EXTENSION FIELDS Part I (Supplement to Section 3.8)

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1 EXTENSION FIELDS Part I (Supplement to Section 3.8) 1 Creating Finite Fields We know a few in nite elds: Q; R and C: We also know in nitely many nite elds: Z p where p is a prime number. We want to determine if there are other nite elds and, if they exist, what they look like. Before going on, we should state the fundamental homomorphism theorem for rings. Theorem 1 (FHT) Let R and S be rings and let : R! S be a ring homomorphism. Then (R) = R= ker : Remark 2 Recall that we proved last semester that the characteristic of a eld is either 0 or prime. Also, if the eld is nite, then the characteristic of the eld is prime. Lemma 3 Let F be a eld. If char(f ) = 0; then F contains a sub eld isomorphic to Q and if char(f ) = p (where p is prime), then F contains a sub eld isomorphic to Z p : Proof. Let F be a eld. De ne : Z! F by (n) = n 1 F : is a ring homomorphism, so (Z) = Z= ker : If char(f ) = 0; then follows that ker = f0g : Thus, (Z) = Z: So, F contains a subring, (Z), isomorphic to Z: Since F is a eld, it must contain the eld of fractions of (Z) : So, since Frac(Z) = Q; we have that F contains a copy of Q: If char(f ) = p; prime, then it follows that ker = pz: So, (Z) = Z=pZ = Z p : Therefore, F contains a subring, (Z) ; isomorphic to Z p : Lemma 4 Every nite eld has order p n for some prime p and some positive integer n: 1

2 Proof. Let E be a nite eld. Then char(e) = p for some prime p: Thus, E contains a sub eld isomorphic to Z p : From this we have that E is a vector space over Z p : Clearly, E is a nite-dimensional vector space of Z p ; since E is nite. Let n = dim Zp (E) : Then, jej = p n : Example 5 Let F = Z 5 [x] = (p) where p = x 3 + x + 1: I claim that F is a eld of order 5 3 : Question: Why is F a eld? Question: Why is jf j = 5 3? Claim 6 Every element of F can be written in the form f + (p) where f = 0 or deg (f) 2. Example 7 Consider the element x 4 + (p) 2 F: We have x 4 + (p) = (x 4 + 5x 2 + 5x)+(p) = [x (x 3 + x + 1) + (4x 2 + 4x)]+(p) = (4x 2 + 4x)+ (p) : Proof. Let g + (p) 2 F: Then g = pf + r for some f; r 2 Z 5 [x] where r = 0 or deg (r) 2: Then g + (p) = (pf + r) + (p) = r + (p) : Conclusion: This gives us that jf j 5 3 : Claim 8 If f + (p) = h + (p) where deg (f) 2 and deg (h) 2; then f = h: Proof. Suppose f + (p) = h + (p) where deg (f) 2 and deg (h) 2: Then f h 2 (p) : So, f h = 0 or deg (f h) deg (p) = 3: Since deg (f h) max fdeg (f) ; deg (h)g 2; we have f h = 0; so f = h: Conclusion: Therefore, jf j = 5 3 : Remark 9 Can we use this idea to create elds of order p n where p is a prime? How many (non-isomorphic) elds are there of order p n? 2

3 2 De nitions and First Examples of Extensions De nition 10 A eld E is an extension eld of a eld F if F E and the operations of F are those of E restricted to F: This is sometimes written as E=F is a eld extension. Lemma 11 If E is an extension eld of F; then E is a vector space over F (under the operations of E): De nition 12 Suppose E is an extension eld of the eld F: We de ne the degree of E over F, denoted [E : F ] to be the dimension of E as a vector space over F: So, [E : F ] = dim F E: De nition 13 Let F be a eld and let E be an extension eld of F: Let a 2 E: 1. F [a] is de ned to be the smallest ring (subring of E) which contains F and a: 2. F (a) is de ned to be the smallest eld (sub eld of E) which contains F and a: Before going any further, we need to verify that F [a] and F (a) actually exist. To see that F [a] exists, note that F [a] = \ fk : K is a subring of E and K contains both F and ag : Since the intersection of subrings is a subring, this (the intersection) is a subring of E and clearly contains F and a and is the smallest such subring of E: We can do something similar to show that F (a) exists, however, another option is the following. Lemma 14 Let F be a eld and let E be an extension eld of F: Let a 2 E: Then F (a) is (isomorphic to) to the eld of quotients of F [a] : 3

