A NOTE ON FINITE FIELDS

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1 A NOTE ON FINITE FIELDS FATEMEH Y. MOKARI The main goal of this note is to study finite fields and their Galois groups. Since I define finite fields as subfields of algebraic closure of prime fields of positive characteristic, I will study the algebraic closure of fields in details and give a proof of their existence and uniqueness. Moreover, as we will see, finite extensions of finite fields are Galois extension. Thus we also study Galois groups of finite extensions and Galois extensions in some details. For more details on these topics I refer the reader to [1], [2] and [3]. I tried this text to be self contained, so I prove almost all my claims. 1. Fields and their characteristic A field is a commutative ring with 1, 1 0, such that its nonzero element have inverse with respect to multiplication of the ring structure. Thus if F is a field, F := F {0} with the operation of multiplication is an abelian group. Note that the only ideals of a filed F are (0) and F. In fact if I is a nonzero ideal of F and if 0 a I, then 1 = a 1 a I. Thus I = F. Example 1.1. (i) The most well known examples of fields are Q, R and C. (ii) Other well known examples are the finite fields F p := Z/pZ, where p is a prime number. (iii) Let p be a prime such that the equation X 2 = 1 does not have any solution in F p (e.g. p = 3 or p = 7). Consider the additive group C p := F p F p = F 2 p. Let 1 := (1, 0) and i := (0, 1). Then any element of C p uniquely can be written as a1 + bi, a, b F p. We define the following operation of multiplication on C p : (a1 + bi).(c1 + di) := (ac bd)1 + (ad + bc)i. It is easy to see that C p with this operation is a field, with 1 Cp = 1. In fact if a1 + bi 0, then (a1 + bi) 1 = a a 2 + b 1 b 2 a 2 + b i. 2 1

2 2 FATEMEH Y. MOKARI So C p is a field with p 2 elements and its construction is very similar to C. Later we will denote this field with F p 2. (iv) Any finite domain R is necessarily a finite field. In fact if a R, then the set {a n : n 0} is finite and so a n = a m for some n m. Thus if n > m, then a n m = 1. This means that a has an inverse a 1 := a n m 1 R. (v) If R is a domain, its field of factions Q(R) is the field Q(R) := {a/s : a, s R, s 0}. In Q(R), a/r = b/s if and only if as = br. The operations of addition and multiplication of Q(R) is defined as usual: a/r + b/s := (as + br)/rs, a/r.b/s := ab/rs. It is clear that if a, r 0, then (a/r) 1 = r/a. We have the natural embedding R Q(R), a a/1. It is easy to see that Q(Z) = Q. (vi) If R is a commutative ring and m is an ideal of R, then R/m is a field if and only if m is a maximal ideal of R. (vii) If F is a filed and F [X] is the ring of polynomials with one variable over F, then for any irreducible polynomial f(x) F [X], the ideal f(x) of F [X] is maximal and thus E := F [X]/ f(x) is a field. This construction is very important for us and we will use it frequently in this note. The ring of integers Z is not a filed but, as we will see now, has very basic connection to fields. Let F be a filed and consider the natural ring homomorphism ϕ : Z F, n n.1. Then ker(ϕ) is an ideal of Z and so is of the form ker(ϕ) = pz, for some nonnegative p Z. If p = 0, i.e. ker(ϕ) = 0, then Z embeds in F and so we can also embed Q in F by the homomorphism ϕ : Q F, a/b ϕ(a)ϕ(b) 1. If p 0, then Z/pZ embeds in F. So Z/pZ is a domain and this is possible only if p is a prime number. Thus we have an embedding of F p in F, ϕ : F p F, m m.1. We say that a field F has characteristic p, and write char(f ) = p, if for any a F, p.a = a } + {{ + a } = 0. p times

3 A NOTE ON FINITE FIELDS 3 The above argument shows that char(f ) either is zero or is a prime number. Thus we have proved the following proposition. Proposition 1.2. Let F be a field. Then the characteristic of F is unique and either char(f ) = 0, which in this case we have an embedding of Q in F, or char(f ) = p is a prime which in this case we have an embedding of F p in F. The finite subgroups of the multiplicative group F are important so we study them here. Lemma 1.3. Let F be a field and let G be a finite subgroup of F. Then G is a cyclic group and G Z/ G Z. In particular if F is a finite field, then F is a cyclic group of order F 1 and thus F Z/( F 1)Z. Proof. Let n be the exponent of G, i.e. the smallest number such that for any g G, g n = 1. Then by the Lagrange Theorem n divides G [1, Corollary 6.11, Chap. 2]. Since g n 1 = 0, the elements of G are roots of the polynomial X n 1. But this polynomial has at most n roots in F, thus G n. Therefore n = G and this means that G is cyclic and hence G Z/ G Z. 2. Field extension We say that F is a subfield of a field E, if F is a filed and is a subring of E. In this case we usually say that E is a field extension of F. Proposition 1.2 shows that any field F contained one of the fields Q or F p, p prime, as subfield. It is also clear from the proposition that Q and F p do not have any nontrivial subfield and in fact they are the only fields with this property. We call them prime fields. Let F be a subfield of a filed E. Then E can be considered as a vector space over F with the scalar product f.a := fa, f F, a E. The dimension of E as F -vector space is called the degree of E over F and we denote it by [E : F ], i.e. [E : F ] := dim F E. If [E : F ] is finite, then we say that E is a finite extension of F and otherwise we say that it is an infinite extension of F. Example 2.1. (i) It is easy to see that [C : R] = 2. In fact {1, i = 1} is a basis of C as vector space over R. (ii) It is well known that π = is transcendental over Q, i.e. it is not the root of any f(x) Q[X] [4, Theorem 4, Chap. 2].

