# calculating the result modulo 3, as follows: p(0) = = 1 0,

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1 Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, , , (9.4.14), Determine whether the following polynomials are irreducible in the rings indicated. For those that are reducible, determine their factorization into irreducibles. The notation F p denotes the finite field Z/pZ, p a prime (a) x 2 + x + 1 in F 2 [x] By Prop the quadratic polynomial p(x) = x 2 +x+1 is irreducible in F 2 [x] iff p(x) has no root in F 2. This we check by substituting all the elements of the field F 2 (there are only 0 and 1) into p(x) and calculating the result modulo 2, as follows: p(0) = = 1 0 and p(1) = = = 1 0. Since p(x) has no roots in F 2, p(x) is irreducible (b) x 3 + x + 1 in F 3 [x]. By Prop the cubic polynomial q(x) = x 3 + x + 1 is irreducible in F 3 [x] iff p(x) has no root in F 3. This we check by substituting all the elements of the field F 3 (there are only 0, 1, and 2) into p(x) and calculating the result modulo 3, as follows: p(0) = = 1 0, p(1) = = = 0, and p(2) = = = 11 = 2 0. Since p(x) has root 1 in F 2, p(x) has linear factor x 1 and factorization p(x) = (x 1)(x 2 + x 1) = (x + 2)(x 2 + x + 2) (note 1 = 2 mod 3) (c) x in F 5 [x]. Let p(x) = x By Prop we can show that p(x) has no linear factor by substituting the elements 0, 1, 2, 3, 4 of F 5 in p(x) and calculating modulo 5: p(0) = 1 0, p(1) = = 2 0, p(2) = = 17 = 2 0, p(3) = = 82 = 2 0, and p(4) = = = 257 = 2 0. It remains to consider factorizations into monic quadratic polynomials. Suppose p(x) = (x 2 + ax + b)(x 2 + cx + d) where a, b, c, d F 5. Then x = p(x) = x 4 + (a + c)x 3 + (b + ac + d)x 2 + (ad + bc)x + bd

2 2 so 0 = a + c, 0 = b + ac + d, 0 = ad + bc, and 1 = bd. From 0 = a + c we have c = a, so 0 = b + a( a) + d and 0 = ad + b( a), so a 2 = b + d and ad = ab. If a 0 then b = d by cancellation from ad = ab, so a 2 = 2b and b 2 = 1 from a 2 = b + d and 1 = bd. Now 1 2 = 1, 2 2 = 4, 3 2 = 9 = 4, 4 2 = 1, so b is 1 or 4. But then either a 2 = 2 or a 2 = 2(4) = 8 = 3, and neither of these equations has a solution in F 5 (since 1 2 = 1, 2 2 = 4, 3 2 = 4, 4 2 = 1). On the other hand, if a = 0 then 0 2 = 0 = b + d, so b = d, so b 2 = bd = 1 = 4, so b = 2, d = 2, a = 0, c = 0, giving a factorization p(x) = x = (x 2 + 2)(x 2 2) = (x 2 + 2)(x 2 + 3). Thus p(x) is a reducible polynomial in F 5 [x] (d) x x in Z[x]. This polynomial has no real-valued roots, since x 4 0 and x 2 0 for all x R, hence x x for all x R. This shows, by Prop. 9.10, that x 4 +10x 2 +1 has no linear factors in Z[x]. We consider factoring x x into quadratic polynomials. If a, b, c, d Z and x x = (x 2 + ax + b)(x 2 + cx + d) = x 4 + (a + c)x 3 + (b + ac + d)x 2 + (ad + bc)x + bd then 0 = a+c, b+ac+d = 10, ad+bc = 0, and bd = 1. From 0 = a+c we get c = a, so, by ad + bc = 0 we have ad ab = 0, hence a = 0 or d = b. If a = 0 then c = 0 and the factorization becomes x x = (x 2 + b)(x 2 + d) = x 4 + (b + d)x 2 + bd so b + d = 10 and bd = 1. These equations have no solution in Z. Indeed, we have either b = d = 1 or b = d = 1 by bd = 1, and neither choice satisfies the equation b + d = 10. If a 0 then b = d and the factorization becomes x x = (x 2 + ax + b)(x 2 ax + b) = x 4 + (2b a 2 )x 2 + b 2

