When \( p \) is a prime integer, then Theorem 2.8 states, in effect, that \( \mathbb{Z}_p \) is a field (and, of course, an integral domain). Here is the analogous result for \( F[x] \) and an irreducible polynomial \( p(x) \).

---

Theorem 5.10

Let \( F \) be a field and \( p(x) \) a nonconstant polynomial in \( F[x] \). Then the following statements are equivalent:

1. \( p(x) \) is irreducible in \( F[x] \).
2. \( F[x]/(p(x)) \) is a field.
3. \( F[x]/(p(x)) \) is an integral domain.

---

Proof of Theorem 5.10

(1) \( \Rightarrow \) (2)

By Theorem 5.7, \( F[x]/(p(x)) \) is a commutative ring with identity, satisfying Axioms 1–10. To prove that \( F[x]/(p(x)) \) is a field, we must verify that every nonzero element in \( F[x]/(p(x)) \) is a unit (Axiom 12). That is, we need to show that for any nonzero \( [a(x)] \in F[x]/(p(x)) \), there exists \( [u(x)] \) such that:

\[ [a(x)][u(x)] = [1] \]

Given that \( [a(x)] \neq [0] \), we know that \( a(x) \neq 0 \ (\text{mod} \ p(x)) \) by Theorem 5.3. Thus, the gcd of \( a(x) \) and \( p(x) \) must be either:

1. \( 1_F \), or
2. A monic associate of \( p(x) \) (since \( p(x) \) is irreducible, it only has monic divisors).

If the gcd were an associate of \( p(x) \), then \( a(x) \) would be divisible by \( p(x) \), which would contradict \( a(x) \neq 0 \ (\text{mod} \ p(x)) \). Therefore, the gcd of \( a(x) \) and \( p(x) \) must be \( 1_F \).

By Theorem 4.8, there exist polynomials \( u(x) \) and \( v(x) \) such that:

\[ a(x)u(x) + p(x)v(x) = 1_F \]

This means:

\[ [a(x)][u(x)] = [1] \]

in \( F[x]/(p(x)) \), as \( p(x)v(x) \) reduces to zero in the quotient ring. Thus, \( [a(x)] \) has an inverse, and \( F[x]/(p(x)) \) is a field, satisfying Axiom 12.

---

(2) \( \Rightarrow \) (3)

This part follows directly from Theorem 3.8, which states that any field is an integral domain.

---

(3) \( \Rightarrow \) (1)

We shall verify statement (2) of Theorem 4.12 to show that \( p(x) \) is irreducible. Suppose that \( b(x) \) and \( c(x) \) are any polynomials in \( F[x] \) and \( p(x) \) divides \( b(x)c(x) \). Then \( b(x)c(x) = 0_F \ (\text{mod} \ p(x)) \). So by Theorem 5.3, and \( p(x) \mid b(x)c(x) \). Then:

\[ [b(x)][c(x)] = [b(x)c(x)] = [0_F] \text{ in } F[x]/(p(x)). \]

Because \( F[x]/(p(x)) \) is an integral domain by (3), we have \( [a(x)] = [0_F] \) or \( [b(x)] = [0_F] \). Thus, \( b(x) = 0_F \ (\text{mod} \ p(x)) \) or \( c(x) = 0_F \ (\text{mod} \ p(x)) \) by Theorem 5.3, which means that \( p(x) \mid b(x) \) or \( p(x) \mid c(x) \) by the definition of congruence. Therefore, \( p(x) \) is irreducible by Theorem 4.12. \( \blacksquare \)

---

Theorem 5.10 can be used to construct finite fields. If \( p \) is prime and \( f(x) \) is irreducible in \( \mathbb{Z}_p[x] \) of degree \( k \), then \( \mathbb{Z}_p[x]/(f(x)) \) is a field by Theorem 5.10. Exercise 7 in Section 5.1 shows that this field has \( p^k \) elements. Finite fields are discussed further in Section 11.6, where it is shown that there are irreducible polynomials of every positive degree in \( \mathbb{Z}_p[x] \) and, hence, finite fields of all possible prime power orders. See Exercise 9 for an example.

Let \( F \) be a field and \( p(x) \) an irreducible polynomial in \( F[x] \). Let \( K \) denote the field of congruence classes \( F[x]/(p(x)) \). By Theorems 5.8 and 5.10, \( F \) is a subfield of the field \( K \). One also says that \( K \) is an extension field of \( F \). Polynomials in \( F[x] \) can be considered to have coefficients in the larger field \( K \), and we can ask about the roots of such polynomials in \( K \). In particular, what can be said about the roots of \( p(x) \) that we started with? Even though \( p(x) \) is irreducible in \( F[x] \), it may have roots in the extension field \( K \).

---

Example 1

The polynomial \( p(x) = x^2 + x + 1 \) has no roots in \( \mathbb{Z}_2 \), and is, therefore, irreducible in \( \mathbb{Z}_2[x] \) by Corollary 4.19. Consequently, \( K = \mathbb{Z}_2[x]/(x^2 + x + 1) \) is an extension field of \( \mathbb{Z}_2 \) by Theorem 5.10. Using the tables for \( K \) in Example 3 of Section 5.2, we see that:

\[ [x]^2 + [x] + 1 = [x + 1] + [x] + 1 = 1 + 1 = 0. \]

This result may be a little easier to absorb if we use a different notation. Let \( \alpha = [x] \). Then the calculation above says that \( \alpha^2 + \alpha + 1 = 0 \); that is, \( \alpha \) is a root in \( K \) of \( p(x) = x^2 + x + 1 \). It's important to note here that you don’t really need the tables for \( K \) to prove that \( \alpha \) is a root of \( p(x) \) because we know that \( x^2 + x + 1 \equiv 0 \ (\text{mod} \ x^2 + x + 1) \). Consequently, \( [x^2 + x + 1] = 0 \) in \( K \), and by the definition of congruence-class arithmetic:

\[ \alpha^2 + \alpha + 1 = [x^2] + [x] + 1 = [x^2 + x + 1] = 0. \]

---

Theorem 5.11

Let \( F \) be a field and \( p(x) \) an irreducible polynomial in \( F[x] \). Then \( F[x]/(p(x)) \) is an extension field of \( F \) that contains a root of \( p(x) \).

