## FIELDS

Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed., © 1996

#### Chapter 6

6.1 Algebraic elements
6.2 Finite and algebraic extensions
6.3 Geometric constructions
6.4 Splitting fields
6.5 Finite fields
6.6 Irreducible polynomials over finite fields

## Algebraic elements

6.1.1. Definition. The field F is said to be an extension field of the field K if K is a subset of F which is a field under the operations of F.

6.1.2. Definition. Let F be an extension field of K and let u F. If there exists a nonzero polynomial f(x) K[x] such that f(u)=0, then u is said to be algebraic over K. If there does not exist such a polynomial, then u is said to be transcendental over K.

6.1.3. Proposition. Let F be an extension field of K, and let u F be algebraic over K. Then there exists a unique monic irreducible polynomial p(x) K[x] such that p(u)=0. It is characterized as the monic polynomial of minimal degree that has u as a root.
Furthermore, if f(x) is any polynomial in K[x] with f(u)=0, then p(x) | f(x).

6.1.4. Definition. Let F be an extension field of K, and let u be an algebraic element of F. The monic polynomial p(x) of minimal degree in K[x] such that p(u)=0 is called the minimal polynomial of u over K. The degree of the minimal polynomial of u over K is called the degree of u over K.

6.1.5. Definition. Let F be an extension field of K, and let u1, u2, . . . , un F. The smallest subfield of F that contains K and u1, u2, . . . , un will be denoted by

K ( u1, u2, . . . , un ).

It is called the extension field of K generated by u1, u2, . . . , un .
If F = K(u) for a single element uF, then F is said to be a simple extension of K.

6.1.6. Proposition. Let F be an extension field of K, and let u F.

(a) If u is algebraic over K, then K(u) K[x]/<p(x)>, where p(x) is the minimal polynomial of u over K.

(b) If u is transcendental over K, then K(u) K(x), where K(x) is the quotient field of the integral domain K[x].
The next proposition is simply a restatement of Kronecker's theorem.

6.1.7. Proposition. Let K be a field and let p(x) K[x] be any irreducible polynomial. Then there exists an extension field F of K and an element u F such that the minimal polynomial of u over K is p(x).

## Finite and algebraic extensions

6.2.1. Proposition. Let F be an extension field of K and let u F be an element algebraic over K. If the minimal polynomial of u over K has degree n, then K(u) is an n-dimensional vector space over K.

6.2.2. Definition. Let F be an extension field of K. If the dimension of F as a vector space over K is finite, then F is said to be a finite extension of K.
The dimension of F as a vector space over K is called the degree of F over K, and is denoted by [F:K].

6.2.3. Proposition. Let F be an extension field of K and let u F. The following conditions are equivalent:

(1) u is algebraic over K;

(2) K(u) is a finite extension of K;

(3) u belongs to a finite extension of K.
6.2.4. Theorem. Let E be a finite extension of K and let F be a finite extension of E. Then F is a finite extension of K, and

[F:K] = [F:E][E:K].

6.2.5. Corollary. Let F be a finite extension of K. Then the degree of any element of F is a divisor of [F:K].

6.2.6 Corollary. Let F be an extension field of K, with algebraic elements u1, u2, . . . , un F. Then the degree of

K ( u1, u2, . . . , un )

over K is at most the product of the degrees of ui over K, for 1 i n.

6.2.7. Corollary. Let F be an extension field of K. The set of all elements of F that are algebraic over K forms a subfield of F.

6.2.8. Definition. An extension field F of K is said to be algebraic over K if each element of F is algebraic over K.

6.2.9. Proposition. Every finite extension is an algebraic extension.

Example. 6.2.3. (Algebraic numbers) Let Q* be the set of all complex numbers u C such that u is algebraic over Q. Then Q* is a subfield of C by Corollary 6.2.7, called the field of algebraic numbers.

6.2.10. Proposition. Let F be an algebraic extension of E and let E be an algebraic extension of K. Then F is an algebraic extension of K.

## Geometric constructions

6.3.1. Definition. The real number a is said to be a constructible number if it is possible to construct a line segment of length |a| by using only a straightedge and compass.

6.3.2. Proposition. The set of all constructible real numbers is a subfield of the field of all real numbers.

6.3.3. Definition. Let F be a subfield of R. The set of all points (x,y) in the Euclidean plane R2 such that x,y F is called the plane of F.
A straight line with an equation of the form ax+by+c = 0, for elements a,b,c F, is called a line in F.
Any circle with an equation of the form x2 + y2 + ax + by + c = 0, for elements a,b,c F, is called a circle in F.

6.3.4. Lemma. Let F be a subfield of R.

(a) Any straight line joining two points in the plane of F is a line in F.

(b) Any circle with its radius in F and its center in the plane of F is a circle in F.
6.3.5. Lemma. The points of intersection of lines in F and circles in F lie in the plane of F(u), for some u F.

6.3.6. Theorem. The real number u is constructible if and only if there exists a finite set u1, u2, . . . , un of real numbers such that

(i) u12 Q,

(ii) ui2 Q(u1,...,ui-1), for i=2,..., and

(iii) u Q(u1,...,un).
6.3.7. Corollary. If u is a constructible real number, then u is algebraic over Q, and the degree of its minimal polynomial over Q is a power of 2.

6.3.9. Theorem. It is impossible to find a general construction for trisecting an angle, duplicating a cube, or squaring a circle.

## Splitting fields

6.4.1. Definition. Let K be a field and let f(x) = a0 + a1 x + · · · + anxn be a polynomial in K[x] of degree n>0. An extension field F of K is called a splitting field for f(x) over K if there exist elements r1, r2, . . . , rn F such that
(i) f(x) = an (x-r1) (x-r2) · · · (x-rn), and

(ii) F = K(r1,r2,...,rn).
In the above situation we usually say that f(x) splits over the field F. The elements r1, r2, . . . , rn are roots of f(x), and so F is obtained by adjoining to K a complete set of roots of f(x).