4 Proof. By de nition, F (a) is the smallest eld that contains both F and a: So, F (a) is a ring that contains both F and a: So, since F [a] is the smallest ring containing F and a; F [a] F (a) : Let K be the eld of quotients of F [a] ; then K is the smallest eld which contains F [a] : So, since F (a) is a eld containing F [a] ; we have K F (a) : Note that K is a eld which contains F [a] ; so K is a eld containing F and a: However, F (a) is the smallest eld containing F and a: Thus, F (a) K: Therefore, F (a) is the eld of quotients of F [a] : The description of F [a] given above is an external description of F [a] and not very helpful in studying F [a] : Our next goal is to get an internal description of F [a] ; i.e. what do the elements of F [a] look like? Theorem 15 Let F be a eld and let E be an extension eld of F: Let a 2 E: Then F [a] = a (F [x]) (where a : F [x]! E is the evaluation homomorphism at a): The proof of this theorem will use the fact that we already know that F [a] exists (by the argument above). Proof. Let S = a (F [x]) : Let r 2 S: Then r = r 0 + r 1 a + + r n a n for some non-negative integer n and some r 0 ; r 1 ; : : : ; r n 2 F: Since, by de nition, F [a] is a ring which contains F and a; we have, by closure, that r 2 F [a] : Therefore, S F [a] : On the other hand, S = a (F [x]) is a subring of E; since a is a homomorphism. Let b 2 F: Then b 2 F [x] and a (b) = b; so b 2 S: Thus, F S: Also, a = a (x) 2 a (F [x]) = S: Thus, a 2 S: So, S is a ring which contains F and a: Therefore, F [a] S: Therefore, F [a] = S: Example 16 Prove Q [i] = fa + bi : a; b 2 Qg and show Q [i] = Q (i) : (Hint: what kind of ring is fa + bi : a; b 2 Qg?) Theorem 17 Let F be a eld and let E be an extension eld of E with a 2 E: Then F [a] = F (a) if and only if F [a] is a eld. Proof. Let F be a eld and let E be an extension eld of E with a 2 E: Suppose F [a] = F (a) : Then, since F (a) is a eld (by de nition), F [a] is a eld. 4

5 On the other hand, suppose F [a] is a eld. Then F [a] is a eld containing F and a: So, since F (a) is the smallest eld containing F and a; we have F (a) F [a] : Also, since F (a) is the smallest eld containing F and a; we have that F (a) is a ring that contains both F and a: So, since F [a] is the smallest ring containing F and a; F [a] F (a) : Therefore, F [a] = F (a) : Example 18 What is Q [] and Q ()? Remark 19 We can adjoin more than one element to a eld. By de nition F (a; b) is the smallest eld containing F and a and b (where a and b are elements of the extension eld E of F ): This is equal to (F (a)) (b) : Example 20 Q p 2; p 3 = Q p 2 + p 3 Proof. Let K = Q p 2; p 3 ; so K is the smallest eld containing Q; p 2 and p 3: Let L = Q p 2 + p 3 ; so L is the smallest eld containing Q and p 2 + p 3: Since K contains p 2 and p 3 and is a eld, p 2 + p 3 2 K: So, K is a eld which contains Q and p 2 + p 3: So, since L is the smallest eld containing Q and p 2 + p 3; we have L K: On the other hand, since p 2 + p 3 2 L; we have p 2 + p L: p p 1 p p p p p p However, = L: So, L: Thus, 2 p 3 2 L: Recall L contains Q; so 1 2 L: Thus, 1 2 p 3 2 L; that is p 3 2 L: 2 2 So, since p 3 2 L and p 2 + p 3 2 L; we have p 2 + p 3 p 3 2 L: So, p 2 2 L: Therefore, L is a eld which contains Q; p 2 and p 3: So, since K is the smallest eld which contains Q; p 2 and p 3; we have K L: Therefore, K = L: That is Q p 2; p 3 = Q p 2 + p 3 : 5

