A NOTE ON FINITE FIELDS


 Dominick Holt
 1 years ago
 Views:
Transcription
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 nonconstant 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 nontrivial 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 nonconstant 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 nonconstant 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 nonconstant f(x) F [X] has a root in E 1. To any nonconstant 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 nonconstant. 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 nonconstant 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 nonconstant and if α f := X f + m E 1, then f(α f ) = 0. Thus any nonconstant 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 nonconstant 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 nontrivial 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 (ntimes) 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 coprime. Therefore L is a finite field with precisely p n elements. Let f(x) F [X] be a nonconstant 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 nonconstant polynomial. Then f(x) has only simple roots if and only if f(x) and its first derivation f (X) are coprime. 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 coprime, 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: PrenticeHall, , 3, 11 [2] Lang, S. Algebra. Revised Third Edition, Graduate Texts in Mathematics 211. SpringerVerlag, New York, [3] Morandi, P. Field and Galois Theory. Graduate Texts in Mathematics 167. SpringerVerlag, New York, [4] Shidlovskii, A. B. Transcendental Numbers. New York: de Gruyter, Fatemeh Yeganeh Mokari, Department of Mathematics (IMECC), University of Campinas, Campinas, Brazil
Galois Theory III. 3.1. Splitting fields.
Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where
More informationIntroduction to finite fields
Introduction to finite fields Topics in Finite Fields (Fall 2013) Rutgers University Swastik Kopparty Last modified: Monday 16 th September, 2013 Welcome to the course on finite fields! This is aimed at
More information1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain
Notes on realclosed fields These notes develop the algebraic background needed to understand the model theory of realclosed fields. To understand these notes, a standard graduate course in algebra is
More information10 Splitting Fields. 2. The splitting field for x 3 2 over Q is Q( 3 2,ω), where ω is a primitive third root of 1 in C. Thus, since ω = 1+ 3
10 Splitting Fields We have seen how to construct a field K F such that K contains a root α of a given (irreducible) polynomial p(x) F [x], namely K = F [x]/(p(x)). We can extendthe procedure to build
More informationIntroduction to Finite Fields (cont.)
Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number
More informationminimal polyonomial Example
Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We
More informationFINITE FIELDS KEITH CONRAD
FINITE FIELDS KEITH CONRAD This handout discusses finite fields: how to construct them, properties of elements in a finite field, and relations between different finite fields. We write Z/(p) and F p interchangeably
More informationIntegral Domains. As always in this course, a ring R is understood to be a commutative ring with unity.
Integral Domains As always in this course, a ring R is understood to be a commutative ring with unity. 1 First definitions and properties Definition 1.1. Let R be a ring. A divisor of zero or zero divisor
More informationQuotient Rings and Field Extensions
Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.
More informationAppendix A. Appendix. A.1 Algebra. Fields and Rings
Appendix A Appendix A.1 Algebra Algebra is the foundation of algebraic geometry; here we collect some of the basic algebra on which we rely. We develop some algebraic background that is needed in the text.
More informationCyclotomic Extensions
Chapter 7 Cyclotomic Extensions A cyclotomic extension Q(ζ n ) of the rationals is formed by adjoining a primitive n th root of unity ζ n. In this chapter, we will find an integral basis and calculate
More informationALGEBRA HW 5 CLAY SHONKWILER
ALGEBRA HW 5 CLAY SHONKWILER 510.5 Let F = Q(i). Prove that x 3 and x 3 3 are irreducible over F. Proof. If x 3 is reducible over F then, since it is a polynomial of degree 3, it must reduce into a product
More informationFACTORING AFTER DEDEKIND
FACTORING AFTER DEDEKIND KEITH CONRAD Let K be a number field and p be a prime number. When we factor (p) = po K into prime ideals, say (p) = p e 1 1 peg g, we refer to the data of the e i s, the exponents
More informationINTRODUCTION TO ARITHMETIC GEOMETRY (NOTES FROM 18.782, FALL 2009)
INTRODUCTION TO ARITHMETIC GEOMETRY (NOTES FROM 18.782, FALL 2009) BJORN POONEN (Please clear your browser s cache before reloading to make sure that you are always getting the current version.) 1. What
More informationUnique Factorization
Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationSOLUTIONS TO PROBLEM SET 3
SOLUTIONS TO PROBLEM SET 3 MATTI ÅSTRAND The General Cubic Extension Denote L = k(α 1, α 2, α 3 ), F = k(a 1, a 2, a 3 ) and K = F (α 1 ). The polynomial f(x) = x 3 a 1 x 2 + a 2 x a 3 = (x α 1 )(x α 2
More informationsome algebra prelim solutions
some algebra prelim solutions David Morawski August 19, 2012 Problem (Spring 2008, #5). Show that f(x) = x p x + a is irreducible over F p whenever a F p is not zero. Proof. First, note that f(x) has no
More informationPROBLEM SET 6: POLYNOMIALS
PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other
More information2. Let H and K be subgroups of a group G. Show that H K G if and only if H K or K H.
