FIELD EXTENSIONS

Proof. K is an abelian group under addition. Elements of K can be multiplied by elements of F . This multiplication satisfies the following properties:

(i)If 1 is the identity element of F then 1k = k for all k K.

(ii)If λ F and k, l K, then λ(k + l) = λk + λl.

(iii)If λ, µ F and k K, then (λ + µ)k = λk + µK.

(iv)If λ, µ F and k K, then (λµ)k = λ(µk).

about the structure of K. The elements of K can be written uniquely as a linear combination of certain elements called basis elements. Furthermore, if the vector space K has finite dimension n over the field F , there will be n basis elements, and the construction of K is particularly simple.

The degree of the extension K of the field F , written [K : F ], is the dimension of K as a vector space over F . The field K is called a finite extension if [K : F ] is finite.

Example 11.3. [C : R] = 2.

Solution. C = {a + ib|a, b R}; therefore, 1 and i span the vector space C over R. Now 1 and i are linearly independent since, if λ, µ R, then λ1 + µi = 0

Example 11.4 is a special case of the following theorem.

Theorem 11.5. If p(x) is an irreducible polynomial of degree n over the field F , and K = F [x]/(p(x)), then [K : F ] = n.

Proof. By Theorem 10.8, K = {a0 + a1x + · · · + an−1xn−1|ai F }, and such expressions for the elements of K are unique. Hence {1, x, x2, . . . , xn−1} is a basis for K over F , and [K : F ] = n.

Theorem 11.6. Let L be a finite extension of K and K a finite extension of F . Then L is a finite extension of F and [L : F ] = [L : K][K : F ].

Proof. We have three fields, F , K, L, with L K F . We prove the theorem by taking bases for L over K, and K over F , and constructing a basis for L over F .

220 11 FIELD EXTENSIONS

Let [L : K] = m and let {u1, . . . , um} be a basis for L over K. Let [K : F ] = n

Solution. The constant polynomials in L are identified with Q. Suppose that K is a field such that L K Q. Then [L : Q] = [L : K][K : Q], by Theorem 11.6. But, by Theorem 11.5, [L : Q] = 3, so [L : K] = 1, or [K : Q] = 1.

If [L : K] = 1, then L is a vector space over K, and {1}, being linearly independent, is a basis. Hence L = K. If [K : Q] = 1, then K = Q. Hence there is no field lying strictly between L and Q.

Given a field extension K of F and an element a K, define F (a) to be the intersection of all subfields of K that contain F and a. This is the smallest subfield of K containing F and a, and is called the field obtained by adjoining a to F .

For example, the smallest field containing R and i is the whole of the complex numbers, because this field must contain all elements of the form a + ib where a, b R. Hence R(i) = C.

In a similar way, the field obtained by adjoining a1, . . . , an K to F is denoted by F (a1, . . . , an) and is defined to be the smallest subfield of F containing a1, . . . , an and F . It follows that F (a1, . . . , an) = F (a1, . . . , an−1)(an).

which is the field of rational functions in R. Any field containing R and x must contain the polynomial ring R[x], and the smallest field containing R[x] is its field of fractions R(x).

ALGEBRAIC NUMBERS

If K is a field extension of F , the element k K is called algebraic over F if there exist a0, a1, . . . , an F , not all zero, such that

a0 + a1k + · · · + ankn = 0.

In other words, k is the root of a nonzero polynomial in F [x]. Elements that are not algebraic over F are called transcendental over F .

the field of rational functions in π with coefficients in Q. Q(π ) must contain all the powers of π and hence any polynomial in π with rational coefficients. Any nonzero element of Q(π ) must have its inverse in Q(π ); thus Q(π ) contains the set of rational functions in π . The number b0 + b1π + · · · + bmπ m is never zero unless b0 = b1 = · · · = bm = 0 because π is not the root of any polynomial with rational coefficients. This set of rational functions in π can be shown to be a subfield of R.