6 3 Algebraic Extensions De nition 21 Let E be an extension eld of a eld F and let a 2 E: 1. We call a algebraic over F if a is the zero of some nonzero polynomial in F [x] : 2. If a is not algebraic over F; we say that it is transcendental over F: 3. E is an algebraic extension of F if every element of E is algebraic over F: 4. Otherwise, E is called a transcental extension of F: Lemma 22 (Hermite 1873) e is transcendental over Q: Lemma 23 (Lindemann 1882) is transcendental over Q ( was found to be irrational in 1767 by Lambert.) Question: Is + e transcendental over Q? Example 24 Show 2 is transcendental over Q: Example 25 Show E = Q p 2 is an algebraic extension of Q: (First, you will need to prove Q p 2 = Q p 2 = a + b p 2 : a; b 2 Q :) Theorem 26 Let F be a eld and let E be an extension eld of F. Let a 2 E be algebraic over F: Then F (a) is isomorphic to F [x] = hpi where p 2 F [x] is an irreducible polynomial which has a as a zero. (Note: such a polynomial exists since a is algebraic over F:) Proof. The evaluation homomorphism a : F [x]! E has already been shown to be a homomorphism. It can be shown that ker a = hpi : (Homework) Therefore, by the FHT, a (F [x]) = F [x] = hpi : Since p is irreducible, F [x] = hpi is a eld. Thus, a (F [x]) is a eld. Recall that a (F [x]) = F [a] (by the external description of F [a] see theorem 15). So, since a (F [x]) is a eld, we have that F [a] is a eld, so F [a] = F (a) : (see theorem 17) Therefore, a (F [x]) = F (a) : So, F (a) is isomorphic to F [x] = hpi : 6

7 Corollary 27 Let F be a eld and let E be an extension eld of F. Let a 2 E be algebraic over F: Let p 2 F [x] be an irreducible polynomial which has a as a zero. If deg (p) = n; then every element of F (a) can be uniquely expressed in the form for some c 0 ; c 1 ; : : : ; c n 1 2 F: c 0 + c 1 a + c 2 a c n 1 a n 1 The corollary above follows from the fact that every element of F [x] = hpi can be written uniquely in the form f +hpi where deg (f) < n or f = 0: Corollary 28 Let F be a eld and let E be an extension eld of F. Let a 2 E be algebraic over F: Let p 2 F [x] be an irreducible polynomial which has a as a zero. If deg (p) = n; then [F (a) : F ] = n: Proof. The rst corollary proves that f1; a; a 2 ; : : : ; a n F (a) as a vector space over F: Thus, [F (a) : F ] = n: 1 g forms a basis for Example 29 Compute Q 3p 2 : Q : Example 30 Compute [Q (i) : Q] : Example 31 Compute Q p i : Q : Example 32 Compute h p p i Q : Q : Example 33 Compute Q p 2 + p 3 : Q : Theorem 34 Let F be a eld and let E be an extension eld of F: Let 2 E where is algebraic over F: Then there is a unique monic irreducible polynomial p 2 F [x] such that is a root of p: Moreover, this polynomial p is of minimal degree of those polynomials in F [x] which have as a root. Finally, if f is a nonzero polynomial in F [x] such that f () = 0; then p divides f: 7

8 Proof. Let F be a eld and let E be an extension eld of F: Let 2 E where is algebraic over F: Let : F [x]! E be the evaluation homomorphism. Then ker = ff 2 F [x] : is a root of fg : We know that ker is an ideal in F [x] : Also, F [x] is a PID. Therefore, ker = (g) for some g 2 F [x] which g is of minimal degree in ker : Let c be the leading coe cient of g: Let p = c 1 g: Then p is a monic polynomial..also, p is an associate of g; so (p) = (g) = ker : Therefore, p is of minimal degree in ker : So, p is of minimal degree of those polynomials in F [x] which have as a root. Suppose f is a nonzero polynomial in F [x] such that f () = 0: Then f 2 ker : So, f 2 (p). Therefore, p divides f: To see that p is irreducible, suppose p = hk for some h; k 2 F [x] where neither h nor k is a unit in F [x] : Then h and k both have degree less than that of p: Therefore, h; k 62 (p) = ker : However 0 = a (p) = a (hk) = a (h) a (k) : So, a (h) = 0 or a (k) = 0: That is, h 2 ker or k 2 ker : This is a contradiction, so p is irreducible. Finally, suppose there is another monic irreducible polynomial q 2 F [x] such that is a root of q: Then q 2 ker ; so p divides q: So, q = pr for some r 2 F [x] : Since q is irreducible and p is not a unit, this means r is a unit. Therefore, p and q are associates. Since p and q are both monic, this means p = q and we have completed the proof. De nition 35 Let F be a eld and let E be an extension eld of F: Let 2 E whe is algebraic over F: The unique monic irreducible polynomial in F [x] which has as a root (as described in the previous theorem) is called the irreducible polynomial for in F or the minimal polynomial for in F: The notation is irr(; F ) or min (; F ). Remark 36 In theorem 26 and the corollaries that follow it, the polynomial p may be chosen to be irr(; F ) : Theorem 37 Let F be a eld and let E be an extension eld of F: Let a 2 E: Then [F (a) : F ] is nite if and only if a is algebraic over F: Proof. Let F be a eld and let E be an extension eld of F: Let a 2 E: 8