Math 307 Abstract Algebra Sample final examination questions with solutions 1. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30 and 40, Determine H. Solution. Since gcd(18,
More informationit is easy to see that α = a
21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore
More information7. Some irreducible polynomials
7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of
More informationF1.3YE2/F1.3YK3 ALGEBRA AND ANALYSIS. Part 2: ALGEBRA. RINGS AND FIELDS
F1.3YE2/F1.3YK3 ALGEBRA AND ANALYSIS Part 2: ALGEBRA. RINGS AND FIELDS LECTURE NOTES AND EXERCISES Contents 1 Revision of Group Theory 3 1.1 Introduction................................. 3 1.2 Binary Operations.............................
More informationGalois theory for dummies
Galois theory for dummies Ruben Spaans May 21, 2009 1 Notes on notation To help avoid vertical figures, I use the notation E/F if E is an extension to the field F. This is the same notation as Wikipedia
More informationThe Dirichlet Unit Theorem
Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if
More informationEXERCISES FOR THE COURSE MATH 570, FALL 2010
EXERCISES FOR THE COURSE MATH 570, FALL 2010 EYAL Z. GOREN (1) Let G be a group and H Z(G) a subgroup such that G/H is cyclic. Prove that G is abelian. Conclude that every group of order p 2 (p a prime
More informationFactorization Algorithms for Polynomials over Finite Fields
Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 20110503 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is
More informationCONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12
CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.
More information(a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9
Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned
More informationField Fundamentals. Chapter 3. 3.1 Field Extensions. 3.1.1 Definitions. 3.1.2 Lemma
Chapter 3 Field Fundamentals 3.1 Field Extensions If F is a field and F [X] is the set of all polynomials over F, that is, polynomials with coefficients in F, we know that F [X] is a Euclidean domain,
More informationModern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)
Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)
More informationModule MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013
Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of
More informationChapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.
Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize
More informationRevision of ring theory
CHAPTER 1 Revision of ring theory 1.1. Basic definitions and examples In this chapter we will revise and extend some of the results on rings that you have studied on previous courses. A ring is an algebraic
More informationChapter 13: Basic ring theory
Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring
More informationMath 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field
Math 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will
More informationABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS
ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS John A. Beachy Northern Illinois University 2014 ii J.A.Beachy This is a supplement to Abstract Algebra, Third Edition by John A. Beachy and William D. Blair
More informationFinite Fields and ErrorCorrecting Codes
Lecture Notes in Mathematics Finite Fields and ErrorCorrecting Codes KarlGustav Andersson (Lund University) (version 1.01316 September 2015) Translated from Swedish by Sigmundur Gudmundsson Contents
More informationAlgebra 3: algorithms in algebra
Algebra 3: algorithms in algebra Hans Sterk 20032004 ii Contents 1 Polynomials, Gröbner bases and Buchberger s algorithm 1 1.1 Introduction............................ 1 1.2 Polynomial rings and systems
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important
More information3. Prime and maximal ideals. 3.1. Definitions and Examples.
