MATH 361: NUMBER THEORY TENTH LECTURE


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1 MATH 361: NUMBER THEORY TENTH LECTURE The subject of this lecture is finite fields. 1. Root Fields Let k be any field, and let f(x) k[x] be irreducible and have positive degree. We want to construct a superfield K of k in which f has a root. To do so, consider the quotient ring R = k[x]/ f, where f is the principal ideal f(x)k[x] of k[x]. That is, R is the usual ring of polynomials over k subject to the additional rule f(x) = 0. Specifically, the ringelements are cosets and the operations are (g + f ) + (h + f ) = (g + h) + f, (g + f )(h + f ) = gh + f. The fact that f is irreducible gives R the structure of a field, not only a ring. The only matter in question is multiplicative inverses. To see that they exist, suppose that g + f f in R. The condition is that g / f, i.e., f g. so (f, g) = 1 (because f is irreducible), and so there exist F, G k[x] such that F f + Gg = 1. That is, f Gg 1, i.e., Gg 1 f, so that Gg + f = 1 + f in R. That is, (G + f )(g + f ) = 1 + f in R, showing that G + f inverts g + f in R. Now use the field R to create a set K of symbols that is a superset of k and is in bijective correspondence with R. That is, there is a bijection σ : R K, σ(a + f ) = a for all a k. Endow K with addition and multiplication operations that turn the set bijection into a field isomorphism. The operations on K thus extend the operations on k. Name a particular element of K, r = σ(x + f ). 1
2 2 MATH 361: NUMBER THEORY TENTH LECTURE Then f(r) = f(σ(x + f )) by definition of r = σ(f(x + f )) since algebra passes through σ = σ(f(x) + f ) by the nature of algebra in R = σ( f ) by the nature of algebra in R = 0 by construction of σ. Thus K is a superfield of k containing an element r such that f(r) = 0. For example, since the polynomial f(x) = X 3 2 is irreducible over Q, the corresponding quotient ring R = Q[X]/ X 3 2 = {a + bx + cx 2 + X 3 2 : a, b, c Q} is a field. And from R we construct a field (denoted Q(r) or Q[r]) such that r 3 = 2. Yes, we know that there exist cube roots of 2 in the superfield C of Q, but the construction given here is purely algebraic and makes no assumptions about the nature of the starting field k to which we want to adjoin a root of a polynomial. 2. Splitting Fields Again let k be a field and consider a nonunit polynomial f(x) k[x]. We can construct an extension field k 1 = k(r 1 ), where r 1 satisfies some irreducible factor of f. Let We can construct an extension field f 2 (X) = f(x)/(x r 1 ) k 1 [X]. k 2 = k 1 (r 2 ) = k(r 1, r 2 ), where r 2 satisfies some irreducible factor of f 2. Continue in this fashion until reaching a field where the original polynomial f factors down to linear terms. The resulting field is the splitting field of f over k, denoted spl k (f). Continuing the example of the previous section, compute that X 3 2 X r = X2 + rx + r 2 in Q(r)[X]. Let s = rt where t 3 = 1 but t 1. Then, working in Q(r, t) we have s 2 + rs + r 2 = r 2 t 2 + r 2 t + r 2 = r 2 (t 2 + t + 1) = r 2 0 = 0, Thus s = rt satisfies the polynomial X 2 + rx + r 2, and now compute that That is, showing that X 2 + rx + r 2 = X rt 2 in Q(r, t)[x]. X rt X 3 2 = (X r)(x rt)(x rt 2 ) Q(r, t)[x], spl Q (X 3 2) = Q(r, t).
3 MATH 361: NUMBER THEORY TENTH LECTURE 3 We already know the finite fields 3. Examples of Finite Fields F p = Z/pZ, p prime. For any positive integer n we can construct the extension field K = spl Fp (X pn X). Furthermore, the roots of X pn X in K form a subfield of K. To see this, check that if a pn = a and b pn = b then (ab) pn = a pn b pn = ab, (a + b) pn = a pn + b pn = a + b, (a 1 ) pn = (a pn ) 1 = a 1 if a 0, ( a) pn = ( 1) pn a pn = a (even if p = 2). (The modulo p result that (a + b) p = a p + b p is sometimes called the freshman s dream.) Thus the splitting field K is exactly the roots of X pn X. The roots are distinct because the derivative (X pn X) = 1 is nonzero, precluding multiple roots. Thus K contains p n elements. We give it a name, F q = spl Fp (X pn X) where q = p n. In fact the fields 4. Exhaustiveness of the Examples F q, q = p n, n 1 are the only finite fields, up to isomorphism. To see this, let K be a finite field. The natural homomorphism Z K, n n 1 K has for its kernel an ideal I = nz of Z such that Z/I K, and so Z/I is a finite integral domain. This forces I = pz for some prime p, and so F p K. Identify F p with its image in K. Then K is a finitedimensional vector space over F p, so that K = p n for some n. Every nonzero element x K satisfies the condition x pn 1 = 1, and so every element x K satisfies the condition x pn = x. In sum, K = F q up to isomorphism where again q = p n.