9 ( =) ) : Suppose [F (a) : F ] is nite. Let n = [F (a) : F ] : Consider f1; a; a 2 ; : : : ; a n g F (a) : Since [F (a) : F ] = n; we have that F (a) is an n-dimensional vector space over F: Thus, since the set f1; a; a 2 ; : : : ; a n g has more than n elements, it must be a linearly dependent set. Thus, there exist a linear combination c c 1 a + c 2 a c n a n = 0 with each c i 2 F and not all of the c i equal to zero. Thus, the polynomial f = c 0 + c 1 x + c 2 x c n x n 2 F [x] is nonzero and has a as a zero. So, by de nition, a is algebraic over F: ((=) : Suppose a is algebraic over F: Then, by de nition, there is an irreducible polynomial, f; in F [x] which has a as a zero. So, by the previous corollary, [F (a) : F ] = deg (f) is nite. Theorem 38 Let E be an extension eld of the eld F: If [E : F ] < 1; then E is algebraic over F: Proof. Let E be an extension eld of the eld F: Suppose [E : F ] = n is nite. Recall: to show that E is algebraic over F; we must show that every element of E is algebraic over F: Let a 2 E: Since [E : F ] = n; we have that E is an n-dimensional vector space over F: Thus, since the set f1; a; a 2 ; : : : ; a n g has more than n elements, it must be a linearly dependent set. Thus, there exist a linear combination c c 1 a + c 2 a c n a n = 0 with each c i 2 F and not all of the c i equal to zero. Thus, the polynomial f = c 0 + c 1 x + c 2 x c n x n 2 F [x] is nonzero and has a as a zero. So, by de nition, a is algebraic over F: Therefore, E is an algebraic extension of F: Note: the converse of this theorem is not true. For example, every element of E = Q p 2; 3p 2; 4p 2; 5p 2; : : : is algebraic over F = Q; however [E : F ] = 1: Theorem 39 Let K be an extension eld of E and let E be an extension eld of F: If [K : E] and [E : F ] are both nite, then [K : F ] = [K : E] [E : F ] : (note: if either of [K : E] or [E : F ] are in nite, then [K : F ] = 1:) 9

10 Proof. Let [K : E] = n and [E : F ] = m: Then there is a basis fk 1 ; k 2 ; : : : ; k n g for K over E and a basis fa 1 ; a 2 ; : : : ; a m g for E over F: Claim: B = fk i a j : 1 i n and 1 j mg is a basis for K over F: Let x 2 K: Then x = b 1 k 1 + b 2 k b n k n for some b i 2 E: Also, each b i 2 E can be written as b i = c i1 a 1 + c i2 a c im a m for some c ij 2 F: Substituting this linear combination for each b i in the rst equation, we get that x is a linear combination of elements of B: Thus, B spans K: Now, suppose P n P m i=1 j=1 r ijk i a j = 0 for some r ij 2 F: Then P n Pm i=1 j=1 r ija j k i = 0 where P m j=1 r ija j 2 E: So, since fk 1 ; k 2 ; : : : ; k n g is a basis over E (and thus linearly independent), we have P m j=1 r ija j = 0 for all i: Thus, since fa 1 ; a 2 ; : : : ; a m g is a basis over F; we have r ij = 0 for all j: Thus, r ij = 0 for all i and j: Therefore, B is linearly independent over F: Therefore, B is a basis for K over F: So, [K : F ] = jbj = nm = [K : E] [E : F ] Example 40 Show that p 5 + p 7 is algebraic over Q: One way to show this would be to nd a polynomial in Q [x] which has p 5 + p 7 as a zero. Here is another approach. We know p 5 + p 7 is algebraic over Q if and only if Q p 5 + p 7 : Q is nite, so we will simply argue that Q p 5 + p 7 : Q is nite. We know Q p 5 + p 7 Q p 5; p 7 ; so Q p 5 + p 7 : Q Q p 5; p 7 ; Q : Also, Q p 5; p 7 : Q is nite because Q p 5 : Q = 2 and Q p 5; p 7 : Q p 5 2; so Q p 5 + p 7 : Q 4: Theorem 41 Let F be a eld and let E be an extension eld of F: The elements of E which are algebraic over F form a eld (a sub eld of E): Proof. Homework. 10

