Homework 10 M 373K by Mark Lindberg (mal4549)

Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients that has each number as a root. I had fun doing this manipulation. x = 7 + 3 2 x 7 = 3 2 (x 7) 3 = 2. (This step adds more roots, but we don t care about them since our number is still a root) x 3 21x 2 + 147x 345 = 0. A quick check verifies that 7 + 3 2 is a root. x = 3 + 5. We ll first take the quadratic with this and its complex conjugate as its roots: (x 3 i 5)(x + 3 i 5) = x 2 2 3x + 8. Now I will find the quadratic that, when multiplied by this, gives integer coefficients. I found it by multiplying term by term, and figuring the next term based on which terms needed to be cancelled in each step. (x 2 2 3x + 8)(x 2 + 2 3x + 8) = x 4 2 3x 3 + 8x 2 + 2 3x 3 12x 2 + 16 3x + 8x 2 16 3x + 64 = x 4 + 4x 2 + 64. Again, a quick check verifies that 3 + 5 is a root. 2. Artin, Chapter 11, Exercise 3.3. Find generators for the kernels of the following maps: (a) R[x, y] R defined by f(x, y) f(0, 0). Let the map be φ. The kernel of φ will be all polynomials whose constant term is 0, as f(0, 0) is always the constant term. This can be expressed as ker(φ) = {xa + yb a, b R[x, y]}, and so the generator is {x, y}. (b) R[x] C defined by f(x) f(2 + i). Let the map be φ. The kernel of φ will be all polynomials f(2 + i) = 0, so with 2 + i as a root. Then the polynomial must have a factor of the form x (2 + i). These can be expressed as ker(φ) = {(x (2 + i))a a R[x]}. Thus, the generator is {x (2 + i)}. (c) Z[x] R defined by f(x) f(1 + 2). Let the map be φ. The kernel of φ will be all polynomials f(1 + 2) = 0, so with 1 + 2 as a root. To get a polynomial in Z[x], we must multiply this by its conjugate root, 1 2. Then the polynomial must have a factor of the form (x (1 + 2))(x (1 2)) = x 2 2x 1. These can be expressed as ker(φ) = {(x 2 2x 1)a a Z[x]}. Thus, the generator is {x 2 2x 1}. (d) Z[x] C defined by x 2 + 3. Let the map be φ. The kernel of φ will be all polynomials such that f(x) = f( 2 + 3) = 0. Thus, following the same procedure as problem 1, 2 + 3 = x x 3 = 2 (x 3) 2 = x 2 2 3x + 3 = 2 x 2 2 3x + 1 = 0. Then, to get a polynomial with coefficients in the integers, we must have, by again, the same procedure as above, (x 2 2 3x + 1)(x 2 + 2 3x + 1) = x 4 2 3x 3 + x 2 + 2 3x 3 12x 2 + 2 3x + x 2 2 3x + 1 = x 4 10x 2 + 1 as a factor. Thus, ker(φ) = {a(x 4 10x 2 + 1) a Z[x]}, and so the generator is {x 4 10x 2 + 1}. (e) C[x, y, z] C[t] defined by x t, y t 2, z t 3. Here, note that under the mapping x 2 y = t 2 t 2 = 0, and x 3 z = t 3 t 3 = 0, and so both of these polynomials are in the kernel of this transformation. Now we will kill them, to consider C[x, y, z]/(x 2 y, x 3 z). Now consider any polynomial f(x, y, z) in this. Then replace all y terms with x 2, and z terms with x 3. Then we can write f as a polynomial just in terms of the variable x. Then since x t, this gets sent to a polynomial simply in terms of t. Let it be g(t). Then if we want to be in the kernel, g(t) = 0, but 1