3 where 2b = a and b 2 = 1. From 2b = a we get b 5, but from b 2 = 1 we get b = ±1, a contradiction. Thus there is no factorization into quadratic factors, and x x is irreducible over Z[x] Prove that the following polynomials are irreducible in Z[x] (a) x 4 4x This polynomial is irreducible by Eisenstein s Criterion, since the prime 2 does not divide the leading coefficient, does divide all the coefficients of the low order terms, namely 4, 0, 0, and 6, but the square of 2 does not divide the constant (b) x x 5 15x 3 + 6x 120 The coefficients of the low order terms are 30, 0, 15, 0, 6, and 120. These numbers are divisible by the prime 3, but 3 2 = 9 does not divide 120 = , so the polynomial is irreducible by Eisenstein s Criterion (c) x 4 + 6x 3 + 4x 2 + 2x + 1 [Substitute x 1 for x.] Let p(x) = x 4 + 6x 3 + 4x 2 + 2x + 1. Then to calculate p(x 1) we first note that (x 1) 4 = x 4 4x 3 + 6x 2 4x + 1 6(x 1) 3 = 6x 3 18x x 6 4(x 1) 2 = 4x 2 8x + 4 2(x 1) = 2x 2 1 = 1 Then we add these up and get p(x 1) = (x 1) 4 + 6(x 1) 3 + 4(x 1) 2 + 2(x 1) + 1 = x 4 + 2x 3 8x 2 + 8x 2 Now x 4 + 2x 3 8x 2 + 8x 2 is irreducible by Eisenstein s Criterion, since the prime 2 divides all the lower order coefficients, but 2 2 = 3

4 4 4 does not divide the constant 2. Any factorization of p(x) would provide a factorization of p(x 1) as well, since if p(x) = a(x)b(x) then x 4 + 2x 3 8x 2 + 8x 2 = p(x 1) = a(x 1)b(x 1), contradicting the irreducibility of p(x 1). Therefore p(x) is itself irreducible in Z[x] (d) (x+2)p 2 p x, where p is an odd prime, in Z[x]. To express (x+2)p 2 p x as a polynomial we expand (x + 2) p according to the binomial theorem. The final constant 2 p cancels with 2 p, so every remaining term has x as a factor, which we cancel, leaving ( ( ) ( ) x p 1 p + 2 x 1) p p x 2 p p 1 p p 1 Every lower order coefficient in this polynomial has the form ( ) 2 k p x k p k 1 = 2 k p (p 1) (p k 1) with 0 < k < p. Each lower order coefficient has p as a factor but does not have p 2 as a factor, so the polynomial is irreducible by Eisenstein s Criterion Find all the monic irreducible polynomials of degree 3 in F 2 [x], and the same in F 3 [x]. All the monic linear (degree 1) polynomials x a are irreducible. For F 2 these irreducible linear polynomials are x and x 1(= x + 1). For F 3 these irreducible linear polynomials are x, x 1(= x + 2), and x 2(= x + 1). In F 2 [x] the only (monic) quadratic polynomials are x 2, x 2 +1, x 2 +x, and x 2 + x + 1. Obviously, x 2 and x 2 + x are reducible since x 2 = xx and x 2 + x = x(x + 1). Less obviously, x has 1 as a root, which leads to the factorization x = (x + 1)(x + 1). Finally, x 2 + x + 1 has no root in F 2 since = 1 0 and = 1 0, so, because it is quadratic, it is irreducible. In F 3 [x] the nine monic quadratic polynomials are x 2, x 2 + 1, x 2 + 2, x 2 + x, x 2 + 2x, x 2 + x + 1, x 2 + 2x + 1, x 2 + x + 2, and x 2 + 2x + 2. Obviously, x 2, x 2 + x, and x 2 + 2x are reducible. Since the remaining

5 polynomials x 2 + 1, x 2 + 2, x 2 + x + 1, x 2 + 2x + 1, x 2 + x + 2, and x 2 + 2x + 2 are quadratic, they are reducible iff they have a root in F 3, so we check each of them for roots: x = = = 2 x = = = 0 x 2 + x = = = 1 x 2 + 2x = = = 0 x 2 + x = = = 2 x 2 + 2x = = = 1 Since x has 1 and 2 as roots, we get the factorization x = (x 1)(x 2) = (x + 2)(x + 1). Since x 2 + x + 1 has 1 as a (multiple) root, we get a factorization x 2 + x + 1 = (x 1)(x 1) = (x + 2)(x + 2). Finally, x + 2x + 1 = (x + 1)(x + 1). The remaining polynomials, which are the only irreducible quadratic polynomials in F 3 [x], are x 2 +1 (used below), x 2 + x + 2, and x 2 + 2x + 2. There are quite a few monic cubic polynomials, so I thought it best to write some code in GAP to do the computations. The following GAP code computes, counts, and prints the irreducible cubic polynomials over the field F p, for p ranging over the first 12 primes. irrcubic:=function(p) local f,c,t,roots,irr; f:=function(c,p) return (c[4]^3+c[1]*c[4]^2+c[2]*c[4]+c[3]) mod p; end; t:=tuples([0..p-1],4); roots:=set(filtered(t,c->f(c,p)=0),c->c{[1..3]}); irr:=difference(tuples([0..p-1],3),roots); Print("The number of irreducible monic cubic "); Print("polynomials over F_",p," is\t",size(irr)); # Print("The irreducible monic cubic "); # Print("polynomials over F_",p," are\n"); # for c in irr do # Print("x^3"); 5