11 4 Homework 1. Prove lemma 11: Let F be a eld and let E be an extension eld of F: Prove that E is a vector space over F: 2. Let I be a nonzero ideal in k [x] where k is a eld. Let p 2 I be an irreducible polynomial. Prove I = (p) : 3. Prove Q p 2 = a + b p 2 : a; b 2 Q : 4. Prove that Q p 2 = Q p 2 : 5. Determine, with proof, Q p 2 : Q : 6. Let K be a nite extension of Z p : Prove char(k) = p: 7. Let K be a eld. (a) Prove that it is not possible for K to have sub elds E and F such that E = Q and F = Z p for some prime p: (b) Prove that it is not possible for K to have sub elds E and F such that E = Z p and F = Z q for some primes p 6= q: 8. Let = 3 q p q p : (a) Verify that p 3 3 = p 3: (b) Determine [Q () : Q] : p p 9. Find a polynomial p in Q [x] so that Q is isomorphic to Q [x] = hpi : 10. Let f = x 3 +x Z 2 [x] : Suppose a is a zero of f in some extension of Z 2 : Then Z 2 (a) is isomorphic to Z 2 [x] = hfi and each element of Z 2 (a) can be written uniquely in the form c 0 +c 1 a+c 2 a 2 where c 0 ; c 1 ; c 2 2 Z 2 : Write each of the following elements of Z 2 (a) in this form. (a) a 3 (b) a 1 11

12 (c) a 50 Prove the following: (d) Q p 2 + p 3 p p = Q (hint: look at p 2 + p 3 2 ) (e) Suppose a+bi 2 C with a; b 2 R and b 6= 0: Prove that R (a + bi) = C: (f) Prove Q (i) 6= C: 11. Complete the following: (a) Compute Q p 2 : Q : (b) Compute Q p 3 : Q : (c) Prove Q p 2 6= Q p 3 : (d) Is Q p 2 isomorphic to Q p 3? (prove or disprove) 12. Prove the following: (a) is transcendental over Q (i) : (You may assume is transcendental over Q:) (b) is algebraic over Q ( 3 ) : 13. Compute the following degrees. (a) Q p 2 p 5 : Q (b) Q p 2 + p 5 : Q (c) Q p 2 + p 5 : Q p 10 (d) Q p 2; p 5; p 10 : Q (e) [Q () : Q ( 3 )] 14. Let F be a eld and let E and E 0 be two extension elds of F with E E 0 : (Hint: see vector spaces homework) (a) Prove [E 0 : F ] [E : F ] : (b) Suppose [E 0 : F ] is nite. Prove [E 0 : F ] = [E : F ] if and only if E 0 = E: 12

13 15. Let = 3p 2 + 3p 4: (a) Prove Q () Q 3p 2 : (b) Compute Q 3p 2 : Q : (c) Based on parts (a) and (b), what can you say (immediately) about [Q () : Q]? (d) Find a monic irreducible polynomial in Q [x] which has as a zero. (e) Compute [Q () : Q] : (f) Is Q () equal to Q 3p 2? 16. Let F be a eld and let E be an extension eld of F: Prove the following. (a) Let a; b 2 E: If a is algebraic over F and b is algebraic over F (a) ; then b is algebraic over F: (b) The elements of E which are algebraic over F form a sub eld of E: (c) Find an example of elements a and b which are transcendental over some eld F such that a + b is algebraic over F: 17. Let F be a eld and let E be an extension eld of F: Let a 2 E and suppose that a is algebraic over F: Also, suppose that c 0 +c 1 x+ +c n x n is an irreducible polynomial in F [x] which has a as a zero. Show that b = p a is algebraic over F by nding a polynomial in F [x] which has b as a zero. 18. Let F = Q and let E = Q p 2; 3p 2; 4p 2; : : : : (a) Prove that E is an algebraic extension of F: (Hint: show E is a sub eld of the eld of algebraic elements over Q:) (b) Prove that [E : F ] = 1: 19. Let F be a eld and let E be an extension eld of F: Suppose [E : F ] = p for some prime p: Show that F (a) = F of F (a) = E for all a 2 E: 20. Let F be a eld and let E be an extension eld of F: Let a; b 2 E: Prove [F (a; b) ; F (a)] [F (b) ; F ] : 13