since we want this for all t, basic algebra (finite polynomials have a finite number of zeros, and here we have an infinite number) says that g(t) 0, and so the kernel only has 1 element now, the 0 polynomial. Thus, x 2 y, x 3 z generates the kernel of C[x, y, z] C[t] on the twisted cubic as given. 3. Artin, Chapter 11, Exercises 3.12 and 3.13. 3.12 Let I and J be ideals of a ring R. Prove that the set I + J of elements of the form x + y, with x I and y J, is an ideal. This ideal is called the sum of the ideals I and J. Since both I and J are non-empty, I + J must be nonempty, and a subset of R. Let a 1, a 2 I + J. Then x 1, x 2 I and y 1, y 2 J, such that a 1 = x 1 + y 1, and a 2 = x 2 + y 2. Then a 1 + a 2 = x 1 + x 2 + y 1 + y 2, and by definition x 1 + x 2 I and y 1 + y 2 J, and thus a 1 + a 2 I + J. Let r R. Then ra 1 = r(x 1 + y 1 ) = rx 1 + ry 1, and by the definition of ideal, rx 1 I and ry 1 J, so rx 1 + ry 1 I + J. Thus, I + J meets all of the conditions of an ideal, and so it is. 3.13 Let I and J be ideals of a ring R. Prove that the intersection I J is an ideal. Show by example that the set of product {xy x Y, y J} need not be an ideal, but that the set of finite sums x v y v of products of elements of I and J is an ideal. This ideal is called the product ideal and is denoted by IJ. Is there a relation between IJ and I J. 0 I and 0 J, and therefore, 0 I J, and therefore, I J is non-empty. Let x 1, x 2 I J. Then x 1, x 2 I and x 1, x 2 J. Thus, x 1 + x 2 I and x 1 + x 2 J, and x 1 + x 2 I J, and therefore the set is closed under addition. Also, rx 1 I and rx 1 J, by definition, and therefore, rx 1 I J, and all conditions are met. Therefore, I J is an ideal. Let the ring be R[x, y]. Let I and J both be (x, y) = {ax + by a, b R}. Then x I, J and y I, J, so x 2, y 2 {ij i I, j J}. Now consider x 2 + y 2. We know that the sum of squares does not factor over the reals, so a, b R[x, y] such that ab = x 2 + y 2, and therefore, it s even more impossible for a, b I, J. Thus, x 2 + y 2 {ij i I, j J}, and the set is not closed under addition, and therefore cannot be an ideal. Let I, J be ideals over a ring R, and define IJ as the set of finite sums as above. 0 I and 0 J, and so 0 0 = 0 IJ, so the set is non-empty. Now consider some element x IJ. Then x = i 1 j 1 + + i n j n for i 1,..., i n I and j 1,..., j n J. Then let r R, then rx = r(i 1 j 1 + + i n j n ) = (ri 1 )j 1 + + (ri n )j n, and because it is an ideal, ri 1,..., ri n I, and therefore, rx IJ. Now consider x 1, x 2 IJ. Then x 1 = i 1 j 1 + + i n j n, and x 2 = i n+1 j n+1 + + i m j m, m > n, for some i 1,..., i m I and j 1,..., j m J. Then x 1 + x 2 = i 1 j 1 + + i n j n + i n+1 j n+1 + + i m j m IJ, and so the set is closed under addition. Therefore, IJ meets all necessary conditions, and is an ideal. Let I and J be ideals of a ring R. Let x IJ. Then i 1,..., i n I and j 1,..., j n J such that x = i 1 j 1 + + i n j n. Then, since i 1, j 1 R, i 1 j 1 J, and i 1 j i J, since they are both ideals, and this extends for any arbitrary i a j a, 1 a n. Thus, i 1 j 1 + + i n j n I and i 1 j 1 + + i n j n J, since ideals are closed under addition, and so x I J. Thus, IJ I J. However, this does not work the other direction. Consider I, J and R as above, (x, y) and R[x, y]. x I and x J, so x I J. But 2