4 4 FATEMEH Y. MOKARI Thus 1, π, π 2,... are linearly independent over Q and this implies that [R : Q] =. (iii) Let Q( 2) := {a+b 2 : a, b Q}. Then Q( 2) is an extension of Q of degree 2. In fact {1, 2} is a basis for this extension. Lemma 2.2. Let E be a field extension of F and K be a field extension of E. Then [K : F ] = [K : E][E : F ]. Proof. In fact if {α i : i I} E is a basis of E over F and {β j : j J} K is a basis of K over E, then {α i β j : i I, j J} is a basis of K over F. There is a common and very important way of constructing finite extensions of a field F. Consider the polynomial ring F [X] with one variable over F and let f(x) F [X] be an irreducible polynomial. Then the ideal f(x) of F [X], generated by f(x), is maximal and so E := F [X]/ f(x) is a filed and we have a natural embedding of F in E, F E = F [X]/ f(x), a a := a + f(x). For simplicity we denote the image of a F in E, i.e. a, again by a. If f(x) is of degree n, i.e. deg(f(x)) = n, then {1, α, α 2,..., α n 1 } is a basis of E as a vector space over F, where α := X = X + f(x). Thus [E : F ] = n. Let E be a field extension of F and let α E. We say that α is algebraic over F if there is a monic polynomial f(x) F [X] such that f(α) = 0, otherwise we say that α is transcendence over F. If among all polynomials g(x) F [X] such that g(α) = 0, f(x) has the minimal degree we say that f(x) is the minimal polynomial of α. Then f(x) is unique and is irreducible. In fact if g(x) F [X] such that g(α) = 0, then f(x) g(x). So if f 1 (X) and f 2 (X) are two minimal polynomials of α, then f 1 (X) f 2 (X) and f 2 (X) f 1 (X). Thus f 1 (X) = f 2 (X). If f(x) is the minimal polynomial of α E, then we have the isomorphism F [X]/ f(x) F [α] := {g(α) : g(x) F [X]}, X α, where F [α] is the smallest subring of E which contains F and α. Thus F [α] is a subfield of E. This implies that F [α] = F (α) := {g(α)h(α) 1 : g(x), h(x) F [X], h(α) 0},

5 A NOTE ON FINITE FIELDS 5 where F (α) is the smallest subfield of E which contains both F and α. Example 2.3. Let X 3 2 Q[X]. Then clearly X 3 2 is irreducible and 3 2 R is its root. Hence we have the following isomorphism of fields: Q[X]/ X 3 2 Q[ 3 2] = Q( 3 2) = {r + s t 3 4 : r, s, t Q}, and [Q( 3 2) : Q] = 3. It worth mentioning that other roots of X 3 2 are ω 3 2 and ω 2 3 2, where ω = e 2πi/3 C, they which do not belong to Q( 3 2) and even they do not belong to R. Example 2.4. The polynomial X 2 + X + 1 F 2 [X] does not have any root in F 2 = {0, 1}, so it is irreducible. Thus the field F := F 2 [X]/ X 2 + X + 1 contains F 2 and [F : F 2 ] = 2. (Note that X 2 + X + 1 is the only irreducible polynomial of degree 2 over F 2.) If α = X + X 2 + X + 1 F, then F = {0, 1, α, α + 1 : α 2 = α + 1}. We denote this field with F 4, since it has four elements. Let F be any other field with four elements. Then by Lemma 1.3, F Z/3Z. Let α be the generator of F as multiplicative group. Then 0, 1, α, α 2 are four different elements of F and so F = {0, 1, α, α 2 }. Since α +1 F can not be equal to any of 0, 1, α, we must have α 2 = α + 1. Now clearly the map F F, 0 0, 1 1, α α, α + 1 α + 1, is an isomorphism of fields. Thus, up to isomorphism, we just have one finite field with four elements. Example 2.5. Since X 2 + X + α F 4 [X] does not have any root in F 4, it is irreducible. Thus E := F 4 [X]/ X 2 + X + α is a field, contains F 4 and [E : F 4 ] = 2. This implies that E has 16 elements and if we put β = X + X 2 + X + α E, then E = { 0, 1, α, α + 1, β, β + 1, αβ, α + β, αβ + 1, α + β + 1, αβ + α, αβ + β, αβ + α + 1, αβ + β + 1, αβ + α + β, αβ + α + β + 1 : α 2 = α + 1, β 2 = α + β }. On the other hands it is not difficult to see that X 4 + X + 1 F 2 [X] is irreducible so [K : F 2 ] = 4, where K is the field K := F 2 [X]/ X 4 + X + 1.

6 6 FATEMEH Y. MOKARI In fact since X 4 +X +1 does not have any in root in F 2, it does not have any linear factor. Moreover it also is not multiplication of two degree two irreducible polynomials, because the only degree two irreducible polynomial in F 2 [X] is X 2 + X + 1 and (X 2 + X + 1) 2 = X 4 + X X 4 + X + 1. Thus X 4 + X + 1 is irreducible. Since x[k : F 2 ] = 4, K has 16 elements. If we put γ = X + X 3 + X + 1, then K = {0, 1, γ, γ + 1, γ 2, γ 2 + 1, γ 2 + γ, γ 2 + γ + 1, γ 3, γ 3 + 1, γ 3 + γ, γ 3 + γ 2, γ 3 + γ + 1, γ 3 + γ 2 + 1, γ 3 + γ 2 + γ, γ 3 + γ 2 + γ + 1 : γ 4 = γ + 1}. We prove that E and K are isomorphic. A direct computation shows that γ 2 + γ K is a root of X 2 + X + 1 F 2 [X]. So L = {0, 1, γ 2 + γ, γ 2 + γ + 1} is a subfield of K and is isomorphism to F 4, which we correspond α to γ 2 + γ. Moreover γ E is the root of the irreducible polynomial X 2 + X + γ 2 + γ L[X], thus we can correspond β to γ. Thus we can construct an isomorphism E K as follow: E K, α γ 2 + γ, β γ. We denote both E and K by F 16. Later we will prove that any field with 16 elements is isomorphic to F 16. With a method similar to the previous example, and some computation, one can show that any field with 16 elements is isomorphic to F 16. We will not prove this here. Later we will prove a more general result. In fact in Section 6, we will show that any two finite fields with the same number of elements are isomorphic. 3. Galois groups Let E be a field extension of F. An F -automorphism σ of E is a ring isomorphism σ : E E such that σ F = id F, i.e. for any a F, σ(a) = a. We denote the group of all F -automorphism of E by Gal(E/F ) and call it the Galois group of E over F. It is reasonable to expect that if [E : F ] <, then Gal(E/F ) is a finite group: To prove this, first note that since [E : F ] <, there are α 1,..., α n E, such that E = F [α 1 ][α 2,..., α n ] = F [α 1,..., α n ] = {f(α 1, α 2,..., α n ) : f(x 1,..., X n ) F [X 1,..., X n ]}.