14 21. Let F be a eld and let E be an extension eld of F: Let a; b 2 E: Suppose [F (a) : F ] = m and [F (b) : F ] = n where m and n are relatively prime. Prove that [F (a; b) : F ] = mn: 14

15 EXTENSION FIELDS Part II (Supplement to Section 3.8) 5 Kronecker s Theorem Theorem 42 (Kronecker s Theorem) Let F be a eld and let f be a nonconstant polynomial in F [x] : Then, there is an extension eld E of F in which f has a zero. Proof. We are actually going to end up building an extension eld E 0 (we ll still call it E) of a eld F 0 (where F 0 is isomorphic to F ): Let F be a eld and let f be a nonconstant polynomial in F [x] : We may assume that f is irreducible over F: (If f is reducible, choose one of its irreducible factors.) Let E = F [x] = hfi : Notice that E is a eld since f is irreducible over F: Let F 0 = fa + hfi : a 2 F g E: Then F 0 is a sub eld of E isomorphic to F. The isomorphism is : F! F 0 de ned by (a) = a + hfi for all a 2 F: (Homework) Now, we need to show that E contains an element which is a zero of f: If f = a n x n + a n 1 x n a 0 2 F [x] ; then under the isomorphism above, we are looking at f = (a n + hfi) x n + (a n 1 + hfi) x n (a 0 + hfi) 2 F 0 [x] : In E, consider the element = x + hfi : Then f () = (a n + hfi) n + (a n 1 + hfi) n (a 0 + hfi) = (a n + hfi) (x + hfi) n + (a n 1 + hfi) (x + hfi) n (a 0 + hfi) = (a n + hfi) (x n + hfi) + (a n 1 + hfi) x n 1 + hfi + + (a 0 + hfi) = (a n x n + hfi) + a n 1 x n 1 + hfi + + (a 0 + hfi) = a n x n + a n 1 x n a 0 + hfi = f + hfi = 0 E : Therefore, E is an extension eld of F (actually F 0 ; an isomorphic copy of F ) which contains a zero for f: 15

16 Example 43 The eld R [x] = hx 2 + 1i is an extension eld of (an isomorphic copy of) R which contains a zero for x 2 + 1: Corollary 44 (Kronecker s Theorem) Let F be a eld and let f be a nonconstant polynomial in F [x] : Then, there is an extension eld E of F in which f can be factored as a product of linear polynomials. That is, f = a (x 1 ) (x 2 ) (x n ) for some a 2 F; 1 ; 2 ; : : : ; n 2 E: 6 Splitting Fields De nition 45 Let F be a sub eld of a eld E and let f 2 F [x] : 1. We say that f splits over E if f can be factored as a product of linear polynomials in E [x] : 2. We say that E is a splitting eld for f if f splits over E but does not split over any proper sub eld of E: Example 46 Let F = Q and let f = x 2 Q p 2 is a splitting eld for f: 2: Then f splits over R; however, Lemma 47 (Splitting Fields Exist) Let F be a eld and let f 2 F [x] : A splitting eld for f over F exists. Proof. By Kronecker s theorem, there is an extension eld E of F in which f splits. So, f = a (x 1 ) (x 2 ) (x n ) for some a 2 F; 1 ; 2 ; : : : ; n 2 E: Then F ( 1 ; 2 ; : : : ; n ) is a splitting eld for f: Theorem 48 (Fields of order p n exist) If p is prime and n is a positive integer, then there is a eld with exactly p n elements. Proof. Let q = p n and let g = x q x 2 Z p [x] : We already know that there exists a eld with exactly 2 elements, Z 2 : So, assume q > 2: By Kronecker s theorem, there is an extension eld K of Z p in which g splits. Notice that we can nd such a K such that K is a nite extension of Z p (see the construction in the proof of Kronecker s theorem). So, char(k) = p (see homework problem #6 from extension elds part I) 16

