Math 418 Spring 2017 Homework 2 8.2.5. Let R = Z[ / -5], and Let I2

Math 418 Spring 2017 Homework 2

√ √ √ √ 0 8.2.5. Let R = Z[ −5], and let I2 = (2, 1 + −5), I3 = (3, 2 + −5), and I3(3, 2 − −5) be ideals of R.

0 (a) Claim: I2, I3, and I3 are not principal ideals.

Proof: By Example 2 in Section 8.1, I3 is not principal. Suppose by way of√ contradiction that I2 were principal. Then there would exist γ ∈ R such that (2, 1+ −5) = (γ). Then there would exist α and β in R such that

2 = αγ, √ 1 + −5 = βγ. √ We can apply the field norm N(a + b −5) = a2 + 5b2, which is multiplicative, to each of these equations, to see 4 = N(α)N(γ), 6 = N(β)N(γ). The first equation implies that N(γ) = 1, 2, or 4. The second implies that N(γ) = 1, 2, 3, or 6. Hence N(γ) can be only 1 or 2. 2 2 It is not possible that N(γ) = 2, since the equation a + 5b = 2 has√ no integer solutions. If N(γ) = 1, then γ = ±1. Hence (γ) = R. Since (γ) = (2, 1 + −√5), this would imply that for every z ∈ R, there exist x, y ∈ R so that√z = 2x + (1 + −5)y. In particular,√ there must exist√x, y ∈ R so that√ 1 = 2x + (1 + −5y. Mutiplying both sides by 1 − −5, we see 1 − −5 = 2(1 − −5)x + 6y. There do not exist x, y ∈ R satisfying this equation, since the right hand side has even coefficients and the left hand side does not. Hence γ does not exist, and I2 is not principal. 0 Now suppose by way of√ contradiction that I3 were principal. Then there would exist δ ∈ R such that (3, 2 − −5) = (δ). Then there would exist α and β in R such that

3 = αδ, √ 2 − −5 = βδ.

We can apply the ﬁeld norm to each of these equations, to see

9 = N(α)N(δ),

1 9 = N(β)N(δ). The equations imply that N(δ) = 1, 3, or 9. As there are no integer solutions to the equation 3 = a2 +5b2, N(δ) can be only 1 or 9. If N(δ) = 9 then N(α) = 1, so α = ±1. This would imply that δ = ±3. But this is not possible, since the second equation has coeﬃcients not divisible by 3 on the left hand side. √ If N(δ) = 1, then δ = ±1. Hence (δ) = R. Since√ (δ) = (3, 2 − −5), this would imply√ that there exist√x, y ∈ R so that√ 1 = 3x + (2 − −5y. Mutiplying both sides by 2 + −5, we see 2 + −5 = 3(2 + −5)x + 9y. There do not exist x, y ∈ R satisfying this equation, since the right hand side has coeﬃcients divisible by 3 and the left hand 0 side does not. Hence δ does not exist, and I3 is not principal.

2 (b) Claim: I2 = (2).

2 2 2 Proof: We will show that (2) ⊆ I2 and that I2 ⊆ (2).√ First, the elements of I2√are all ﬁnite sums of elements of the form αβ for α = 2x+(1+ −5)y and β = 2z+(1+ −5)w in R. We can write √ √ αβ = 4xz + 2(1 + −5)(yz + xw) + (−4 + 2 −5)yw = 2r,

where √ r = (2xz + yz + xw − 2yw) + (yz + xw + yw) −5 2 as an element of (2). Hence I2 ⊆ (2). On the other hand, we can write √ √ −(1 + −5)2 + 2(1 + −5) − 22 = 2,

2 2 2 and so 2 ∈ I2 . Hence (2) ⊆ I2 . This shows that I2 = (2).

√ √ 0 2 0 (c) Claim: I2I3 = (1 − −5), I2I3 = (1 + −5), and I2 I3I3 = (6). √ √ Proof: First,√ we will show that I2I3 = (1 − −5) by showing that I2I3 ⊆ (1 − −5) and (1− −5) ⊆ I2I3.√ The elements of I2I3 are ﬁnite√ sums of elements of the form αβ, where α = 2x + (1 + −5)y and √β = 3z + (2 + √−5)w. Therefore,√ I2I3 is√ generated by the√ elements√ 2 · 3 = 6, 2(2√ + −5) = 4 + 2 −5, 3(1 + −5) = 3 + 3 −5, and (1 + −5)(2 + −5) = −3 + 3 −5. Since √ √ 6 = (1 − −5)(1 + −5), √ √ √ 4 + 2 −5 = (1 − −5)(−1 + −5), √ √ √ 3 + 3 −5 = (1 − −5)(−2 + −5),