4 4 MATH 361: NUMBER THEORY TENTH LECTURE 5. Containments of Finite Fields A natural question is: For which m, n Z + do we have F p m F p n? Assuming that the containment holds, the larger field is a vector space over the smaller one, and so p n = (p m ) d = p md for some d, showing that m n. Conversely, if m n then X pn 1 1 X pm 1 1 = p n 1 p m 1 1 i=0 X (pm 1)i. (Note that p m 1 p n 1 by the finite geometric sum formula; specifically, their quotient is n/m 1 j=0 p mj.) So and so In sum, X pm 1 1 X pn 1 1, F p m F p n. F p m F p n m n. For example, F 27 is not a subfield of F Cyclic Structure For any prime power q = p n, the unit group F q is cyclic. The proof is exactly the same as for F p = (Z/pZ). Either we quote the structure theorem for finitelygenerated abelian groups, or we note that for any divisor d of q 1, (q 1)/d 1 X q 1 1 = (X d 1) X di, and the first factor on the right side has at most d roots while the second factor has at most q 1 d roots. But the left side has a full contingent of q 1 roots in F q. Now factor q 1, q 1 = r er. For each prime factor r, X re 1 has r e roots and X re 1 1 has r e 1 roots, showing that there are φ(r e ) roots of order r e. Thus there are φ(q 1) elements of order q 1 in F q. That is, F q has φ(q 1) generators. 7. Examples To construct the field of 9 = 3 2 elements we need an irreducible polynomial of degree 2 over F 3. The polynomial X will do. Thus up to isomorphism, i=0 F 9 = F 3 [X]/ X Here we have created F 9 by adjoining a square root of 1 to F 3. To construct the field of 16 = 2 4 elements we need an irreducible polynomial of degree 4 over F 2. (Note that the principle no roots implies irreducible is valid only for polynomials of degree 2 and 3. For example, a quartic polynomial can have two quadratic factors.)
5 MATH 361: NUMBER THEORY TENTH LECTURE 5 The polynomial X 4 + X + 1 works: it has no roots, and it doesn t factor into two quadratic terms because (X 2 + ax + 1)(X 2 + bx + 1) = X 4 + (a + b)x 3 + abx 2 + (a + b)x + 1. Thus, again up to isomorphism, F 16 = F 2 [X]/ X 4 + X + 1 = F 2 (r) where r 4 + r + 1 = 0 = {a + br + cr 2 + dr 3 : a, b, c, d F 2 }. 8. A Remarkable Polynomial Factorization Fix a prime p and a positive integer n. To construct the field F p n, we need an irreducible monic polynomial of degree n over F p. Are there such? How do we find them? For any d 1, let MI p (d) denote the set of monic irreducible polynomials over F p of degree d. We will show that X pn X = f(x) in F p [X]. d n f MI p(d) To establish the identity, consider a monic irreducible polynomial f of degree d 1. Working in F p n, whose elements form a full contingent of roots of X pn X, we show that f(x) X pn X in F p [X] f(x) has a root in F p n. For the = direction, we have f(x)g(x) = X pn X for some g(x), and so f(α)g(α) = 0 for every α F p n. But deg(g) = p n deg(f) < p n, and so g(α) 0 for at least one such α, for which necessarily f(α) = 0. For the = direction, let α be a root of f(x) in F p n. Substitute α for X in the relation X pn X = q(x)f(x) + r(x) (where r = 0 or deg(r) < deg f) to get r(α) = 0. Thus α is a root of gcd(f(x), r(x)), but since f(x) is irreducible, this greatest common divisor is 1 unless r = 0. So r = 0 and thus f(x) X pn X as desired. With the displayed equivalence established, again let f be a monic irreducible polynomial of degree d 1. Adjoining any root of f to F p gives a field of order p d. Thus f(x) X pn X in F p [X] a field of order p d lies in F p n. By the nature of finite field containments, it follows that for any monic irreducible polynomial f over F p of degree d 1, f(x) X pn X in F p [X] d n. Also, no monic irreducible f(x) of degree d 1 can divide X pn X twice, because the derivative of the latter polynomial never vanishes. The desired product formula in the box follows. (This argument can be simplified if one works in a given algebraic closure of F p, but constructing the algebraic closure requires Zorn s Lemma.) To count the monic irreducible polynomials over F p of a given degree, take the degrees of both sides of the identity X pn X = f(x) in F p [X] d n f MI p(d)