x cannot be expressed as a sum of products of terms, since it can only be x 1, and 1 I or J. Thus, I J IJ in general, and all we can say is that IJ I J. 4. Playing with finite fields: Recall that finite subgroups of the multiplicative subgroups of fields are cyclic. Given a ring R, if R is cyclic, call a generator of R a primitive root for R (this generalizes the number theory terminology when R = Z/n). By the above, when R = F is a finite field, F is cyclic, hence there exists a primitive root of F. Recall from class that F p k can be realized as F p [x]/p (x) F p [x] where P (x) is a polynomial of degree k with coefficients in F p which is irreducible. Problem: Find primitive roots for the fields F 8 = F2 [x]/(x 3 + x + 1) F[x] (with elements a+bx+cx 2 for a, b, c F 2 ) and for F 9 = F[x]/(x 2 +1) (with elements a+bx for a, b F 3 ). To do it, you just need to show that some polynomial has multiplicative order equal to 7 in the first case and 8 in the second case. Hint: I will do the baby case of F 4 in class on Thursday! F 8 = F2 [x]/(x 3 + x + 1) F 2 [x]: First, we must show that x 3 + x + 1 is irreducible in F 2 [x] = Z/2[x]. Consider the elements of Z/2 x 0 (mod 2): x 3 + x + 1 0 + 0 + 1 = 1 0 (mod 2). x 1 (mod 2): x 3 + x + 1 1 + 1 + 1 = 3 1 0 (mod 2). Therefore, there are no zeroes of x 3 + x + 1 in Z/2. By the result from problem 8, the only unit in F 2 [x] is 1, and therefore the polynomial x 3 +x+1 has neither any proper factors nor is a unit, and is therefore irreducible. Thus, the isomorphism holds, by the identity from class. Now we have that x 3 + x + 1 = 0, so x 3 + x = 1 and x 3 = 1 x. Then consider the element x, and take x 7. By an earlier homework, if x 7 1, then order(x) 7, so order(x) = 1 or 7, and x 1, so order(x) 1. Now note that x 7 = x x 3 x 3 = x( 1 x)( 1 x) = x 3 + 2x 2 + x = 1 1, because 2 0 and 1 1 in F 2. Therefore, x has order 7, and F 8 = (F 2 [x]/(x 3 + x + 1) F 2 [x]) has 7 elements, and therefore, we have found a primitive root, by definition. F 9 = F3 [x]/(x 2 + 1) F 3 [x]: First, we must show that x 2 + 1 is irreducible in F 3 [x] = Z/3[x]. Consider the elements of Z/3 x 0 (mod 3): x 2 + 1 0 + 1 = 1 0 (mod 3). x 1 (mod 3): x 2 + 1 1 + 1 = 2 0 (mod 3). x 2 (mod 3): x 2 + 1 4 + 1 = 5 2 1 0 (mod 3). Therefore, there are no zeroes of x 2 +1 in Z/2. By the result from problem 8, the only unit in F 3 [x] are 1 and 2, and therefore the polynomial x 2 + 1 has neither any proper factors nor is a unit, and is therefore irreducible. Thus, the isomorphism holds, by the identity from class. Now we have that x 2 + 1 = 0 or x 2 = 1. Consider the element x + 2. x + 2 1. (x + 2) 2 = x 2 + 4x + 4 = x 2 + 1 + x = x 1, (x + 2) 4 = x 2 = 1 1, and (x + 2) 8 = ( 1) 2 = 1, and since the only numbers which divide 8 are 1, 2, 4, and we showed that x + 2, (x + 2) 2, (x + 2) 4 are not 1, and so x + 2 has order 8. Since F 9 = F 3 [x]/(x 2 + 1) F 3 [x] has 8 elements, and therefore, we have found a primitive root, by definition. 3

5. Chinese Remainder Theorem revisited: Recall from Homework 5, problem 4 the isomorphism of groups Z/r Z/s = Z/rs when gcd(r, s) = 1. There were two isomorphisms we defined: ϕ : Z/r Z/s Z/rs given by ϕ(ā, b) = sa + rb, and ψ : Z/rs (Z/r Z/s) given by ā (ā, ā). Show that ψ is a ring homomorphism, and hence an isomorphism of rings, but not ϕ. Hence, conclude that ϕ cannot be the inverse of ψ (we already pointed this out in Homework 5, problem 4). Bonus: More generally, if I, J R are ideals such that I + J = R, then the map R/I R/J R/(I J) given by (a + I, b + J) a + b + (I J) is an isomorphism, broadly