2 and √ √ −3 + 3 −5 = −3(1 − −5), √ we can see that I2I3 ⊆ (1 − −5). On the other hand, √ √ √ 1 − −5 = 2(2 + −5) − 3(1 + −5). √ √ Hence (1 − −5) ⊆ I2I3. This proves I2I3 = (1 − −5). √ √ 0 0 Now√ we will show that I2I3 = (1 + −5) by showing that I2I3 ⊆ (1 + −5) and 0 0 (1 + −5) ⊆ I2I3. The√ elements of I2I3 are ﬁnite√ sums of elements of the form αβ, 0 where α = 2x + (1 + −5)y and √β = 3z + (2 − √−5)w. Therefore,√ I2I3 is√ generated by the√ elements√ 2 · 3 = 6, 2(2√ − −5) = 4 − 2 −5, 3(1 + −5) = 3 + 3 −5, and (1 + −5)(2 − −5) = 7 + −5. Since √ √ 6 = (1 + −5)(1 − −5), √ √ √ 4 − 2 −5 = (1 + −5)(−1 − −5), √ √ 3 + 3 −5 = 3(1 + −5),

and √ √ √ 7 + −5 = (1 + −5)(2 − −5), √ 0 we can see that I2I3 ⊆ (1 + −5). On the other hand, √ √ √ 1 + −5 = −2(3) + (1 + −5)(2 − −5). √ √ 0 0 Hence (1 + −5) ⊆ I2I3. This proves I2I3 = (1 + −5). √ √ 0 0 2 0 Since √I2I3 = (1√− −5) and I2I3 = (1 + −5) are principal, I2I3I2I3 = I2 I3I3 = ((1 − −5)(1 + −5)) = (6).

√ 8.3.5. Let R = Z[ −n] where n is a squarefree integer greater than 3. √ √ (a) Claim: The elements 2, −n, 1 + −n are irreducible in R.

Proof: In each case we will show that the element is irreducible by supposing it factors into a product αβ for some α, β ∈ R, and then showing one of α, β must be a unit. First consider a factorization 2 = αβ. We apply the norm function to both sides to get the equation

4 = N(α)N(β).

3 √ √ For any element a + b −n ∈ R, N(a + b −n) = a2 + bn2 cannot be 2, since 2 is not a square and n > 2. Therefore one of α and β has norm 1 and the other has norm 4. But the only elements with norm 1 in R are the units 1, −1. So α or β is a unit. This shows 2 is irreducible in R.

Now suppose we factor √ −n = αβ for some α, β ∈ R. We apply the norm function to both sides to see

n = N(α)N(β).

√ √ 2 2 For any element a +√b −n ∈ R, either b = 0 or N(a + b −n) = a + nb ≥ n. If 2 b = 0, then N(a + b −√5) = n cannot divide n since n is squarefree, unless a = ±1. If b 6= 0, then N(a + b −n) divides n if and only if b = ±1 and√ a = 0. Therefore, if N√(α)N(β) = n, then one of α, β is a unit and the other is ± −n. This proves that −n is irreducible in R.

Finally, suppose we factor √ 1 + −n = αβ for some α, β ∈ R. If we apply the norm function to each side we get the equation

1 + n = N(α)N(β). √ √ Let α = a + b −n and β = c + d −n. Since N(α) divides 1 + n, a2 + nb2 ≤ 1 + n. So b = 0, ±1. If b = ±1 then a = ±1 and N(α) = 1 + n, so N(β) = 1 and β is a unit. If 2 √ √ b = 0 then a divides 1 + n, and 1 + −n =√ac + ad −n. But then a must divide 1 as well, so a = ±1 is a unit. This proves 1 + −n is irreducible in R.

(b) R is not a UFD.

Proof: By Proposition 8.12, if R is a UFD, then√ all irreducible√ elements are prime. We will show that at least one of the elements√ −n, 1 + −n is not prime, which will show that R is not a UFD. Suppose that −n is prime. Then ﬁrst, we can conclude that n is√ prime in Z√. Otherwise, there would√ exist a, b ∈ N√>1 so that ab = n. And then (a + −n)(b + −n) = (ab − n) + (a + b) −n =√ (a + b) −n would√ be an element of R which n divides, although n divides neither a + −n nor b + −n.

Since n is prime and √n > 3, n + 1 is not prime. We√ can ﬁnd√c, d ∈ N>1 so that cd = 1 + n. But 1 + −n divides 1 + n, since (1√ + −n)(1 − −n) = 1 + n, and 1 + sqrt−n does not divide c or d. Therefore 1 + −n is not prime. Since R contains an irreducible element which is not prime, it is not a UFD.