17 De ne E = f 2 K : g () = 0g : Notice that E is simply the set of all roots of g in K: Also, a 2 E i a q a = 0 i a q = a: The symbolic derivative of g is g 0 = qx q 1 1: So, g 0 = p n x q 1 1 = 1 (since we are in Z p [x]): Therefore, gcd (g; g 0 ) = 1: It can be shown that a polynomial, f; has no repeated roots i gcd (f; f 0 ) = 1: (Homework). Therefore, g has no repeated roots. This means, g has precisely q = p n roots. That is jej = q = p n : Claim 49 E is a eld (and thus, an example of a eld of order p n ): We know jej = p n : Thus, E 6= ;: Let a; b 2 E: Then a q = a and b q = b: Thus, (ab) q = a q b q = ab: So, ab 2 E: Thus, E is closed under multiplication. Also, by applying the binomial theorem and using the fact that char(k) = p; we have (a b) q = a q b q = a b: Thus, E is closed under subtraction. Finally, suppose a 6= 0 (we must show that a 1 2 E): Since a 2 E; we have a q = a: So, since a 6= 0; we have a q 1 = 1; by cancellation. Thus, (a) (a q 2 ) = 1: So, a 1 = a q 2 : Since q > 2; we have q 2 > 0; so, since E is closed under multiplication, a q 2 2 E: Thus, a 1 2 E: Therefore, E is a eld with exactly p n elements. Notation 50 A nite eld of order q = p n is denoted F q : Question: Does this notation make sense? What if there are non-isomorphic elds of order q? De nition 51 Let E be a nite eld. Recall that the multiplicative group E = U (E) is a cyclic group. An element 2 E is called a primitive element of E if is a generator for the cyclic group E : Corollary 52 For every prime p and every positive integer n; there exists an irreducible polynomial g 2 Z p [x] with deg (g) = n: 17

18 Proof. Let E be an extension eld of Z p with p n elements. Let be a primitive element of E: It is easy to show that Z p () = E: By theorem 26, Z p () = Z p [x] = hgi where g is an irreducible polynomial in Z p [x] with as a root. Also, by a corollary to theorem 26, [Z p () : Z p ] = deg (g) : However, [Z p () : Z p ] = [E : Z p ] = n; since jej = n: Therefore, deg (g) = n: 7 Homework 1. Prove the following fact used in the proof of Kronecker s Theorem (theorem 42): Let F be a eld and let p be an irreducible polynomial in F [x] : Show that fa + hpi : a 2 F g is a sub eld of F [x] = hpi which is isomorphic to F: 2. Prove the following fact used in the proof of theorem 26: Let p be an irreducible polynomial in F [x] : Suppose E is an extension eld of F and that a 2 E such that a (p) = 0 where a : F [x]! E is an evaluation homomorphism. Prove that ker a = hpi : 3. Let f = x Z 7 [x] : Let I = hfi : Let R be the factor ring Z 7 [x] =I: (a) Prove that every element of R can be written in the form p + I where p is an element of Z 7 [x] and deg (p) 2 or p = 0: (b) Prove that if p and q are elements of Z 7 [x] and deg (p) 2 and deg (q) 2 and p + I = q + I; then p = q: (c) Use parts (a) and (b) to explain why R has 343 = 7 3 elements. (d) Prove that R is a eld. (e) Let a be a zero of f in some extension eld of Z 7 : How do we know such an a exists? (f) Use parts (a) and (b) to determine a description of the elements of Z 7 (a) : (g) Determine, with explanation, the number of elements in Z 7 (a) : (h) Determine [Z 7 (a) : Z 7 ] : 18

19 EXTENSION FIELDS Part III (Supplement to Section 3.8) 8 Splitting Fields Are Unique Remark 53 Our goal now is to show that the notation F q makes sense, that is, that there is exactly one eld of order q for each q = p n : The major step in doing this is to show that splitting elds are unique. Lemma 54 Let k and k 0 be elds and let : k! k 0 be an isomorphism. Let : k [x]! k 0 [x] be the induced mapping g (x) = a 0 + a 1 x + + a n x n 7! g (x) = (a 0 ) + (a 1 ) x + + (a n ) x n : Then is an isomorphism.and (r) = (r) for all r 2 k: Furthermore, if g is irreducible in k [x] ; then g is irreducible in k 0 [x] : Proof. Homework. Lemma 55 Let k and k 0 be elds and let : k! k 0 be an isomorphism. Let f 2 k [x] : Then, there is an isomorphism : k [x] = (f)! k 0 [t] = (f ) such that (r + (f)) = (r) + (f ) for each r 2 k and (x + (f)) = t + (f ) : Proof. The isomorphism is p + (f) 7! p + (f ) : Theorem 56 Let k and k 0 be elds and let : k! k 0 be an isomorphism. Let p be an irreducible polynomial in k [x] and let z be a root of p in some extension of k: Then k (z) is isomorphic to k 0 (z 0 ) where z 0 is a root of p in some extension of k 0 : Moreover, this isomorphism, ; satis es the following: 1. (z) = z 0 ; and 2. (a) = (a) for every a 2 k: Proof. Let z be a root of p 2 k [x] where z lies in some extension eld K of k: Let I = ff 2 k [x] : f (z) = 0g : It is easy to show that I is a proper ideal in k [x] : Notice p 2 I: So, since p is irreducible, I = (p) : (See homework problem #2 from part I.) 19

