Irreducibles which are not prime in some quadratic integer rings

Let n be a squarefree integer greater than 3, and let R = \mathbb{Z}[\sqrt{\text{-}n}].

  1. Prove that 2, \sqrt{\text{-}n}, and 1 + \sqrt{\text{-}n} are irreducible in R.
  2. Prove that R is not a unique factorization domain.
  3. Give an explicit ideal in R that is not principal.

We begin with some general remarks.

Let R be an integral domain. A divisibility norm (nonstandard term) on R is a mapping N : R \rightarrow \mathbb{N} such that N(ab) = N(a)N(b) for all a,b \in R. That is, N is a multiplicative semigroup homomorphism, or a poset homomorphism with respect to divisibility. We say that N is bounded if N(0) = 0 and N(1) = 1. We can see that if N is a bounded divisibility norm on a domain R and u \in R is a unit with inverse v, then N(u)N(v) = 1, so that N(u) = 1.

Let n be squarefree greater than 3. We claim that N(a+b\sqrt{\text{-}n}) = a^2 + nb^2 is a bounded divisibility norm. To see this, let \alpha = a_1 + a_2 \sqrt{\text{-}n} and \beta = b_1 + b_2 \sqrt{\text{-}n}.

N(\alpha\beta)  =  N((a_1 + a_2 \sqrt{\text{-}n})(b_1 + b_2 \sqrt{\text{-}n}))
 =  N((a_1b_1 - na_2b_2) + (a_1b_2 + a_2b_1)\sqrt{\text{-}n})
 =  (a_1b_1 - na_2b_2)^2 + n(a_1b_2 + a_2b_1)^2
 =  a_1^2b_1^2 + na_1^2b_2^2 + na_2^2b_1^2 + n^2a_2^2b_2^2
 =  (a_1^2 + na_2^2)b_1^2 + (a_1^2 + na_2^2)nb_2^2
 =  (a_1^2 + na_2^2)(b_1^2 + nb_2^2)
 =  N(\alpha)N(\beta)

So N is a divisibility norm. Certainly N is bounded because N(1) = 1^2 = 1 and N(0) = 0^2 = 0. In particular, if \zeta = u_1 + u_2 \sqrt{\text{-}n} is a unit in R, then N(\zeta) = u_1^2 + nu_2^2 = 1. Since n > 3, we have u_2 = 0, and then u_1 = \pm 1. So the units in R are precisely \pm 1. Conversely, if N(\zeta) = 1, then \zeta = \pm 1 is a unit.

  1. Suppose 2 = \alpha\beta. Then N(\alpha)N(\beta) = 4. Suppose N(\alpha) = 2. With \alpha = a + b\sqrt{\text{-}n}, we have a^2 + nb^2 = 2. Since n > 3, we have b = 0. Then a^2 = 2, a contradiction. So either N(\alpha) = 1, and \alpha is a unit, or N(\beta) = 1 so that \beta is a unit. Thus 2 is irreducible in R.

    Suppose \sqrt{\text{-}n} = \alpha\beta. Then N(\alpha)N(\beta) = n; we may assume without loss of generality that N(\alpha) < N(\beta), with strictness because n is squarefree. Suppose \alpha = a+b\sqrt{\text{-}n}. Then a^2 + nb^2 divides n properly. Then b = 0, and thus a^2 divides n. Since n is squarefree, a = \pm 1 and thus \alpha is a unit. So \sqrt{\text{-}n} is irreducible.

    Suppose 1 + \sqrt{\text{-}n} = \alpha\beta. Then N(\alpha)N(\beta) = n+1. Letting \alpha = a + b\sqrt{\text{-}n}, a^2 + nb^2 divides n+1. If |b| \geq 2, then since n > 3 we have a contradiction since N(\alpha) is too large. If |b| = 1, then a^2 + n divides 1+n, and we have |a| = 1. in either case, we have N(\alpha) = n+1, so that N(\beta) = 1 and thus \beta is a unit. Suppose b = 0, and let \beta = c + d\sqrt{\text{-}n}. Then a^2(c^2 + nd^2) = (ac)^2 + n(ad)^2 = n+1. If |ad| \geq 2, we have a contradiction. If |ad| = 1, then in particular, a = \pm 1. Then \alpha is a unit. If |ad| = 0, then we have d = 0. In this case we have ac = \sqrt{\text{-}n}, a contradiction. Thus 1 + \sqrt{\text{-}n} is irreducible.

