## 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.