7 A NOTE ON FINITE FIELDS 7 Also note that if σ, τ Gal(E/F ), such that σ(α i ) = τ(α i ), 1 i n, then σ = τ: σ(f(α 1,..., α n )) = f(σ(α 1 ),..., σ(α n )) = f(τ(α 1 ),..., τ(α n )) = τ(f(α 1,..., α n )). Let f i (X) F [X] be the minimal polynomial of α i and let σ Gal(E/F ). Then f i (σ(α i )) = σ(f i (α i )) = 0, and so σ(α i ) is also a root of f i (X). This implies that for each i any element of the set {σ(α i ) : σ Gal(E/F )} is a root of f i (X) and so it must be a finite set. This shows that, for each i, there are only finitely many possibilities for the image of α i, under the elements of Gal(E/F ) and so there must be only finitely many F -automorphism of E. Therefore Gal(E/F ) is finite. Example 3.1. Consider the finite field F 4 = {0, 1, α, α + 1 : α 2 = α + 1} from Example 2.4, which is an extension of F 2 and [F 4 : F 2 ] = 2. It is easy to see that the identity homomorphism id F4 : F 4 F 4 and the homomorphism σ : F 4 F 4 defined by r +sα r +s(α+1), r, s F 2, are F 2 -automorphisms. Since 2 Gal(F 4 /F 2 ) [F 4 : F 2 ] = 2, we have Gal(F 4 /F 2 ) = {id F4, σ}. Example 3.2. Consider the field Q( 3 2) as an extension of Q. Let σ Gal(Q( 3 2)/Q). Since X 3 2 Q[X] is the minimal polynomial of 3 2, σ( 3 2) Q( 3 2) is also a root of X 3 2. But as we have seen in Example 2.3, other roots of this polynomial are not in Q( 3 2), thus σ = id Q( 3 2). This implies that Gal(Q( 3 2)/Q) = {id Q( 3 2) } is the trivial group. Therefore 1 = Gal(Q( 3 2)/Q) < [Gal(Q( 3 2) : Q] = 3. In the following we give a general result that compares the Galois group Gal(E/F ) and the degree [E : F ] of the finite extension E over F. But first we have to prove the following lemma. Lemma 3.3 (Dedekind s Lemma). Let G be a group and let F be a field. Let ϕ i : G F, 1 i n, be distinct group homomorphisms. Then the set {ϕ i : 1 i n} is linearly independent over F, i.e. if for any g G, n i=1 a iϕ i (g) = 0, where a i F, then a i = 0 for all i. Proof. Assume that the lemma is false. Let k be the minimum number of ϕ i s that are linearly dependent. By rearranging the ϕ i, we may assume that {ϕ 1,..., ϕ k } are linearly dependent. Thus there are a i

8 8 FATEMEH Y. MOKARI F, all nonzero, such that for any g G, k i=1 a iϕ i (g) = 0. Let h G such that ϕ 1 (h) ϕ 2 (h). For any g G, we have k k a i ϕ 1 (h)ϕ i (g) = ϕ 1 (h) a i ϕ i (g) = 0 and i=1 k a i ϕ i (gh) = i=1 i=1 k (a i ϕ i (h))ϕ i (g) = 0. Now subtracting these two formulas, for any g G, we have k (a i (ϕ 1 (h) ϕ i (h)))ϕ i (g) = 0. i=2 But this contacts minimality of k. So the lemma is true. i=1 Theorem 3.4. Let E be a finite extension of F. Then Gal(E/F ) is a finite group and Gal(E/F ) [E : F ]. Proof. We already have seen that Gal(E/F ) is finite. Let Gal(E/F ) = {σ 1,..., σ n } and let m := [E : F ] < n. Let α 1,..., α m be a basis of E as vector space over F. Then the matrix A := σ 1 (α 1 ) σ 1 (α 2 )... σ 1 (α m ) σ 2 (α 1 ) σ 2 (α 2 )... σ 2 (α m ) σ n (α 1 ) σ n (α 2 )... σ n (α m ) has rank(a) m < n. So the of rows A are linearly dependent over E. Thus there are a i, not all zero, such that for any 1 j m, k i=1 a iσ i (α j ) = 0. If G = E, then for g G, there are c i F such that g = m j=1 c jα j. Thus n a i σ i (g) = i=1 = = n ( m ) a i σ i c j α j i=1 n i=1 j=1 j=1 m a i c j σ i (α j ) j=1 i=1 m ( n ) c j a i σ i (α j ) = 0. Now the Dedekind s lemma implies that all a i s are zero. contradiction. Thus Gal(E/F ) [E : F ]. This is a