Proof

Let \( K = F[x]/(p(x)) \). Then \( K \) is an extension field of \( F \) by Theorems 5.8 and 5.10. Let \( p(x) = a_nx^n + \dots + a_1x + a_0 \), where each \( a_i \) is in \( F \) and, hence, in \( K \). Let \( \alpha = [x] \) in \( K \). We shall show that \( \alpha \) is a root of \( p(x) \). By the definition of congruence-class arithmetic in \( K \):

\[ a_n\alpha^n + \dots + a_1\alpha + a_0 = a_n[x]^n + \dots + a_1[x] + a_0 \] \[ = [a_nx^n + \dots + a_1x + a_0] \] \[ = [p(x)] = 0_F \ (\text{Because } p(x) = 0_F \ (\text{mod} \ p(x))). \]

Therefore, \( \alpha \) is a root of \( p(x) \). \( \blacksquare \)

---

Corollary 5.12

Let \( F \) be a field and \( f(x) \) a nonconstant polynomial in \( F[x] \). Then there is an extension field \( K \) of \( F \) that contains a root of \( f(x) \).

Proof

By Theorem 4.14, \( f(x) \) has an irreducible factor \( p(x) \) in \( F[x] \). By Theorem 5.11, \( K = F[x]/(p(x)) \) is an extension field of \( F \) that contains a root of \( p(x) \). Since every root of \( p(x) \) is a root of \( f(x) \), \( K \) contains a root of \( f(x) \). \( \blacksquare \)

---

The implications of Theorem 5.11 run much deeper than might first appear. Throughout the history of mathematics, the passage from a known number system to a new, larger system has often been greeted with doubt and distrust. In the Middle Ages, some mathematicians refused to acknowledge the existence of negative numbers. When complex numbers were introduced in the seventeenth century, this skepticism persisted for nearly a century—because some mathematicians would not accept the idea that there could be a number whose square is \( -1 \), that is, a root of \( x^2 + 1 \). One cause for these difficulties was the lack of a suitable framework in which to view the situation. Abstract algebra provides such a framework. Theorem 5.11 and its corollary, then, take care of the doubt and uncertainty.

It is instructive to consider the complex numbers from this point of view. Instead of asking about a number whose square is \( -1 \), we ask, "Is there a field containing \( \mathbb{R} \) in which the polynomial \( x^2 + 1 \) has a root?" Since \( x^2 + 1 \) is irreducible in \( \mathbb{R}[x] \), Theorem 5.11 tells us that the answer is yes: \( K = \mathbb{R}[x]/(x^2 + 1) \) is an extension field of \( \mathbb{R} \) that contains a root of \( x^2 + 1 \), namely \( \alpha = [x] \). In the field \( K \), \( \alpha \) is an element whose square is \( -1 \). But how is the field \( K \) related to the field of complex numbers introduced earlier in the book?

As is noted in Example 5 of Section 5.1, every element of \( K = \mathbb{R}[x]/(x^2 + 1) \) can be written uniquely in the form \( [ax + b] \) with \( a, b \in \mathbb{R} \). Since we are identifying each element \( r \in \mathbb{R} \) with the element \( [r] \) in \( K \), we see that every element of \( K \) can be written uniquely in the form:

\[ [a + bx] = [a] + [b][x] = a + b\alpha. \]

Addition in \( K \) is given by the rule:

\[ (a + b\alpha) + (c + d\alpha) = [a + bx] + [c + dx] = [(a + bx) + (c + dx)] \] \[ = [(a + c) + (b + d)x] = [a + c] + [b + d][x] \]

so that:

\[ (a + b\alpha)(c + d\alpha) = (a + c) + (b + d)\alpha. \]

Multiplication in \( K \) is given by the rule:

\[ (a + b\alpha)(c + d\alpha) = [a + bx][c + dx] = [(a + bx)(c + dx)] \] \[ = [ac + (ad + bc)x + bdx^2] \] \[ = ac + (ad + bc)\alpha + bd\alpha^2. \]

However, \( \alpha \) is a root of \( x^2 + 1 \), and so \( \alpha^2 = -1 \). Therefore, the rule for multiplication in \( K \) becomes:

\[ (a + b\alpha)(c + d\alpha) = (ac - bd) + (ad + bc)\alpha. \]

If the symbol \( \alpha \) is replaced by the symbol \( i \), then these rules become the usual rules for adding and multiplying complex numbers. In formal language, the field \( K \) is isomorphic to the field \( \mathbb{C} \), with the isomorphism \( f \) being given by \( f(a + b\alpha) = a + bi \).

Up to now we have taken the position that the field \( \mathbb{C} \) of complex numbers was already known. The field \( K \) constructed above then turns out to be isomorphic to the known field \( \mathbb{C} \). A good case can be made, however, for not assuming any previous knowledge of the complex numbers and using the preceding example as a definition instead. In other words, we can define \( \mathbb{C} \) to be the field \( \mathbb{R}[x]/(x^2 + 1) \). Such a definition is obviously too sophisticated to use on high-school students, but for mature students it has the definite advantage of removing any lingering doubts about the legitimacy of the complex numbers and their arithmetic. Had this definition been available several centuries ago, the introduction of the complex numbers might have caused no stir whatsoever.