6 6 MATH 361: NUMBER THEORY TENTH LECTURE to get p n = d n d MI p (d). By Möbius inversion, MI p (n) = 1 µ(n/d)p d n d n (where µ is the Möbius function). The sum on the right side is positive because it is a basep expansion with top term p n and the coefficients of the lower powers of p all in {0, ±1}. So MI p (n) > 0 for all n > 0. That is, there do exist monic irreducible polynomials of every degree over every field F p. For example, taking p = 2 and n = 3, X 8 X = X(X 7 1) = X(X 1)(X 6 + X 5 + X 4 + X 3 + X 2 + X + 1) = X(X 1)(X 3 + X 2 + 1)(X 3 + X + 1) mod 2, a product of two linear factors and two cubic factors. And the counting formula from the previous section gives the results that it must, MI 2 (1) = 1 1 µ(1)21 = 2, MI 2 (3) = 1 3 (µ(1)23 + µ(3)2 1 ) = 1 (8 2) = 2. 3 Similarly, taking p = 3 and n = 2, X 9 X = X(X 8 1) = X(X 4 + 1)(X 2 + 1)(X + 1)(X 1) = X(X 1)(X + 1)(X 2 + 1)(X 2 + X 1)(X 2 X 1) mod 3, and MI 3 (1) = 1 1 µ(1)31 = 3, MI 3 (2) = 1 2 (µ(1)32 + µ(3)3 1 ) = 1 (9 3) = Common Errors The finite field F q where q = p n is neither of the algebraic structures Z/qZ and (Z/pZ) n as a ring. As a vector space, F q = F n p, but the ring (multiplicative) structure of F q is not that of Z/qZ or of (Z/pZ) n. The finite field F p m is not a subfield of F p n unless m n, in which case it is. 10. Primes in Extensions We return to a question from the very first lecture: Does a given odd prime p factor or remain prime in the Gaussian integer ring Z[i]? Equivalently, is the quotient ring Z[i]/pZ[i] = Z[i]/ p an integral domain? Compute, Z[i]/ p Z[X]/ p, X F p [X]/ X Thus the question is whether X is reducible or irreducible in F p [X], which is to say whether the Legendre symbol ( 1/p) is 1 or 1. By Euler s Criterion,
7 MATH 361: NUMBER THEORY TENTH LECTURE 7 ( 1/p) = ( 1) (p 1)/2, so the 1 mod 4 primes p factor in Z[i] while the 3 mod 4 primes p don t. Returning to quotient ring structure, we have shown that Z[i]/ p F p [X]/ X r F p [X]/ X + r, p = 1 mod 4, where r 2 = 1 mod p. Take, for example, p = 5, and create two ideals of Z[i] modeled on the two polynomial ideals in the previous display with r = 2. I 1 = 5, i 2, I 2 = 5, i + 2. Their product is generated by the pairwise products of their generators, I 1 I 2 = 5 2, 5(i 2), 5(i + 2), 5. Each generator of I 1 I 2 is a multiple of 5 in Z[i], and 5 is a Z[i]linear combination of the generators. That is, the methods being illustrated here have factored 5 as an ideal of Z[i], I 1 I 2 = 5Z[i]. The previous example overlooks the elementwise factorization 5 = (2 i)(2 + i) willfully. Here is another example. For any odd prime p 19, Z[ 19]/ p Z[X]/ p, X 2 19 F p [X]/ X 2 19, and this this ring decomposes if (19/p) = 1. For example, (19/5) = 1 because 2 2 = 4 = 19 mod 5, and so we compute that in Z[ 19], 5, , = 5 2, 5( 19 2), 5( ), 15 = 5Z[ 19]. Here we have factored the ideal 5Z[ 19] without factoring the element 5 itself. Similarly, letting q be an odd prime and Φ q (X) the qth cyclotomic polynomial (the smallest monic polynomial over Z satisfied by ζ q ), compute for any odd prime p q, Z[ζ q ]/ p Z[X]/ p, Φ q (X) F p [X]/ Φ q (X). Thus if g Φ q (X) = ϕ i (X) ei in F p [X] then the Sun Ze Theorem gives the corresponding ring decomposition g Z[ζ q ]/ p F p [X]/ ϕ i (X) ei, i=1 i=1 and plausibly this decomposition somehow indicates the decomposition of p in Z[ζ q ]. Factoring Φ q (X) in F p [X] is a finite problem, so these methods finiteize the determination of how p decomposes in Z[ζ q ].
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