9 A NOTE ON FINITE FIELDS 9 4. Galois extensions A finite extension of E over F is called a Galois extension if Gal(E/F ) = [E : F ]. Example 4.1. (i) The field of complex numbers C is a Galois extension of R with the Galois group Gal(C/R) = {id C, σ}, where σ is the conjugation map, i.e. σ(a + ib) = a ib. (ii) Consider the field E = Q(ω, 3 2), which is the smallest subfield of C containing Q, ω := e 2πi/3 and 3 2. Let F = Q(ω). Since ω is a root of the irreducible polynomial X 2 + X + 1 Q[X], we have F = Q[ω] Q[X]/ X 2 + X + 1. Thus [F : Q] = 2. On the other hand X 3 2 F [X] is irreducible, so E = F ( 3 2) = F [ 3 2] F [X]/ X 3 2, which from it we have [E : F ] = 3. Therefore [E : Q] = [E : F ][F : Q] = 3.2 = 6. Note that 3 2, ω 3 2 and ω are all roots of X 3 2. Let σ and τ be the following elements of Gal(E/Q): σ : E E, ω ω 2, , τ : E E, ω ω, 3 2 ω 3 2. Then it is easy to see that id E, σ, τ, στ, στ 2, τσ are distinct elements of Gal(E/Q). Since [E : Q] = 6, thus by Theorem 3.4, these elements are all elements of Gal(E/Q) and thus Gal(E/Q) = 6 = [E : Q]. Hence E is a Galois extension of Q. Note that Gal(E/Q) is a nonabelian group of order 6. Since, up to isomorphism, the only nonabelian group of order 6 is S 6, we have Gal(E/Q) S 6, where for any n, S n is the symmetric group of order n. (iii) From Example 3.1, we see that Gal(F 4 /F 2 ) = 2 = [F 4 : F 2 ]. Thus F 4 is a Galois extension of F 2. Let E be a finite extension of F and set F Gal(E/F ) := {a E : σ(a) = a for any σ Gal(E/F )}. It is easy to see that F Gal(E/F ) is a subfield of E and it contains F.

10 10 FATEMEH Y. MOKARI Theorem 4.2. Let E be a finite extension of F. Then E is a Galois extension of F if and only if F Gal(E/F ) = F. Proof. Let L := F Gal(E/F ). First assume that E is a Galois extension of F. By definition it is clear that Gal(E/L) = Gal(E/F ). Now by Theorem 3.4, Gal(E/L) [E : L] [E : F ] = Gal(E/F ) = Gal(E/L). So [E : L] = [E : F ], which implies that F = L. To prove the converse, let F = L and let Gal(E/F ) = n. To prove the claim, it is sufficient to show that any n + 1 elements of E are linearly dependent, because then [E : F ] = dim F E n = Gal(E/F ) and this combined with Theorem 3.4 gives the result. Let Gal(E/F ) = {σ 1 = id E, σ 2,..., σ n } and let {x 1,..., x n+1 } be a subset of E. Consider the system of equations σ 1 (x 1 )z 1 + σ 1 (x 2 )z σ 1 (x n+1 )z n+1 = 0 σ 2 (x 1 )z 1 + σ 2 (x 2 )z σ 2 (x n+1 )z n+1 = 0. σ n (x 1 )z 1 + σ n (x 2 )z σ n (x n+1 )z n+1 = 0. Clearly this system of equations has a nontrivial solution (z 1,..., z n+1 ) in E n+1. We show that this system has a solution in F n+1. Let (z 1,..., z n+1 ) be a solution which among all the solutions of the system has the minimal number of nonzero coefficients. Let r be this minimal number. By rearranging the unknowns of the system and also the coefficients of the solution, we may assume that z 1, z 2,..., z r are nonzero and the remaining z i s are zero. By multiplying the system with zr 1, we may assume that z r = 1. Thus (z 1,..., z r 1, 1, 0,..., 0) E n+1 is our chosen solution of the system. Take σ Gal(E/F ) and apply it to the above system. Since σgal(e/f ) = {σ σ 1, σ σ 2,..., σ σ n } = Gal(E/F ), we see that (σ(z 1 ), σ(z 2 )..., σ(z r 1 ), 1, 0,..., 0) is also a solution of the system. Thus (z 1 σ(z 1 ), z 2 σ(z 2 ),..., z r 1 σ(z r 1 ), 1 1, 0,..., 0) also is a solution of our system. But this contradicts the minimality of r, unless z i σ(z i ) = 0 for all i. Hence σ(z i ) = z i for any 1 i n + 1. However this is true for any σ Gal(E/F ), thus the chosen z i s belong to L = F...

11 A NOTE ON FINITE FIELDS 11 Thus we have found a solution (z 1,..., z n+1 ) F n+1 of the above system and if we put these in the first equation of the system we have x 1 z 1 + x 2 z x n+1 z n+1 = 0. Thus the set {x 1,..., x n+1 } E is linearly dependent. This completes the proof of the theorem. In Section 6 we will show that any finite extension if finite fields is Galois. 5. Algebraically closed fields We say that a field E is an algebraic extension of a filed F if F E and any element of E is algebraic over F. It is easy to see that any finite extension is an algebraic extension. In fact if E is a finite extension of F of degree [E : F ] = n, then for any α E the set {1, α, α 2..., α n } is linearly dependent. Thus there are a i F, 0 i n, such that n i=0 a iα i = 0. So if f(x) = n i=0 a ix i F [X], then f(α) = 0. A field K is called algebraically closed if any non-constant polynomial f(x) E[X] has a root in E. Thus any polynomial over such field decomposes to linear factors, i.e. for f(x) E[X] there exist a, α 1,... α n E such that f(x) = a(x α 1 )(X α 2 ) (X α n ), where n = deg(f(x)). Algebraically closed fields do not have any non-trivial algebraic extension. This follows from the fact that the only irreducible polynomials over such fields are the linear polynomials. In fact if L is algebraic over F and if α L, then the minimal polynomial f(x) K[X] of α is linear. So if f(x) = X a, then α = a K. Example 5.1. (i) By the Fundamental Theorem of Algebra [1, Theorem 9.1, Chap. 13] we know that C is algebraically closed. (ii) Let Q be the set of all algebraic elements of C over Q. Then Q is a field, is algebraic over Q and is algebraically closed. In fact if α, β Q, then Q[α] is finite over Q and Q[α, β] = Q[α][β] is finite over Q[α]. Thus Q[α, β] is finite, and so algebraic, over Q. Since α + β, αβ, α 1 Q[α, β], α 0, they are algebraic over Q and hence they belong to Q. This shows that Q is a field and, by definition, is algebraic over Q. Now let f(x) = X n +a n 1 X n 1 + +a 0 Q[X] be non-constant and let β 1,..., β n be its roots in C. Let F := Q[a 0,..., a n 1 ] Q. Since each a i is algebraic over Q, F is a finite extension of Q. On the other