generalizing the Chinese Remainder Theorem (cf. Problem 3 above). First, we will show that ϕ is not a ring homomorphism. To be a ring homomorphism, in addition to preserving addition (ϕ(a 1 + a 2, b 1 + b 2 ) = ϕ(a 1, b 1 )ϕ(a 2, b 2 )), the property we proved already, a ring homomorphism must also preserve multiplication. However, but ϕ(a 1 a 2, b 1 b 2 ) = ϕ(a 1 a 2, b 1 b 2 ) = sa 1 a 2 + rb 1 b 2 ϕ(a 1, b 1 )ϕ(a 2, b 2 ) = (sa 1 + rb 1 ) (sa 2 + rb 2 ) = (sa 1 + rb 1 ) (sa 2 + rb 2 ) = s 2 a 1 a 2 + sra 1 b 2 + rsb 1 a 2 + r 2 b 1 b 2 = s 2 a 1 a 2 + r 2 b 1 b 2 Where the last two steps can occur because we are working in a commutative ring, and rs = sr 0 (mod rs). However, sa 1 a 2 + rb 1 b 2 s 2 a 1 a 2 + r 2 b 1 b 2, because s 2 s (mod rs) in general, so this is not a ring homomorphism. Now let us examine ψ. and ψ(a + b) = (a + b, a + b) = (a, a) + (b + b) = ψ(a) + ψ(b) ψ(ab) = (ab, ab) = (a, a) (b, b) = ψ(a) ψ(b) and finally, ψ(1) = (1, 1), and so the map preserves addition and multiplication, in addition to the multiplicative identity, and therefore, it is a ring homomorphism. Thus, since we already showed that it was a group isomorphism, we can extend this to say it is a ring isomorphism as well. The inverse of an isomorphism is, by definition, an isomorphism, and therefore, since ϕ is not even a homomorphism, much less an isomorphism, it cannot be the case that ϕ is the inverse of ψ. 4

6. (a) Using the ring isomorphism Z/rs Z/r Z/s, show that we have an isomorphism (Z/rs) = (Z/r) (Z/s) of groups of units (i.e., invertible elements under multiplication). This allows you to compute (Z/n) by decomposing n as a product of prime powers, and reduces the problem to (Z/p n ). (b) Show that (Z/n) cannot be cyclic (i.e., there is no primitive root modulo n) unless n is either a prime power or twice a prime power. Bonus for those who like number theory: Show that, conversely, (Z/p k ) is cyclic of order p k 1 (p 1) if p 3 (so there is a primitive root modulo p k for p 3), whereas (Z/2 k ) = Z/2 Z/2 k 2 (so there is not a primitive root modulo 2 k for k 3). (This together with the assigned part allows you to get the whole structure of (Z/m) for arbitrary m.) Hint: Observe first that the cyclic multiplicative subgroup of Z/p k generated by 1 + p k 1 is {1 + p k 1 a 0 a p 1}. Inductively, you can then prove that the cyclic multiplicative subgroup of Z/p k generated by 1 + p j is {1 + p j a 0 a p k j 1} (the inductive step will require at some point that (p, j) (2, 1), but that s ok since we assumed p 3). Applied to j = 1 gives that 1 + p generates everything congruent to one modulo p, which is therefore a cyclic group of (Z/p k ) of order p k 1. Finally, take a primitive root modulo p, say 1 a p 1, and consider a pk 1, which must have order p 1, so that a pk 1 (p 1) is a primitive root (of order (p 1)p k 1 ) in Z/p k 1 Z. The same proof works in the case p = 2 to show (Z/2 k ) = Z/2 Z/2 k 2. (a) Note: We assume that gcd(r, s) = 1 again, because this is not true otherwise. Let ψ : (Z/rs) (Z/r) (Z/s) be given by ψ : a (a, a). First, we will show that this is a homomorphism: ψ(ab) = (ab, ab) = (a, a)(b, b) = ψ(a)ψ(b). Therefore, ψ is a homomorphism. Now we will show ψ is an isomorphism. The identity of (Z/r) (Z/s) is (1, 1). Now consider any a such that ψ(a) = (1, 