k is an extension and α ∈K is algebraic over k, then there is a unique monic irreducible polynomial p(x) ∈k[x] having α as a root. Moreover, if I = (p(x)), then k[x]/I ∼= k(α); indeed, there exists an isomorphism ϕ : k[x]/I →k(α) with ϕ(x + I) = α and ϕ(c + I) = c for all c ∈k. (ii) If α′ ∈K is another root of p(x), then there is an isomorphism θ : k(α) →k(α′) with θ(α) = α′ and θ(c) = c for all c ∈k.

Let k ⊆E ⊆K be fields, with E a finite extension of k and K a finite extension of E. Then K is a finite extension of k, and [K : k] = [K : E][E : k].

If k is a field and f (x) ∈k[x], then there exists a field K containing k as a subfield and with f (x) a product of linear polynomials in K[x].

Let k be a field, and let f (x) ∈k[x]. Then a splitting field of f (x) over k exists.

If p is a prime and n is a positive integer, then there is a field having exactly pn elements.

For every prime p and every integer n ≥1, there exists an irreducible polynomial g(x) ∈Fp[x] of degree n. In fact, if α is a primitive element of Fpn, then its minimal polynomial g(x) = irr(α, Fp) has degree n. Remark. An easy modification of the proof replaces Fp by any finite field. ◀

says that E = F3[x]/(p(x)) is given by E = {a + bα : where α2 + 1 = 0}. Quotient Rings and Finite Fields Similarly, if F = F3[x]/(q(x)), then F = {a + bβ : where β2 + β −1 = 0}. These two fields are isomorphic, for the map ϕ : E →F (found by trial and error), defined by ϕ(a + bα) = a + b(1 −β), is an isomorphism. ◀ We are now going to solve the isomorphism problem for finite fields.

, so that the inductive hypothesis gives an isomorphism →: E →E′ that extends Δϕ, and hence ϕ. •

If k is a field and f (x) ∈k[x], then any two splitting fields of f (x) over k are isomorphic via an isomorphism that fixes k pointwise.

The polynomial x2+1 ∈R[x] is irreducible, and so K = R[x]/(x2+1) is a field extension K/R of degree 2. If β is a root of x2 + 1, then β2 = −1; moreover, every element of K has a unique expression of the form a + bβ, where a, b ∈R. Clearly, this is another construction of C (which we have been viewing as the points in the plane equipped with a certain addition and multiplication). Here is a natural way to construct an isomorphism K →C. Consider the evaluation map ϕ : R[x] →C given by ϕ : f (x) ↦f (i). First, ϕ is surjective, for a + ib = ϕ(a + bx) ∈im ϕ. Second, ker ϕ = { f (x) ∈R[x] : f (i) = 0}, the set of all polynomials in R[x] having i as a root. We know that x2 + 1 ∈ker ϕ, so that (x2 + 1) ⊆ker ϕ. For the reverse inclusion, take g(x) ∈ker ϕ. Now i is a root of g(x), and so gcd (g, x2 + 1) ̸= 1 in C[x]; therefore, gcd (g, x2 + 1) ̸= 1 in R[x]. Irreducibility of x2 + 1 in R[x] gives x2 + 1 | g(x), and so g(x) ∈(x2 + 1), Therefore, ker ϕ = (x2 + 1). The first isomorphism theorem now gives R[x]/(x2 + 1) ∼= C. ◀ The easiest way to multiply in C is to first treat i as a variable and then to impose the condition i2 = −1. To compute (a + bi)(c + di), first write ac + (ad + bc)i + bdi2, and then observe that i2 = −1. More generally, if β is a root of an irreducible p(x) ∈k[x], then the proper way to multiply (b0 + b1β + · · · + bn−1βn−1)(c0 + c1β + · · · + cn−1βn−1) in the quotient ring k[x]/(p(x)) is to regard the factors as polynomials in β, multiply them, and then impose the condition that p(β) = 0. A first step in classifying fields involves their characteristic; that is, describing prime fields. A next step considers whether the elements are algebraic over the prime field.

On the other hand, we may construct a field of order 4 as the quotient F = F2[x]/(q(x)), where q(x) ∈F2[x] is the irreducible polynomial x2+x +1. By Proposition 3.117(v), F is a field consisting of all a + bz, where z = x + (q(x)) is a root of q(x) and a, b ∈I2. Since z2+z+1 = 0, we have z2 = −z−1 = z+1; moreover, z3 = zz2 = z(z+1) = z2+z = 1. Let us look at the first two in more detail.