12 12 FATEMEH Y. MOKARI hand each β i is algebraic over F. This implies that E := F [β 1,..., β n ] is finite over F. But [E : Q] = [E : F ][F : Q] <, thus E is finite over Q which implies that each β i is algebraic over Q. Therefore β 1,..., β n Q and thus Q is algebraically closed. The field Q is called the algebraic closure of Q. In the following we will generalize this concept. In the next theorem will show that any field can be embedded in an algebraically closed field. But for that we need the following simple lemma. Lemma 5.2. Let F be a field and let f 1 (X),..., f r (X) F [X] be nonconstant polynomials. Then there is a finite extension E of F such that each f i (X) has a root in E. Proof. The proof is by induction on r. First let r = 1. Let g 1 (X) F [X] be an irreducible factor of f 1 (X). Now if E 1 := F [α] = F [X]/ g 1 (X), α := X + g 1 (X), then [E 1 : F ] = deg(g 1 (X)) < and α E 1 is a root of g 1 (X) F [X]. Thus α also is a root of f 1 (X). Now assume that the claim is true for any r 1 non-constant polynomials over any field. Let E 1 be a finite extension of F such that f 1 (X) has a root. Then since f 2 (X),..., f r (X) E 1 [X], by induction there is a finite extension E of E 1 such that any f i (X), 2 i r, has a root E. Now [E : F ] = [E : E 1 ][E 1 : F ] < and by construction any f i (X) has a root in E. The next theorem is very known and is very fundamental in the subject of Algebra. Theorem 5.3. Any Field can be embedded in an algebraically closed filed. Proof. (Artin) Let F be a field. We wish to construct an algebraically closed field K that contains F. First we construct a field E 1 that any non-constant f(x) F [X] has a root in E 1. To any non-constant f(x) F [X] we associate a letter X f. Let S be the set of all the letters X f such that f(x) is non-constant. Consider the polynomial ring F [S] = F [X f : X f S] and let I be the ideal of F [S] generated by all f(x f ) F [S]. We prove that I F [S]: If I = F [S], then there are non-constant polynomials f 1 (X f1 ),..., f n (X fn ) I and g 1,..., g n F [S], such that g 1 f 1 (X f1 ) + + g n f n (X fn ) = 1.

13 A NOTE ON FINITE FIELDS 13 For simplicity we put X i := X fi. Since the number of g i s are finite, they have only finitely many variables X 1,..., X N in them (with N n). Thus the above formula has the following form g 1 (X 1,..., X N )f 1 (X 1 ) + + g n (X 1,..., X N )f n (X n ) = 1. By Lemma 5.2, there exists a finite extension F 1 of F such that all the polynomials f 1 (X),..., f n (X) have a root in F 1. Let α i F 1 be the root of f i (X), 1 i n and for n < i N, we put α i = 0. If we substitute the α i s, 1 i N, in the above equation, we get 0 = 1, which is a contradiction. Thus I F [S]. Let m be a maximal ideal of F [S] that contains I and let E 1 := F [S]/m. Then we have the natural embedding F E 1, a a + m. We denote the image of a F in E 1 again by a. Now if f(x) F [X] is non-constant and if α f := X f + m E 1, then f(α f ) = 0. Thus any non-constant polynomial of F [X] has a root in E 1. Now inductively we can construct a sequence of fields F := E 0 E 1 E 2 E 3... such that every non-constant polynomial of E n [X] has a root in E n+1. Now put K := n 1 E n. Clearly K is a field and contains F. Now if h(x) K[X] is nonconstant, then there is an n 1 such that h(x) E n [X] which has a root in E n+1 K. This completes the proof. Proposition 5.4. Let E be an algebraic extension of F and let F K, where K is an algebraically closed field. Then there is an embedding σ : E K such that σ F = id F. If E is algebraically closed and if K is algebraic over F, then σ is an isomorphism of E onto K. Proof. Let S be the set of all pairs (L, τ) such that L is a subfield of E, contains F and τ is an embedding of L in K such that τ F = id F. If (L 1, τ 1 ), (L 2, τ 2 ) S, we say (L 1, τ 1 ) (L 2, τ 2 ), if L 1 L 2 and τ 2 L1 = τ 1. With this relation S is a partially ordered set. Since (F, id F ) S, S. Moreover if {L i, τ i } i I is a chain of elements of S, then (L, τ) S with L = i I L i and τ Li = τ i for any i I, is an upper bound for the chain. Thus by the Zorn lemma S has a maximal element (L, σ). If L E, take an element α E L. Since E is algebraic over F, it is also algebraic over L. Let f(x) L[X] be