1). Now this means that a 1 (mod r) and a 1 (mod s). Then we have by a property of number theory that a 1 (mod rs) (s a 1, r a 1 s = (a 1)x, r = (a 1)y rs = (a 1)(a 1)xy rs a 1). Therefore, a = 1, and so the kernel of ψ is 1. This shows by an earlier homework that ψ is injective. Now consider the relative size of the groups. Any number which is relatively prime to r will have an inverse in Z/r, and therefore, be a unit and in (Z/r), by some basic properties of number theory (I can turn in all of my number theory HW if you want...). Then the same applies for (Z/s). This means that there are φ(r) units in Z/r, and φ(s) units in Z/s, or equivalently Z/s = φ(s) and Z/r = φ(r), where φ is the Euler Totient function, and φ(a) is the number of numbers less than a which are relatively prime to it. Similarly Z/rs = φ(rs). Since gcd(r, s) = 1, a basic property of number theory (Again, I ve proved this on a number theory HW at some point. See Schedler s wife s website.) says that φ(rs) = φ(r)φ(s), and therefore, Z/rs = Z/s Z/r, and therefore, the groups are the same size (Number of Cartesian products is product of number of elements in the first group times the number of elements in the second group.). Thus, ψ is an injective homomorphism between two groups of equal size, and therefore, by an earlier homework, ψ is an isomorphism. (b) Any positive integer n can be expressed as 2 r p s m, for gcd(p, 2) = 1 and gcd(p, m) = 1, and gcd(2, m) = 1 by taking the prime factorization, extracting the powers of two, and then one other prime and its powers, and then multiplying the rest back 5

together to get m. Then by part (a), we have that (Z/n) = (Z/2 r p s m) = (Z/2 r ) (Z/p s m) = (Z/2 r ) (Z/p s ) (Z/m). Then as noted above, and seen before, to be cyclic, at most one of the three factors can be an even cyclic group, otherwise the Klein four-group would be a subgroup, which cannot happen in a cyclic group. Now let us examine what it would mean for each of the groups given to have odd order, using the Euler Totient function φ as seen in part (a): 2 r : Z/2 r = φ(2 r ) = 2 r 1 (2 1) = 2 r 1. For 2 r 1 to be odd, we must have that r 1 = 0, or r = 1, and thus, if it exists, for it not to be an even cyclic group, this group must be of the form Z/2. p s : Z/p s = φ(p s ) = p s 1 (p 1). But p must be a prime not equal to 2, so p must be odd, and therefore, p 1 must be even, so if this group exists, it can never have odd order. m: Z/m = φ(m). Now write m in its unique prime factorization, and note that it will have each prime factor raised to a power, and since each prime factor to a power is relatively prime to all of the others, we can use the fact that φ(ab) = φ(a)φ(b), where a and b are the primes to the powers, and each of these must be even if they exist, by the same reasoning as above, and so if this group exists, it can never have odd order. Then if (Z/n) is cyclic, since we always pull out a prime factor for the second group, this means that m = 1, and so the third group never exists, and also, we either have that the first group does not exist, or it is only Z/2. Thus, we have that either n = p k or n = 2p k. 7. Artin, Chapter 11, Exercise 4.3. Identify the following rings. Note: Identify the rings means to rewrite them in terms we already know; for example, Z[x]/(x 2 d), we