4 √ √ √ √ 0 8.3.8. Let R = Z[ −5], and let I2 = (2, 1 + −5), I3 = (3, 2 + −5), and I3(3, 2 − −5) be ideals of R. √ √ (a) Claim: The elements 2,√3, 1+ −5,√1− −5 are irreducible in R, and no pair is associate. Hence 6 = 2 · 3 = (1 + −5)(1 − −5) has two factorizations in R.

√ Proof: Let N be the ﬁeld norm on Z[ −5]. First, suppose 2 = αβ for some α, β ∈ R. Then 4 = N(α)N(β). There are no integer solutions to the equation 2 = s2 + 5b2, and so one of N(α) and N(β) must be 1. Hence one of N(α) and N(β) is a unit, and 2 is irreducible. Similarly, if 3 = αβ for some α, β ∈ R, 9 = N(α)N(β). Since 3 = a2 + 5bs has no integer solutions, one of α and√ β must be a unit.√ Hence 3 is irreducible in R. By the 2 2 same argument, since N(1+ −5) == N(1− −5) = 6,√ and as we’ve seen, 2 = a +5b 2 2 and 3 = a + 5b have√ no integer√ solutions, if αβ = 1 ± −5, one of α and β must be a unit. Hence 1 + −5 and 1 − −5 are irreducible in R.

√If two element√ α, β ∈ R are associate, then N(α) = N(√β). Hence no√ pair of 2, 3, 1 + −5, 1− −5 could√ possibly be associate except for 1+ −5 and 1− −5. However, if 1+ −5 √ 2 1 √ we compute √ in Q( −5), we get + −5, which is not in R. Hence no pair of 1− −5 3 3 √ √ these irreducibles are associate, and the factorizations 6 = 2 · 3 = (1 + −5)(1 − −5) are distinct.

0 (b) Claim: I2, I3, and I3 are prime ideals of R.

Proof: First, consider R/I2. If this is an integral domain,√ then I2 is a prime ideal. First, consider√ the quotient√ ring R/(2). Every a + b −5 ∈ R is congruent to one of 4 element: 0, 1, −5, 1 + √−5, depending on whether a and b are even.√ Now, consider I2/(2). Every 2x + (1 + −5)y ∈ I2 is congruent to either 0 or 1 + −5 in I2/(2). By ∼ the Third Isomorphism Theorem, R/I2 = (R/(2))/(I2/(2)). We now know that R/(2) is a ring with four elements, and I2/(2) is a ring with 2 elements. Hence, R/I2 is a ring with 2 elements. Since R has an identity, R/I2 will as well. So R/I2 is the ring with the two elements 0 and 1. This is an integral domain, so I2 is prime. √ Now consider R/I3. First, consider√ the quotient ring R/(3). All elements a + b −5 are equivalent to the elementa ¯ + ¯b −5 in R/(3), wherea ¯ is the class of a in Z/3Z. Hence, R/(3) has 9√ elements. Now consider the ring I3/(3). All elements of I3 are of the form 3x + (2√ + −5)y for√ some x, y ∈ R. Hence, all elements of I3 are equivalent to one of 0, 2 + −5, or 1 + 2 −5 in I3/(3). ∼ By the Third Isomorphism Theorem of Rings, R/I3 = (R/(3)/(I3/(3)). By the previous argument, this will be a commutative ring with a 1, and with 3 elements. We can choose

5 √ representative elements 0, 1, −5. No pair of nonzero elements in this ring multiply to 0, so this is an integral domain. Hence I3 is a prime ideal of R. 0 Finally consider R/I3. Again, consider the quotient ring R/(3). We saw before that 0 0 R/(3) has√ 9 elements. Now consider the ring I3/(3). All elements of I3 are of the form 0 3x + (2 √− −5)y for√ some x, y ∈ R. Hence, all elements of I3 are equivalent to one of 0 0, 2 + 2 −5, or 1 + −5 in I3/(3). 0 ∼ 0 By the Third Isomorphism Theorem of Rings, R/I3 = (R/(3)/(I3/(3)). By the previous argument, this will be a commutative√ ring with a 1, and with 3 elements. We can choose representative elements 0, 1, −5. No pair of nonzero elements in this ring multiply to 0 0, so this is an integral domain. Hence I3 is a prime ideal of R.

√ √ (c) Claim: (6) = (2)(3) = (1+ −5)(1− −5) are the same factorization of (6) into prime ideals.

Proof:√ Since√ 6 = 2 · 3, the product of principal ideals√ (2)(3) = (6). Similarly,√ (1 + 2 0 0 −5)(1 − −5) = (6). But (2) = I2 , (3) = I3I3, (1 + −5) = I2I3, and (1 − −5) = I2I3. Hence these are the same factorizations of (6) into the product of prime ideals.