14 14 FATEMEH Y. MOKARI the minimal polynomial of α and consider σ(f(x)) σ(l)[x] K[X]. Since f(x) is irreducible over L, σ(f(x)) is irreducible over σ(l). But K is algebraically closed, so σ(f(x)) has a root β in K. Define the natural homomorphism n n σ α : L[α] K, g(α) = a i α i σ(g)(β) = σ(a i )β i. i=0 Clearly this is an embedding, because L[α] L[X]/ f(x) σ(l)[x]/ σ(f(x)) σ(l)[β] K. This implies that (L[α], σ α ) S, which contradicts the maximality of (L, σ). Therefore we must have L = E. Thus we have an embedding σ : E K such that σ F = id F. Now if E is algebraically closed, then σ(e) is also algebraically closed. Since K is algebraic over F, it is also algebraic over σ(e). Therefore K = σ(e), because algebraically closed field do not have non-trivial algebraic extensions. Let F be a subfield of E. We say that E is an algebraic closure of F if we have the following two conditions: (i) E is algebraic over F, (ii) E is algebraically closed. Theorem 5.5. Let F be a field. Then an algebraic closure of F exist and is unique up to F -isomorphism. Proof. Let K be an algebraically closed field that containing F. By 5.3 such a field exists. Let L be the set of all elements of K that are algebraic over F. By an argument similar to one in Example 5.1(ii), we can show that L is a field, contains F and is algebraically closed. Thus an algebraic closure of F exist. Let L 1 and L 2 be two algebraic closure of F. Then by Proposition 5.4, there is an embedding σ : L 1 L 2 such that σ F = id F. Since L 2 is algebraic over F, again by Proposition 5.4, σ is an isomorphism. Therefore algebraic closure of F exist and is unique up to isomorphism. Since any two algebraic closure of F are F -isomorphism, we can talk about the algebraic closure of F and we denote it by F. Corollary 5.6. Let E be an algebraic extension of a field F. Then E = F i=0 Proof. This follows immediately from Proposition 5.4.

15 A NOTE ON FINITE FIELDS Finite fields In the rest of this note we will study finite fields and their finite extensions. Let F be a finite field. Then F has prime characteristic char(f ) = p and thus F p embeds in F (Proposition 1.2). Since [F : F p ] F <, F is a F p -vector space of finite dimension. Let [F : F p ] = dim Fp F = n. Then as a F p -vector space, F F n p = F p F p (n-times) and so F = p n. Hence we have the isomorphism of additive groups (F, +) (F n p, +). On the other hand, by Lemma 1.3, F is a cyclic group of order F 1 = p n 1. Hence for any a F, a pn 1 = 1. Multiplying the both side of this equation with a we see that a pn = a. So any elements of F is a root of the polynomial X pn X F p [X]. This fact suggests a way to construct a finite field with p n elements, for any prime p any positive integer n. Let p be a prime and let F p be the algebraic closure of F p (see Theorem 5.5). Let L be the set of all roots of the polynomial X pn X F p [X] in F p. If a, b, c L, c 0, then (a + b) pn = a pn + b pn = a + b, (ab) pn = a pn b pn = ab, (c 1 ) pn = (c pn ) 1 = c 1. Thus L is a subfield of F p and contains F p. Moreover all roots of X pn X are distinct. This follows from the following simple lemma, since f(x) = X pn X and f (X) = 1 are co-prime. Therefore L is a finite field with precisely p n elements. Let f(x) F [X] be a non-constant polynomial, where F is a field. Then we say that α F is a simple root of f(x) if X α f(x) but (X α) 2 f(x). Lemma 6.1. Let F be a field and let f(x) F [X] be a non-constant polynomial. Then f(x) has only simple roots if and only if f(x) and its first derivation f (X) are co-prime. Proof. The proof is easy and we leave it to the reader. Let K be any other field with p n elements. Let the cyclic group K is generated by β, i.e. K = β (Lemma 1.3). Clearly K is the smallest subfield of K containing both F p and β. Thus K = F p [β]. Let g(x) F p [X] be the minimal polynomial of β. Then we have the isomorphism K = F p [β] F p [X]/ g(x),

16 16 FATEMEH Y. MOKARI and therefore n = [K : F p ] = deg(g(x)). Note that since β is a root of X pn X, g(x) X pn X. Let α F p be any root of g(x) and consider the F p -homomorphism F p [X]/ g(x) F p, X + g(x) α. This homomorphism is injective and thus we have an injective F p - homomorphism ψ : K = F p [β] = {h(β) : h(x) F p [X]} F p, h(β) h(α). Since all elements of K are roots of X pn X, all elements of ψ(k) also are roots of this polynomial. Thus ψ(k) L. Since we have K = ψ(k) = L = p n, K ψ(k) = L. Thus any field with p n element is isomorphic to L and so we have proven the following theorem. Theorem 6.2. Let p be a prime and let n be a positive integer. Then, up to isomorphism, there is a unique finite field with p n elements. We denote this field with F p n and it can be considered as the set of all roots of X pn X in F p. Let F be a subfield of F p n. Then F = p m for some positive integer m Z and we have n = [F p n F p ] = [F p n : F ][F : F p ] = m[f p n : F ]. This implies that m n. Thus if F p n has a subfield isomorphic to F p m, then m n. Now let m, n be positive integers such that m n. We have seen that F p n can be considered as a subfield of F p, as the set of all roots of X pm X F p [X] in F p (Theorem 6.2). Let F p m be the set of all roots of X pm X F p [X] in F p. Since m n, any root of X pm X is also a root of X pn X. This follows from the following fact: If m n, then X pm X X pn X. In fact if n = mr, then p n 1 = p mr 1 = (p m ) r 1 = (p m 1)t for some positive t Z and so X pn X = X(X pn 1 1) = X(X t(pm 1) 1) = X(X pm 1 1)h(X) = (X pm X)h(X),