get an isomorphism Z[x]/(x 2 d) Z[ d] sending x to d (and which is the identity on Z). (a) Z[x]/(x 2 3, 2x + 4). We will look at what it means to kill off these elements. That is, let x 2 3 = 0, or x 2 = 3, and 2x + 4 = 0, or 2x = 4. We wish to combine these to see if we can find any simplifications, so we want an x 2 term in the second equation. We multiply both sides by x to get 2x 2 = 2(3) = 6 = 4x = 2(2x) = 2( 4) = 8. Since we have really been working with congruence classes so far, this means than 6 = 8, or 0 = 2. Then we can work in Z/2Z, and so 2x + 4 = 0, and x 2 3 = x 2 + 1 = x 2 2x + 1 = (x 1) 2. Thus, Z[x]/(x 2 3, 2x + 4) = (Z/2Z)/((x 1) 2 ). (b) Z[i]/(2+i). Again, let s see what happens when 2+i = 0. This means than i = 2, or i 2 = 1 = 4. Again, since we are using congruence classes, this means than 1 = 4, or 0 = 5. Therefore, Z[i]/(2 + i) = Z/5Z. (c) Z[x]/(6, 2x 1). We have that 6 = 0, and 2x 1 = 0, or 2x = 1. Then 3(2x) = 6x = 0 = 3(1) = 3, and so 3 = 0. Working in Z/3Z, we have that 2x 1 = 2x + 2, and so Z[x]/(6, 2x 1) = (Z/3Z)[x]/(2x + 2). (d) Z[x]/(2x 2 4, 4x 5). Here, we have that 2x 2 = 4, and 4x = 5. Squaring the second yields that 16x 2 = 8(2x 2 ) = 8(4) = 32 = 5 2 = 25, or 0 = 7. Thus, we can work in 6

Z/7Z. Here, 2(4x 5)(2x 1) = 16x 2 20x 8x 10 2x 2 21x + 4 2x 2 + 4 and therefore 4x 5 2x 2 + 4, and so Z[x]/(2x 2 4, 4x 5) = (Z[x]/6Z)/(4x 5). (e) Z[x]/(x 2 + 3, 5). Here, it s fairly obvious that 5 = 0 means that we are working in Z/5Z, and x 2 + 3 is unchanged, so Z[x]/(x 2 + 3, 5) = (Z/5Z)[x]/(x 2 + 3). Note that x 0: x 2 + 3 3 (mod 5). x 1: x 2 + 3 4 (mod 5). x 2: x 2 + 3 7 2 (mod 5). x 3: x 2 + 3 12 2 (mod 5). x 4: x 2 + 3 19 4 (mod 4). And thus x 2 + 3 has no zeroes, and thus no factors, simplest form above. (mod 5), and we have found the 8. Artin, Chapter 11, Exercise 7.2. Let R be a domain. Prove that the polynomial ring R[x] is a domain, and identify the units in R[x]. Proof by Contradiction: Since R is a domain, by definition, a, b R, ab = 0 a = 0 or b = 0. Now consider two non-zero polynomials, f(x), g(x) R[x], with f(x)g(x) = 0. Since they are non-zero, f(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0, and g(x) = b m x m + b m 1 x m 1 + + b 1 x + b 0, a n, b m 0, for some n, m. Then f(x)g(x) = a n b m x n+m + (a n b m 1 +a n 1 b m )xn + m 1+ +(a 0 b 1 +a 1 b 0 )x+a 0 b 0 = 0. Since this must be true for all x, it must be the case that each of the coefficients, a n b m, a n b m 1 + a n 1 b m,..., a 0 b 1 + a 1 b 0, a 0 b 0 = 0 (Else our polynomial would have at most m + n < zeroes, which is a contradiction.). Consider a n b m = 0. Here, a n, b n R, a domain, so either a n = 0 or b m = 0. This contradicts the definition of our polynomials, and therefore, it cannot be the case that both of our polynomials is non-zero. Thus, f(x)g(x) = 0 f(x) = 0 or g(x) = 0, and so, by definition, R[x] is a domain. Now I will demonstrate the units. Let f(x), g(x) be units, so f(x)g(x) = 1, and be written as before. Now since f(0)g(0) = a 0 b 0 = 1, it must mean that a 0, b 0 are units, and a 0, b 0 0. Also, it means that, if n 1 or m 1, and therefore, at least one of the polynomials is non-constant, then f(x)g(x) a 0 x 0 = a n b m x n+m + (a n b m 1 + a n 1 b m )xn + m 1 + + (a 0 b 1 + a 1 b 0 )x = 0, and this product still has at least