6.4.2. Theorem. Let f(x) K[x] be a polynomial of degree n>0. Then there exists a splitting field F for f(x) over K, with [F:K] n!.

6.4.3. Lemma. Let : K -> L be an isomorphism of fields. Let F be an extension field of K such that F = K(u) for an algebraic element u F. Let p(x) be the minimal polynomial of u over K. If v is any root of the image q(x) of p(x) under , and E=L(v), then there is a unique way to extend to an isomorphism : F -> E such that (u) = v and (a) = (a) for all a K.

6.4.5. Theorem. Let f(x) be a polynomial over the field K. The splitting field of f(x) over K is unique up to isomorphism.

## Finite fields

6.5.1. Proposition. Let F be a finite field of characteristic p. Then F has pn elements, for some positive integer n.

If F is any field, then the smallest subfield of F that contains the identity element 1 is called the prime subfield of F. If F is a finite field, then its prime subfield is isomorphic to Zp, where p=chr(F) for some prime p.

6.5.2. Theorem. Let F be a finite field with k = pn elements. Then F is the splitting field of the polynomial xk-x over the prime subfield of F.

Example 6.5.1. [Wilson's theorem] Let p > 2 be a prime number. Then

(p-1)! -1 (mod p).

6.5.3. Corollary. Two finite fields are isomorphic if and only if they have the same number of elements.

6.5.4. Lemma. Let F be a field of prime characteristic p, let n Z+, and let k = pn. Then

{ a F | ak = a }

is a subfield of F.

6.5.5. Proposition. Let F be a field with pn elements. Each subfield of F has pm elements for some divisor m of n. Conversely, for each positive divisor m of n there exists a unique subfield of F with pm elements.

6.5.6. Lemma. Let F be a field of characteristic p. If n is a positive integer not divisible by p, then the polynomial xn-1 has no repeated roots in any extension field of F.

6.5.7. Theorem. For each prime p and each positive integer n, there exists a field with pn elements.

6.5.8. Definition. Let p be a prime number and let n Z+. The field (unique up to isomorphism) with pn elements is called the Galois field of order pn, denoted by GF(pn).

6.5.9. Lemma. Let G be a finite abelian group. If a G is an element of maximal order in G, then the order of every element of G is a divisor of the order of a.

6.5.10. Theorem. The multiplicative group of nonzero elements of a finite field is cyclic.

6.5.11. Theorem. Any finite field is a simple extension of its prime subfield.

6.5.12. Corollary. For each positive integer n there exists an irreducible polynomial of degree n over GF(p).

## Irreducible polynomials over finite fields

6.6.1. Theorem. Let F = GF(q), where q = pn, and let k = qm. The irreducible factors of xk-x in F[x] are precisely the monic irreducible polynomials in F[x] whose degree is a divisor of m.

Convention: In the notation d | n and d | n we will assume that d | n refers to the positive divisors of n.

6.6.2. Definition. If d is a positive integer, we define the Moebius function (d) as follows:

(1) = 1;

(d) = 1 if d has an even number of prime factors (each occurring only once);

(d) = -1 if d has an odd number of prime factors (each occurring only once);

(d) = 0 if d is divisible by the square of a prime.
6.6.3. Proposition. If m, n Z+ and gcd(m,n)=1, then

(mn) = (m) (n).

If R is a commutative ring, then a function f : Z+ -> R is said to be a multiplicative function if

f(mn) = f(m)f(n),

whenever gcd(m,n)=1.

6.6.4. Proposition. Let R be a commutative ring, and let f : Z+ -> R be a multiplicative function. If F : Z+ -> R is defined by

F(n) = d | n f(d),

for all n Z+, then F is a multiplicative function.

6.6.5. Proposition. For any positive integer n,

d | n (d) = 1 if n = 1, and

d | n (d) = 0 if n > 1.

6.6.6. Theorem. [Moebius Inversion Formula] Let R be a commutative ring, and let f : Z+ -> R be any function. If the function F : Z+ -> R is defined by

F(n) = d | n f(d), for all n Z+,

then

f(m) = n | m ( m/n ) F(n), for all m Z+.

6.6.7. Theorem. [Moebius Inversion Formula (2)] Let R be a commutative ring, and let g : Z+ -> R be any function. If the function G : Z+ -> R is defined by

G(n) = d | n g(d), for all n Z+,

then

g(m) = n | m G(n)k, for all m Z+,

where k = ( m/n ).

6.6.8. Definition. The number of irreducible polynomials of degree m over the finite field GF(q), where q is a prime power, will be denoted by Iq(m).

The following formula for Iq(m) is due to Gauss.

6.6.9. Theorem. For any prime power q and any positive integer m,

Iq(m) = (1/m) d | m ( m / d ) qd.

6.6.10. Corollary. For all positive integers m and all prime powers q we have Iq(m)1.

6.7.1. Definition. Let n be a positive integer, and let a be an integer such that na. Then a is called a quadratic residue modulo n if the congruence x2a(mod n) is solvable, and a quadratic nonresidue otherwise.
When n is a prime, we write =1 if a is a quadratic residue modulo n and =-1 if a is a quadratic nonresidue modulo n. The symbol is called the Legendre symbol.

6.7.2. Proposition. [Euler's Criterion] If p is an odd prime, and if a Z with p a, then

6.7.3. Theorem. [Quadratic Reciprocity] Let p, q be distinct odd primes. Then

6.7.4. Theorem. Let p be an odd prime. Then
(i) , and

(ii) , where k = (p2 - 1)/8.