20 The evaluation homomorphism z : k [x]! k (z) is surjective (onto k (z)) and ker ( z ) = I = (p) : So, k (z) = k [x] = (p) : The isomorphism is : k [x] = (p)! k (z) where (g + (p)) = g (z) : By lemma 54, p is irreducible in k 0 [t] : Therefore, as above, we have an isomorphism : k 0 [t]= (p )! k 0 (z 0 ) given by (f + (p )) = f (z 0 ) : Similarly, there is an isomorphism : k 0 [t]= (p ) By lemma 55, there is an isomorphism : k [x] = (p)! k 0 [t] =(p ) such that (r + (p)) = (r) + (p ) for every r 2 k: We now have k (z) 1! k [x] = (p)! k 0 [t]= (p )! k 0 (z 0 ) : The isomorphism we are looking for is = 1 : The function is a composition of isomorphism, so it is an isomorphism. Also, it can be easily veri ed that has the desired properties. Lemma 57 Let f 2 k [x] where k is a eld and let E be a splitting eld for f over k: Let : k! k 0 be an isomorphism of elds, let : k [x]! k 0 [x] be the induced isomorphism g (x) = a 0 + a 1 x + + a n x n 7! g (x) = (a 0 ) + (a 1 ) x + + (a n ) x n ; and let E 0 be a splitting eld for f over k 0 : Then, there is an isomorphism : E! E 0 extending : That is j k = : Proof. Let d = [E : k] : We will prove this theorem by induction on d: Suppose d = 1: Then E = k and f is a product of linear polynomials in k [x] : It follows that f is a product of linear polynomials in k 0 [x] and thus E 0 = k 0 : We can let = : If d > 1; then f has a root z such that z 2 E and z 62 k: Let p be an irreducible polynomial in k [x] which has z as a root. Since z 62 k; we have deg (p) > 1: Also, by one of the corollaries to theorem 26, we have [k (z) : k] = deg (p) : Let z 0 2 E 0 be a root of p 2 k 0 [x] : From theorem 56, we have the isomorphism : k (z)! k 0 (z 0 ). Note: j k = : Note f 2 k [x] ; so, since k k (z) ; we have f 2 (k (z)) [x] : Then E is a splitting eld for f over k (z) : 20

21 Similarly, E 0 is a splitting eld for f over k (z 0 ) : Note [E : k (z)] [k (z) : k] = [E : k] : So, since [k (z) : k] > 1; we have [E : k] > [E : k (z)] : Let F = k (z) and let F 0 = k 0 (z 0 ) : Then, we now have the following: 1. f 2 F [x] 2. E is the splitting eld for f over F: 3. : F! F 0 is an isomorphism of elds 4. E 0 is the splitting eld for f over F 0 5. [E : F ] < d So, by the inductive hypothesis, we have that there is an isomorphism : E! E 0 extending : Since extends ; this means that extends and we have proven the result. Theorem 58 (Uniqueness of splitting elds) Let k be a eld and let f 2 k [x] : Then, any two splitting elds of f over k are isomorphic via an isomorphism which xes k: Proof. Use the lemma above with : k! k being the identity isomorphism. Corollary 59 Any two nite elds having exactly p n elements are isomorphic. Proof. Let E be a eld with q = p n elements. Then, the multiplicative group E = U (E) has q 1 elements. So, by Lagrange s theorem a q 1 = 1 for every a 2 E : Thus, every element of E is a root of f = x q x 2 Z p [x] and so E is a splitting eld for f over Z p : So, by the uniqueness of splitting elds, this corollary is proven. 21

22 9 Homework 1. Prove that Z 3 [x] = (x 3 x 2 + 1) = Z 3 [x] = (x 3 x 2 + x + 1) 2. Write addition and multiplication tables for F 8 ; the eld with 8 elements. 3. Is F 4 a sub eld of F 8? (Prove or disprove). 4. For any prime p; prove that if F p n is a sub eld of F p m; then njm: 22