Those readers acquainted with the theory of infinite sets can prove that the set of rational polynomials, Q[x], is countable. Since each polynomial has only a finite number of roots in C, there are only a countable number of real or complex numbers algebraic over Q. Hence there must be an uncountable number of real and complex numbers transcendental over Q.

Example 11.10. Is cos(2π/5) algebraic or transcendental over Q?

Solution. We know from De Moivre’s theorem that

(cos 2π/5 + i sin 2π/5)5 = cos 2π + i sin 2π = 1.

Taking real parts and writing c = cos 2π/5 and s = sin 2π/5, we have

Theorem 11.11. Let α be algebraic over F and let p(x) be an irreducible polynomial of degree n over F with α as a root. Then

F (α) F x /(p(x)),

= [ ]

and the elements of F (α) can be written uniquely in the form

c0 + c1α + c2α2 + · · · + cn−1αn−1 where ci F.

Proof. Define the ring morphism f : F [x] → F (α) by f (q(x)) = q(α). The kernel of f is an ideal of F [x]. By Corollary 10.3, all ideals in F [x] are principal; thus Kerf = (r(x)) for some r(x) F [x]. Since p(α) = 0, p(x) Kerf , and so r(x)|p(x). Since p(x) is irreducible, p(x) = kr(x) for some nonzero element k of F . Therefore, Kerf = (r(x)) = (p(x)).

By the morphism theorem,

Now, by Theorem 10.17, F [x]/(p(x)) is a field; thus Imf is a subfield of F (α) that contains F and α. Since Imf cannot be a smaller field than F (α), it follows

The unique form for the elements of F (α) follows from the isomorphism above and Theorem 10.8.

Corollary 11.12. If α is a root of the polynomial p(x) of degree n, irreducible over F , then [F (α): F ] = n.

Proof. By Theorems 11.11 and 11.5, [F (α) : F ] = [F [x]/(p(x)) : F ] = n.

is irreducible over Q, by Eisenstein’s criterion (Theorem 9.30).

Lemma 11.13. Let p(x) be an irreducible polynomial over the field F . Then F has a finite extension field K in which p(x) has a root.

Proof. Let p(x) = a0 + a1x + a2x2 + · · · + anxn and denote the ideal (p(x)) by P . By Theorem 11.5, K = F [x]/P is a field extension of F of degree n whose elements are cosets of the form P + f (x). The element P + x K is a root of p(x) because

a0 + a1(P + x) + a2(P + x)2 + · · · + an(P + x)n

=a0 + (P + a1x) + (P + a2x2) + · · · + (P + anxn)

=P + (a0 + a1x + a2x2 + · · · + anxn) = P + p(x)

=P + 0,

and this is the zero element of the field K.

Theorem 11.14. If f (x) is any polynomial over the field F , there is an extension field K of F over which f (x) splits into linear factors.

Proof. We prove this by induction on the degree of f (x). If deg f (x) 1, there is nothing to prove.

Suppose that the result is true for polynomials of degree n − 1. If f (x) has degree n, we can factor f (x) as p(x)q(x), where p(x) is irreducible over F . By Lemma 11.13, F has a finite extension K in which p(x) has a root, say α.

By the induction hypothesis, the field K has a finite extension, K, over which g(x) splits into linear factors. Hence f (x) also splits into linear factors over K and, by Theorem 11.6, K is a finite extension of F .

Let us now look at the development of the complex numbers from the real numbers. The reason for constructing the complex numbers is that certain equations, such as x2 + 1 = 0, have no solution in R. Since x2 + 1 is a quadratic polynomial in R[x] without roots, it is irreducible over R. In the above manner, we can extend the real field to

R[x]/(x2 + 1) = {a + bx|a, b R}.

to be R(i) and, by Theorem 11.11, there is an isomorphism

ψ : R[x]/(x2 + 1) → R(i)