17 A NOTE ON FINITE FIELDS 17 where h(x) F p [X]. This proves the claim. Going back to our discussion, we see that we have F p m F p n. Thus F p n has a unique subfield with p m elements. We gather these results in the following theorem. Theorem 6.3. The field F p m can be embedded in the field F p n if and only if m n. In this case F p m can be considered as the set of all roots of X pm X F p [X] in F p n. Now we wish to study the irreducible polynomials over finite fields. First we prove the following lemma. Lemma 6.4. Let F be a finite field. Then for any positive integer n, there is an irreducible polynomial of degree n in F [X]. Proof. Since any finite field is isomorphic to some F p m, we may assume that F = F p m. Since m mn, F p m can be considers as a subfield of F p mn (Theorem 6.3). Let β be the generator of F p (Lemma 1.3). Since mn F p mn is the smallest subfield of F p mn that contains both β and F p m, we have F p mn = F p m[β]. Let f(x) F p m[x] be the minimal polynomial of β. Then we have deg(f(x)) = [F p mn : F p m] = [F p mn : F p ]/[F p m : F p ] = mn/m = n. Thus f(x) F p n[x] is an irreducible polynomial of degree n. The following theorem gives a precise decomposition of X pn X F p [X] to its irreducible factors. Theorem 6.5. Any irreducible polynomial of degree n in F p [X] is a factor of X pn X F p [X]. Moreover the irreducible factors of X pn X in F p [X] are precisely the irreducible polynomials whose degree divide n. Proof. Let f(x) F p [X] be an irreducible polynomial of degree n. Then E := F p [X]/ f(x) is a field containing F p and [E : F p ] = deg(f(x)) = n. Thus E has p n elements and E = F p [β], where β = X + f(x). Since f(β) = 0, f(x) is the minimal polynomial of β. On the other hand E F p n so any element of E, including β, is a root of X pn X. Therefore f(x) divides X pn X. To proof the second part, first let g(x) be an irreducible polynomial of degree m such that m n. Then by the first part of the theorem g(x) X pm X. We also have seen in above that X pm X X pn X. This implies that g(x) X pn X. Now let h(x) be an irreducible factor of X pn X of degree m. Since X pn X decomposes completely in

18 18 FATEMEH Y. MOKARI F p n[x], h(x) has a root γ in F p n. Then clearly F p [γ] F p [X]/ h(x) and so [F p [γ] : F p ] = deg(h(x)) = m. Now from the equality n = [F p n : F p ] = [F p n : F p [γ]][f p [γ] : F p ] = m[f p n : F p [γ]], we see that m divides n. This completes the proof of the theorem. In fact the above theorem can be generalized to all finite field in the following sense. Proposition 6.6. Let F be a finite field with q = p r elements. Then any irreducible polynomial of degree n in F [X] is a factor of X qn X F [X]. Moreover the irreducible factors of X qn X in F [X] are precisely the irreducible polynomials whose degree divide n. Proof. The proof is the same as the proof of Theorem 6.5, replacing q with p and F with F p. Example 6.7. Let α F p and let f(x) F p [X] be the minimal polynomial of α. If n = deg(f(x)), then by Theorem 6.5, f(x) X pn X and thus α is the root of X pn X. This implies that α F p n F p and thus F p = F p n. n 1 Note that if F is any finite field of characteristic p, then F = F p. This follows from Corollary 5.6. Example 6.8. Let X 2n X F 2 [X]. Then by Theorem 6.5 an irreducible polynomial f(x) divides X 2n X if and only deg(f(x)) n. (i) If n = 2, then deg(f(x)) = 1 or deg(f(x)) = 2. The polynomials X and X + 1 are the only irreducible polynomials of degree one and X 2 + X + 1 is the only irreducible polynomial of degree two. Thus X 4 X = X(X + 1)(X 2 + X + 1). (ii) If n = 3, then deg(f(x)) = 1 or deg(f(x)) = 3. Note that since 2 3, no irreducible polynomial of degree 2 divides X 8 X. Again X and X + 1 are the only irreducible polynomials of degree one. Since X 3 + X + 1 and X 3 + X do not have any root in F 2, they are irreducible. So X, X + 1, X 3 + X + 1 and X 3 + X divide X 8 X. Since the sum of degrees of these polynomials is 8, we have X 8 X = X(X + 1)(X 3 + X + 1)(X 3 + X 2 + 1). (iii) If n = 4, then deg(f(x)) = 1, deg(f(x)) = 2 or deg(f(x)) = 3. Again X and X + 1 are the only irreducible polynomials of degree one and X 2 + X + 1 is the only irreducible polynomial of degree two. With an argument similar to the one in Example 2.5, we see that the degree

19 A NOTE ON FINITE FIELDS 19 four polynomials X 4 + X + 1, X 4 + X and X 4 + X 3 + X 2 + X + 1 are irreducible. Thus as the argument in (ii) we have X 16 X =X(X + 1)(X 2 + X + 1)(X 4 + X + 1)(X 4 + X 3 + 1) (X 4 + X 3 + X 2 + X + 1). Now we wish to study the Galois group of finite extension if finite fields. Consider the following F p -automorphism of F p n, σ : F p n F p n, a a p. It is easy to see that σ n = id Fp n, and that σ 0 = id Fp n, σ 1, σ 2,..., σ n 1 are n distinct elements of Gal(F p n/f p ). Thus by Theorem 3.4, and so Thus we showed that F p n n Gal(F p n/f p ) [F p n : F p ] = n Gal(F p n/f p ) = [F p n : F p ]. is a Galois extension of F p and Gal(F p n/f p ) = σ Z/nZ. This fact can easily be generalized to all finite fields. Let E be a finite field with p n elements and let F be its subfield with p m elements. Then by Theorem 6.3 m n and F is the set of all roots of X pm X F p [X] in E. If τ is the following F -automorphism of E, τ : E E, a a pm, then τ n/m = id E and τ 0 = id E, τ 1,..., τ (n/m) 1 are distinct element of Gal(E/F ). Thus Gal(E/F ) = [E : F ] = n/m. This implies that E is a Galois extension of F and Gal(E/F ) τ Z/(n/m)Z. Thus we have proved the following theorem. Theorem 6.9. Let E be a finite field of characteristic p and let F be a subfield of E. Then E is a Galois extension of F and their Galois group is a finite cyclic group of order [E : F ], generated by τ : E E, a a pm, where m = [F : F p ].