COMMUTATIVE ALGEBRA 5 3.1. Definitions and Examples. 3. Prime and maximal ideals Definition. An ideal P in a ring A is called prime if P A and if for every pair x, y of elements in A\P we have xy P. Equivalently,
More informationZORN S LEMMA AND SOME APPLICATIONS
ZORN S LEMMA AND SOME APPLICATIONS KEITH CONRAD 1. Introduction Zorn s lemma is a result in set theory that appears in proofs of some nonconstructive existence theorems throughout mathematics. We will
More information3 1. Note that all cubes solve it; therefore, there are no more
Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if
More informationFACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z
FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization
More informationFactoring of Prime Ideals in Extensions
Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree
More informationGROUPS ACTING ON A SET
GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for
More information11 Ideals. 11.1 Revisiting Z
11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationA number field is a field of finite degree over Q. By the Primitive Element Theorem, any number
Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and
More information1 Finite Fields and Function Fields
1 Finite Fields and Function Fields In the first part of this chapter, we describe the basic results on finite fields, which are our ground fields in the later chapters on applications. The second part
More informationON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS
ON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 19967 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More information5.1 Commutative rings; Integral Domains
5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following
More informationLinear Maps. Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007)
MAT067 University of California, Davis Winter 2007 Linear Maps Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007) As we have discussed in the lecture on What is Linear Algebra? one of
More informationFACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set
FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly,
More informationa 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)
ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x
More informationFactoring Polynomials
Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent
More informationDIVISORS AND LINE BUNDLES
DIVISORS AND LINE BUNDLES TONY PERKINS 1. Cartier divisors An analytic hypersurface of M is a subset V M such that for each point x V there exists an open set U x M containing x and a holomorphic function
More informationLinear Algebra I. Ronald van Luijk, 2012
Linear Algebra I Ronald van Luijk, 2012 With many parts from Linear Algebra I by Michael Stoll, 2007 Contents 1. Vector spaces 3 1.1. Examples 3 1.2. Fields 4 1.3. The field of complex numbers. 6 1.4.
More informationGalois Theory. Richard Koch
Galois Theory Richard Koch April 2, 2015 Contents 1 Preliminaries 4 1.1 The Extension Problem; Simple Groups.................... 4 1.2 An Isomorphism Lemma............................. 5 1.3 Jordan Holder...................................
More informationSolutions A ring A is called a Boolean ring if x 2 = x for all x A.
1. A ring A is called a Boolean ring if x 2 = x for all x A. (a) Let E be a set and 2 E its power set. Show that a Boolean ring structure is defined on 2 E by setting AB = A B, and A + B = (A B c ) (B
More informationGROUP ALGEBRAS. ANDREI YAFAEV
GROUP ALGEBRAS. ANDREI YAFAEV We will associate a certain algebra to a finite group and prove that it is semisimple. Then we will apply Wedderburn s theory to its study. Definition 0.1. Let G be a finite
More information2.5 Gaussian Elimination
page 150 150 CHAPTER 2 Matrices and Systems of Linear Equations 37 10 the linear algebra package of Maple, the three elementary 20 23 1 row operations are 12 1 swaprow(a,i,j): permute rows i and j 3 3
More information1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]
1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not
More informationWinter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com
Polynomials Alexander Remorov alexanderrem@gmail.com Warmup Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).
More information50. Splitting Fields. 50. Splitting Fields 165
50. Splitting Fields 165 1. We should note that Q(x) is an algebraic closure of Q(x). We know that is transcendental over Q. Therefore, p must be transcendental over Q, for if it were algebraic, then (
More information6. Fields I. 1. Adjoining things
6. Fields I 6.1 Adjoining things 6.2 Fields of fractions, fields of rational functions 6.3 Characteristics, finite fields 6.4 Algebraic field extensions 6.5 Algebraic closures 1. Adjoining things The general
More informationMA106 Linear Algebra lecture notes
MA106 Linear Algebra lecture notes Lecturers: Martin Bright and Daan Krammer Warwick, January 2011 Contents 1 Number systems and fields 3 1.1 Axioms for number systems......................... 3 2 Vector
More informationIntroduction to Algebraic Geometry. Bézout s Theorem and Inflection Points
Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a
More informationORDERS OF ELEMENTS IN A GROUP
ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since
More information1 Lecture: Integration of rational functions by decomposition
Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.
More informationABSTRACT ALGEBRA. Romyar Sharifi
ABSTRACT ALGEBRA Romyar Sharifi Contents Introduction 7 Part 1. A First Course 11 Chapter 1. Set theory 13 1.1. Sets and functions 13 1.2. Relations 15 1.3. Binary operations 19 Chapter 2. Group theory
More informationCHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY
January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.
More informationAn Advanced Course in Linear Algebra. Jim L. Brown
An Advanced Course in Linear Algebra Jim L. Brown July 20, 2015 Contents 1 Introduction 3 2 Vector spaces 4 2.1 Getting started............................ 4 2.2 Bases and dimension.........................