the x n+m term, because n + m 1. Then by the same argument as above, this means that a n b m = 0 and therefore, either a n = 0 or b m = 0, which contradicts the fact that these are polynomials of degree n and m respectively, and therefore, it must be the case that n = m = 0, and both polynomials are constant. Therefore, the only units in R[x] are constants. Then we see that for any unit u R, u, u 1 R[x], and so u is still a unit, and all units of R are units of R[x]. Similarly, if a R is not a unit, then it has no inverse, and there are no constant terms in R[x] that are not in R, and so a R[x] has no inverse, and is not a unit. Therefore, the units of R[x] are the units of R. 9. Artin, Chapter 12, Exercise 2.1. Factor the following polynomials into irreducible factors in F y [x]. Note: you should not need to use anything from Chapter 12 beyond what was in Thursday s lecture (4/23/15), just the definitions in (12.2.1). 7

(a) x 3 + x 2 + x + 1, p = 2. We see that since 3 1 (mod 2), when working in Z/2 = F 2, x 3 + x 2 + x + 1 = x 3 + 3x 2 + 3x + 1 = (x + 1) 3, the irreducible factorization. (b) x 2 3x 3, p = 5. Again with congruences, we have that x 2 3x 3 = x 3 3x + 2 = (x 2)(x 1), the irreducible factorization. (c) x 2 + 1, p = 7. Consider all perfect squares (mod 7): x 0 (mod 7): Then x 2 0 0 (mod 7). x 1 (mod 7): Then x 2 1 6 (mod 7). x 2 (mod 7): Then x 2 2 5 (mod 7). x 3 (mod 7): Then x 2 9 5 (mod 7). x 4 (mod 7): Then x 2 16 5 (mod 7). x 5 (mod 7): Then x 2 25 3 (mod 7). x 6 (mod 7): Then x 2 36 6 (mod 7). Thus, having exhausted all possibilities, we see that there are no integers x such that x 2 1 (mod 7), and so there are no numbers which x 2 + 1 factors into in Z/7, so it is, by definition, irreducible. 10. Artin, Chapter 12, Exercise 5.1. Factor the following into primes in Z[i]. You should only need the following results: A unit in the Gaussian integers is one of 1, 1, i, or i, and the Gaussian primes (or equivalently irreducibles) are exactly the numbers of the following forms: p Z an integer prime with p 3 (mod 4), a + bi for a, b Z and a 2 + b 2 = p an integer prime with p 1 (mod 4), 1 + i, Associates of the above. (a) 1 3i. This can be written as 2 3i + (i) 2, and factored as a quadratic in i, so i 2 3i + 2 = (i 2)(i 1)Then i 1 = i(i + 1), and as stated above, i + 1 is prime. Then i(i 2) = 1 2i. We see that ( 1) 2 + ( 2) 2 = 1 + 4 = 5 1 (mod 4), and 5 is an integer prime, so 1 2i is prime. Thus, 1 3i = (i + 1)( 1 2i) is the prime factorization. (b) 10 = 9 + 1 = (1 + 3i)(1 3i). We see that 1 + 3i = 2 + 3i + (i) 2 = (i + 2)(i + 1). As seen above, i + 2 and i + 1 are primes. Also, the previous problem showed that 1 3i is (i + 1)( 1 2i), so 10 = (i + 1)( 1 2i)(i + 2)(i + 1) is the prime factorization. (c) 6 + 9i. 6 + 9i = (3)(2 + 3i). Here, 3 is an integer prime and 3 3 (mod 4). Also, 2 2 + 3 2 = 4 + 9 = 13, and 13 is an integer prime and 13 1 (mod 4). Thus, both 3 and (2+3i) are primes, and therefore, 6+9i = (3)(2 + 3i) is the prime factorization. (d) 7 + i. Here, 7 + i = 6 + i 1( 1) = i 2 + i + 6 = ( i + 3)(i + 2) = i(1 + 3i)(i + 2) = (1 2i)(i + 1)(i + 2), and as seen above, i + 1 and i + 2 are prime, and since (1) 2 + ( 2) 2 = 1 + 4 = 5 1 (mod 4) is prime, 1 2i is prime as well, so 7 + i = (1 2i)(i + 1)(i + 2), is the prime factorization. 8