20 20 FATEMEH Y. MOKARI 7. Norm and trace Let E be a finite Galois extension of F. We define the norm and the trace of the extension as follow: N E/F : E F, α σ(α), Tr E/F : E F, α σ Gal(E/F ) σ Gal(E/F ) σ(α). Note that for any τ Gal(E/F ) = {σ 1,..., σ n }, we have τgal(e/f ) = {τ σ 1,..., τ σ n } = Gal(E/F ) and thus ( τ(n E/F (α)) = τ σ Gal(E/F ) and ( τ(tr E/F (α)) = τ σ Gal(E/F ) Theorem 4.2 implies that ) σ(α) = ) σ(α) = σ Gal(E/F ) σ Gal(E/F ) N E/F (α), Tr E/F (α) F Gal(E/F ) = F. τ σ(α) = N E/F (α) τ σ(α) = Tr E/F (α). The following properties for norm easily follows from the definition: (i) N E/F (αβ) = N E/F (α)n E/F (β), for any α, β E, so N E/F is a group homomorphism. (ii) N E/F (α) = α [E:F ], and for any α F. (iii) N L/E N E/F = N L/F, where L is a Galois extension of E. The following properties for trace easily follows from the definition: (i) Tr E/F (aα + bβ) = atr E/F (α) + btr E/F (β), for any α, β E and a, b F, so Tr E/F is a F -linear transformation and clearly it is surjective. (ii) Tr E/F (α) = [E : F ]α, for any α F. (iii) Tr L/E Tr E/F = Tr L/F, where L is a Galois extension of E. Let F q be the finite field with q = p m elements. If E is a finite extension of F q, then E is isomorphism to F q n for some positive integer n (Theorem 6.3). Hence we may assume that E = F q n, and so [F q n : F q ] = n. By Theorem 6.9, Gal(F q n/f q ) is cyclic of order n and is generated by σ : F q n F q n, α α q,

21 A NOTE ON FINITE FIELDS 21 so Gal(F q n/f q ) = {id Fq n, σ,..., σ n 1 }. Now if α F q n, then we have the following explicit formulas N Fq n/f q (α) = αα q α q2... α qn 1 = α 1+q+q2 + +q n 1, Tr Fq n/f q (α) = α + α q + α q2 + + α qn 1. We have seen that for any finite Galois extension E of a field F, Tr E/F is surjective. This is not true in general for the norm map N E/F. But easily can be proved that N E/F is surjective when E and F are finite fields. To prove this we may assume F = F q, q = p m for a prime p. Then E F q n for some n (Theorem 6.3). Let α, be the generator of the cyclic group F q (Lemma 1.3). Then n 1 = α qn 1 = ( α qn 1 + +q+1 ) q 1 Since α is of order q n 1, α qn 1 + +q+1 is of order q 1. But N Fq n/f q (α) = α qn 1 + +q+1 F q. Since F q is a cyclic group of order q 1, N Fq n/f q (α) should generate F q, i.e. F q = N Fq n/f q (α). But the norm is a homomorphism of groups, it must be surjective. Proposition 7.1. Let q = p m, p a prime, and let f(x, Y ) = Y qn 1 + +q 2 +q+1 X qn X q2 + X q + X F q [X, Y ]. Then f(x, Y ) has q 2n 1 roots in (F q n) 2 = F q n F q n. Proof. Let (x, y) (F q n) 2 be a solution of f(x, Y ). Fix x. If x qn x q2 + x q + x = 0, then y = 0. First we count the number of (x, 0) (F q n) 2 such that f(x, 0) = 0. Let (x, 0) be such a point. Then Since Tr Fq n/f q Tr Fq n/f q (x) = x qn x q2 + x q + x = 0. : F q n F q is a surjective F q -linear transformation, dim Fq (ker(tr Fq n/f q )) = dim Fq (F q n) dim Fq (F q ) = n 1. Thus ker(tr Fq n/f q ) = q n 1 and this implies that {(x, 0) (F q n) 2 : f(x, 0) = 0} = q n 1. Now let x qn x q2 + x q + x 0, where x F q n. Consider the one variable polynomial g(y ) = Y qn 1 + +q 2 +q+1 x qn x q2 + x q + x F q n[y ].

22 22 FATEMEH Y. MOKARI We have g (Y ) = (q n q 2 + q + 1)Y qn 1 + +q 2 +q = Y qn 1 + +q 2 +q. Hence (g(y ) and g (Y )) are co-prime, which implies that all the roots of g(y ) in F q n are different (Lemma 3.1). If g(y) = 0, then We have y qn = yy qn 1 y qn 1 + +q 2 +q+1 = x qn x q2 + x q + x. = y ( y qn 1 + +q 2 +q+1 ) q 1 = y ( x qn x q2 + x q + x ) q 1 = y ( x qn x q2 + x q + x ) 1( x q n x q2 + x q + x ) q = y ( x qn x q2 + x q + x ) 1( x q n + + x q3 + x q2 + x q) = y ( x qn x q2 + x q + x ) 1( x q n x q2 + x q + x ) = y. Just we should remind that for any x F q n, x qn = x, so x qn + + x q3 + x q2 + x q = x + x qn x q3 + x q2 + x q = x qn x q3 + x q2 + x q + x. Now by Theorem 6.2, y F q n. Thus we have shown that for any fixed x F q n such that x qn x q2 + x q + x 0, we have q n q 2 + q + 1 elements y F q n such that f(x, y) = 0. Thus the number of roots of f(x, Y ) in (F q n) 2 is equal to q n 1 + (q n q n 1 )(q n q 2 + q + 1) = q 2n 1. References [1] Artin, M. Algebra. Englewood Cliffs, NJ: Prentice-Hall, , 3, 11 [2] Lang, S. Algebra. Revised Third Edition, Graduate Texts in Mathematics 211. Springer-Verlag, New York, [3] Morandi, P. Field and Galois Theory. Graduate Texts in Mathematics 167. Springer-Verlag, New York, [4] Shidlovskii, A. B. Transcendental Numbers. New York: de Gruyter, Fatemeh Yeganeh Mokari, Department of Mathematics (IMECC), University of Campinas, Campinas, Brazil

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