6 6 # if c[1]=1 then Print(" + x^2");fi; # if c[1] in [2..p-1] then Print(" + ",c[1],"x^2");fi; # if c[2]=1 then Print(" + x");fi; # if c[2] in [2..p-1] then Print(" + ",c[2],"x");fi; # if c[3]=1 then Print(" + 1");fi; # if c[3] in [2..p-1] then Print(" + ",c[3]);fi; # Print("\n"); # od; Print("\n"); return irr; end; LogTo("roots.output"); for p in Primes{[1..12]} do irrcubic(p); od; Running the GAP code without printing the polynomials produces the data in the following table, which shows the number n p of irreducible monic cubic polynomials over F p for the first 12 primes. p n p By uncommenting the Print commands we get lists of irreducible monic cubic polynomials. The 2 irreducible monic cubic polynomials in F 2 [x] are x 3 + x + 1 x 3 + x The 8 irreducible monic cubic polynomials over F 3 [x] are x 3 + 2x + 1 x 3 + x 2 + 2x + 1 x 3 + 2x + 2 x 3 + 2x x 3 + x x 3 + 2x 2 + x + 1 x 3 + x 2 + x + 2 x 3 + 2x 2 + 2x + 2

7 7 The 40 irreducible monic cubic polynomials over F 5 are x 3 + x + 1 x 3 + 2x 2 + 2x + 2 x 3 + x + 4 x 3 + 2x 2 + 2x + 3 x 3 + 2x + 1 x 3 + 2x 2 + 4x + 2 x 3 + 2x + 4 x 3 + 2x 2 + 4x + 4 x 3 + 3x + 2 x 3 + 3x x 3 + 3x + 3 x 3 + 3x x 3 + 4x + 2 x 3 + 3x 2 + x + 1 x 3 + 4x + 3 x 3 + 3x 2 + x + 2 x 3 + x x 3 + 3x 2 + 2x + 2 x 3 + x x 3 + 3x 2 + 2x + 3 x 3 + x 2 + x + 3 x 3 + 3x 2 + 4x + 1 x 3 + x 2 + x + 4 x 3 + 3x 2 + 4x + 3 x 3 + x 2 + 3x + 1 x 3 + 4x x 3 + x 2 + 3x + 4 x 3 + 4x x 3 + x 2 + 4x + 1 x 3 + 4x 2 + x + 1 x 3 + x 2 + 4x + 3 x 3 + 4x 2 + x + 2 x 3 + 2x x 3 + 4x 2 + 3x + 1 x 3 + 2x x 3 + 4x 2 + 3x + 4 x 3 + 2x 2 + x + 3 x 3 + 4x 2 + 4x + 2 x 3 + 2x 2 + x + 4 x 3 + 4x 2 + 4x Construct fields of each of the following orders: (a) 9, (b) 49, (c) 8, (d) 81. (you may exhibit these as F [x]/(f(x)) for some F and f). [Use Exercises 2 and 3 in Section 2.]

8 8 By Exercise 9.2.3, F [x]/(f(x)) is a field when f F [x] is irreducible, and by Exercise 9.2.2, F [x]/(f(x)) has q n elements if F = q and deg(f) = n. (a) To get a field of order 9, we choose F = F 3 so that q = F 3 = 3, and we choose an irreducible quadratic, so that n = 2 and the order of the field F [x]/(f(x)) is q n = 3 2 = 9, as desired. For f we may use x 2 + 1, which is an irreducible quadratic polynomial in F 3 [x] by Exercise 9.4.5, solved above. (b) To get a field of order 49, we choose F = F 7 so that q = F 7 = 7, and we let f be any irreducible quadratic in F 7 [x]. Then n = 2 and the order of the field F [x]/(f(x)) is q n = 7 2 = 49. One choice that works is f(x) = x By the way, the quadratic irreducible polynomials in F 7 [x] are x x x x 2 + x + 3 x 2 + x + 4 x 2 + x + 6 x 2 + 2x + 2 x 2 + 2x + 3 x 2 + 2x + 5 x 2 + 3x + 1 x 2 + 3x + 5 x 2 + 3x + 6 x 2 + 4x + 1 x 2 + 4x + 5 x 2 + 4x + 6 x 2 + 5x + 2 x 2 + 5x + 3 x 2 + 5x + 5 x 2 + 6x + 3 x 2 + 6x + 4 x 2 + 6x + 6 (c) To get a field of order 8, we choose F = F 2 so that q = F 2 = 2, and we choose an irreducible cubic for f, so that n = 3 and the order of the field F [x]/(f(x)) is q n = 2 3 = 8. For f we may use x 3 + x + 1 or x 3 + x (the two irreducible monic cubics in F 2 [x], as shown above). (d) To get a field of order 81, we choose F = F 3 so that q = F 3 = 3, and we choose an irreducible quartic polynomial f in F 3 [x], so that n = 4 and the order of the field F [x]/(f(x)) is q n = 3 4 = 81. Let f(x) = x 4 + x + 2. Then f has no linear factors because it has no roots in F 3, since = 2, = 1, and = 2.