  2. First, we prove the following lemma.

    Lemma: For n a squarefree integer greater than 3 and R = \mathbb{Z}[\sqrt{\text{-}n}], \sqrt{\text{-}n} divides a+b\sqrt{\text{-}n} if and only if n divides a. Proof: Note that \sqrt{\text{-}n}(x+y\sqrt{\text{-}n}) = -ny + x\sqrt{\text{-}n}. Conversely, nx + y\sqrt{\text{-}n} = \sqrt{\text{-}n}(y-x\sqrt{\text{-}n}). \square

    Suppose n is a composite integer; say n = ab where a,b < n. Using the lemma, \sqrt{\text{-}n} divides (a + \sqrt{\text{-}n})(b + \sqrt{\text{-}n}), but does not divide a + \sqrt{\text{-}n} or b + \sqrt{\text{-}n}. Thus \sqrt{\text{-}n} is not prime in R.

    Now suppose n is a prime integer; since n > 3, n+1 is a composite integer. Say n+1 = ab with 0 < a,b < n+1. Note that (1 + \sqrt{\text{-}n})(1 - \sqrt{\text{-}n}) = n+1 = ab. But if a(u + w \sqrt{\text{-}n}) = au + aw\sqrt{\text{-}n} = 1 + \sqrt{\text{-}n}, then we have a = 1, a contradiction. Thus n is not prime in R.

  3. Suppose n is prime. Since n > 3, we have n \geq 5, and moreover n is odd. Suppose the ideal (1 + \sqrt{\text{-}n}, 1 - \sqrt{\text{-}n}) = (\zeta) is principal, with \zeta = u + v \sqrt{\text{-}n}. Note that (1 + \sqrt{\text{-}n}) + (1 - \sqrt{\text{-}n}) = 2 \in (\zeta), so that N(\zeta_ = u^2 + nv^2 divides N(2) = 4. If |v| \geq 1, then N(\zeta) is larger than 4. Thus v = 0, and since u^2 divides 4, up to units we have \zeta \in \{1,2\}. If \zeta = 1, then we have (1 + \sqrt{\text{-}n})(a_1 + a_2 \sqrt{\text{-}n}) + (1 - \sqrt{\text{-}n})(b_1 + b_2 \sqrt{\text{-}n}) = 1 for some a_i, b_i. Multiplying this out and comparing coefficients, we have a_1 + b_1 + nb_2 - na_2 = 1 and a_1 + a_2 + b_2 - b_1 = 0. We wish to show that these equations have no solution in the integers. To see this, reduce both equations mod 2. Since n is odd, we have the sum a_1 + a_2 + b_1 + b_2 simultaneously congruent to 0 and 1 mod 2, a contradiction. Thus \zeta \neq 1. Suppose instead that \zeta = 2. But then 2(x+y\sqrt{\text{-}n}) = 2x + 2y\sqrt{\text{-}n} = 1 + \sqrt{\text{-}n} for some x,y, a contradiction. Thus the ideal (1 + \sqrt{\text{-}n}, 1 - \sqrt{\text{-}n}) is not principal after all.

    Now suppose n is composite; let n = pq, where p is prime. Suppose the ideal (p, \sqrt{\text{-}n}) = (\zeta) is principal, with \zeta = u + v\sqrt{\text{-}n}. Then N(\zeta) = u^2 + nv^2 divides n. If |v| \geq 2, then N(\zeta) is too large. If |v| = 1, then u^2 + n divides n, and we have u = 0. But now, N(\zeta) = n divides N(p) = p^2; this contradicts either the compositeness or the squarefreeness of n. Thus we have v = 0, so that u^2 divides n. Since n is squarefree, u = 1, and thus (p, \sqrt{\text{-}n}) = R. In particular, for some \alpha = a_1 + a_2 \sqrt{\text{-}n} and \beta = b_1 + b_2 \sqrt{\text{-}n}, we have p\alpha + \sqrt{\text{-}n} \beta = 1. Simplifying and comparing coefficients, we see that pa_1 - nb_2 = 1; this is a contradiction because p divides n. Thus (p, \sqrt{\text{-}n}) is not principal after all.

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Comments

  • Shay Logan  On February 15, 2011 at 11:03 pm

    I think there are some typos in the third paragraph of part two. For example, I’m fairly certain the last bit should say that n isn’t prime, etc.

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