More informationUNIT 2 MATRICES  I 2.0 INTRODUCTION. Structure
UNIT 2 MATRICES  I Matrices  I Structure 2.0 Introduction 2.1 Objectives 2.2 Matrices 2.3 Operation on Matrices 2.4 Invertible Matrices 2.5 Systems of Linear Equations 2.6 Answers to Check Your Progress
More informationThe Division Algorithm for Polynomials Handout Monday March 5, 2012
The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,
More informationCommutative Algebra Notes Introduction to Commutative Algebra Atiyah & Macdonald
Commutative Algebra Notes Introduction to Commutative Algebra Atiyah & Macdonald Adam Boocher 1 Rings and Ideals 1.1 Rings and Ring Homomorphisms A commutative ring A with identity is a set with two binary
More information4. CLASSES OF RINGS 4.1. Classes of Rings class operator Aclosed Example 1: product Example 2:
4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets
More informationr + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.
Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in
More informationRESULTANT AND DISCRIMINANT OF POLYNOMIALS
RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results
More information4.1 Modules, Homomorphisms, and Exact Sequences
Chapter 4 Modules We always assume that R is a ring with unity 1 R. 4.1 Modules, Homomorphisms, and Exact Sequences A fundamental example of groups is the symmetric group S Ω on a set Ω. By Cayley s Theorem,
More informationfg = f g. 3.1.1. Ideals. An ideal of R is a nonempty ksubspace I R closed under multiplication by elements of R:
30 3. RINGS, IDEALS, AND GRÖBNER BASES 3.1. Polynomial rings and ideals The main object of study in this section is a polynomial ring in a finite number of variables R = k[x 1,..., x n ], where k is an
More informationGalois Fields and Hardware Design
Galois Fields and Hardware Design Construction of Galois Fields, Basic Properties, Uniqueness, Containment, Closure, Polynomial Functions over Galois Fields Priyank Kalla Associate Professor Electrical
More informationMOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu
Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing
More informationFurther linear algebra. Chapter I. Integers.
Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides
More informationH/wk 13, Solutions to selected problems
H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.
More informationChapter 10. Abstract algebra
Chapter 10. Abstract algebra C.O.S. Sorzano Biomedical Engineering December 17, 2013 10. Abstract algebra December 17, 2013 1 / 62 Outline 10 Abstract algebra Sets Relations and functions Partitions and
More information13 Solutions for Section 6
13 Solutions for Section 6 Exercise 6.2 Draw up the group table for S 3. List, giving each as a product of disjoint cycles, all the permutations in S 4. Determine the order of each element of S 4. Solution
More informationThe Mathematics of Origami
The Mathematics of Origami Sheri Yin June 3, 2009 1 Contents 1 Introduction 3 2 Some Basics in Abstract Algebra 4 2.1 Groups................................. 4 2.2 Ring..................................
More informationGroup Theory. Contents
Group Theory Contents Chapter 1: Review... 2 Chapter 2: Permutation Groups and Group Actions... 3 Orbits and Transitivity... 6 Specific Actions The Right regular and coset actions... 8 The Conjugation
More informationa 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.
Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given
More information5. Linear algebra I: dimension
5. Linear algebra I: dimension 5.1 Some simple results 5.2 Bases and dimension 5.3 Homomorphisms and dimension 1. Some simple results Several observations should be made. Once stated explicitly, the proofs
More informationTHE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS
THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear
More informationT ( a i x i ) = a i T (x i ).
Chapter 2 Defn 1. (p. 65) Let V and W be vector spaces (over F ). We call a function T : V W a linear transformation form V to W if, for all x, y V and c F, we have (a) T (x + y) = T (x) + T (y) and (b)
More informationProofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate.
Advice for homework: Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate. Even if a problem is a simple exercise that doesn t
More informationLemma 5.2. Let S be a set. (1) Let f and g be two permutations of S. Then the composition of f and g is a permutation of S.
Definition 51 Let S be a set bijection f : S S 5 Permutation groups A permutation of S is simply a Lemma 52 Let S be a set (1) Let f and g be two permutations of S Then the composition of f and g is a
More information(x + a) n = x n + a Z n [x]. Proof. If n is prime then the map
22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find
More informationGalois theory. a draft of Lecture Notes of H.M. Khudaverdian. Manchester, Autumn 2006 (version 16 XII 2006)
Galois theory a draft of Lecture Notes of H.M. Khudaverdian. Manchester, Autumn 2006 (version 16 XII 2006) Contents 0.1................................... 2 0.2 Viète Theorem..........................
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationBasics of Polynomial Theory
3 Basics of Polynomial Theory 3.1 Polynomial Equations In geodesy and geoinformatics, most observations are related to unknowns parameters through equations of algebraic (polynomial) type. In cases where
More information