9 Suppose f(x) = (x 2 + ax + b)(x 2 + cx + d) where a, b, c, d F 3. Then x 4 + x + 2 = x 4 + (a + c)x 3 + (b + ac + d)x 2 + (ad + bc)x + bd, which implies that 0 = a + c, 0 = b + ac + d, 1 = ad + bc, and 2 = bd. From 0 = a + c we have c = a. Substituting a for c in the equations 0 = b + ac + d and 1 = ad + bc (and simplifying) yields a 2 = b + d and 1 = a(d b). By the latter equation, a = a 2 (d b), so by a 2 = b + d we have a = (d b)(b + d) = d 2 b 2. From bd = 1 we know b 0 and d 0. The only nonzero element that is a square in F 3 is 1, since 1 2 = 1 and 2 2 = 1 in F 3. Therefore a = b 2 d 2 = 1 1 = 0, but then 1 = a(d b) = 0(d b) = 0, a contradiction. This shows that f is actually irreducible, so F 3 [x]/(f) is a field with 81 elements in it Show that R[x]/(x 2 + 1) is isomorphic to the field of complex numbers. Let p(x) = x and let P = (x 2 + 1) be the principal ideal of R[x] generated by p(x). Obviously p(x) has no real roots since, for every r R, r Since p(x) is quadratic and has no real roots, p(x) is irreducible in R[x] by Prop Therefore R[x]/P is a field by Exercise (or the upcoming Prop. 9.15). The element x = x + P R[x]/P is a solution to x 2 = 1 and plays the same role as the imaginary i = 1 in C, because x 2 = (x + P ) 2 = x 2 + P = x 2 (x 2 + 1) + P = 1 + P. The elements of R[x]/P are cosets of the form a + bx + P, a, b, R. Define a map ϕ : R[x]/P C by ϕ(a + bx + P ) = a + b 1. To show ϕ is well-defined, let us assume that a + bx + P = c + dx + P for some a, b, c, d R. Then a c+(b d)x P, hence a c+(b d)x = p(x)q(x) for some q(x) R[x]. If b d 0 then 1 = deg(a c + (b d)x) = deg(p(x)q(x)) = deg(p(x)) + deg(q(x)) = 2 + deg(q(x)) 2, a contradiction, so b = d. But then a c P, hence a c = p(x)q (x) for some q (x) R[x]. If a c 0 then 0 = deg(a c) = deg(p(x)q (x)) = deg(p(x)) + deg(q (x)) = 2 + deg(q (x)) 2, a contradiction, so a = c. 9

10 10 Thus, each element of R[x]/P has the form a + bx + P for uniquely determined reals a, b R. Next are calculations that show ϕ is a ring homomorphism because it preserves differences and products. ϕ((a + bx + P ) (c + dx + P )) = ϕ((a c) + (b d)x + P ) = (a c) + (b d) 1 = (a + b 1) (c + d 1) = ϕ(a + bx + P ) ϕ(c + dx + P ) ϕ((a + bx + P )(c + dx + P )) = ϕ((a + bx)(c + dx) + P ) = ϕ(ac + (ad + bc)x + bdx 2 + P ) = ϕ(ac + (ad + bc)x + bdx 2 + ( bd)(x 2 + 1) + P ) = ϕ(ac + (ad + bc)x + bdx 2 bdx 2 bd + P ) = ϕ(ac bd + (ad + bc)x + P ) = ac bd + (ad + bc) 1 = ac + (ad + bc) 1 + bd( 1) 2 = (a + b 1)(c + d 1) = ϕ(a + bx + P )ϕ(c + dx + P ) Finally, to see that ϕ is injective, we note that kernel of ϕ is trivial since if ϕ(a + bx + P ) = 0 then a + b 1 = 0, hence a = b = 0, so a + bx + P = P = 0 R[x]/P.

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