Compute the discriminant of an ideal in an algebraic integer ring

Let K = \mathbb{Q}(\sqrt{-5}) and let \mathcal{O} = \mathbb{Z}[\sqrt{-5}] be the ring of integers in K. Let A = (3,1+\sqrt{-5}) and B = (2). Exhibit bases for A and B, compute \mathcal{O}/A and \mathcal{O}/B, and compute the discriminant of A and B.

We claim that \{3, 1+\sqrt{-5}\} is a basis for A over \mathbb{Z}. It is clear that (3, 1+\sqrt{-5})_\mathbb{Z} \subseteq A, and if \zeta = 3(a+b\sqrt{-5}) + (1+\sqrt{-5})(h+k\sqrt{-5}), then evidently \zeta = 3(a-b-2k) + (1 + \sqrt{-5})(3b+k+h). So (3, 1+\sqrt{-5})_\mathbb{Z} = A. Now if a,b \in \mathbb{Z} and 3a + (1+\sqrt{-5})b = 0, then b = 0, so that a = 0. Hence \{3, 1+\sqrt{-5}\} is a basis for A over \mathbb{Z}. The discriminant of A is then \mathsf{det}^2 \left[ \begin{array}{cc} 3 & 1+\sqrt{-5} \\ 3 & 1-\sqrt{-5} \end{array} \right] = -180.

We now claim that A is proper. To see this, suppose to the contrary that 1 \in A; then 1 = 3a + (1+\sqrt{-5})b for some integers a and b; comparing coefficients, we have b = 0 and 3a+b = 1. But then 3a = 1, which has no integer solutions; thus A is proper.

Now let a+b\sqrt{-5} be arbitrary in \mathcal{O} and let a-b \equiv k mod 3, where k \in \{0,1,2\}. Now a+b\sqrt{-5} = a-b+b(1+\sqrt{-5}) \equiv a-b \equiv k mod A, so that \mathcal{O}/A = \{\overline{0}, \overline{1}, \overline{2}\}. Since A is proper, 0 \not\equiv 1 and 2 \not\equiv 1 mod A. If 2 \equiv 0, then we have 3-2 = 1 \in A, a contradiction; so 2 \not\equiv 0 mod A. Thus \mathcal{O}/A \cong \mathbb{Z}/(3).

We claim now that \{2,2\sqrt{-5}\} is a basis for B over \mathbb{Z}. Certainly (2,2\sqrt{-5})_\mathbb{Z} \subseteq B, and if \zeta = 2(a+b\sqrt{-5}) \in (2), then \zeta = 2a + 2\sqrt{-5}b. So (2,2\sqrt{-5})_\mathbb{Z} = B. Moreover, if 2a + 2b\sqrt{-5} = 0, then a = b = 0. So \{2,2\sqrt{-5}\} is a basis for B over \mathbb{Z}. The discriminant of B is then \mathsf{Det}^2 \left[ \begin{array}{cc} 2 & 2\sqrt{-5} \\ 2 & -2\sqrt{-5} \end{array} \right] = -320.

We claim that B is proper. Indeed, if 1 \in B, then 1 = 2a for some integer a, a contradiction.

Now let a+b\sqrt{-5} be arbitrary in \mathcal{O}. Say a \equiv a_0 and b \equiv b_0 mod 2, where a_0, b_0 \in \{0,1\}. Then a+b\sqrt{-5} \equiv a_0 + b_0 \sqrt{-5} mod B. Thus \mathcal{O}/B = \{\overline{0}, \overline{1}, \overline{\sqrt{-5}}, \overline{1+\sqrt{-5}}\}. Since B is proper, we have 0 \not\equiv 1 and \sqrt{-5} \not\equiv 1+\sqrt{-5} mod B. If \sqrt{-5} \in B, then we have 1 = 2b for some integer b, a contradiction. Thus \sqrt{-5} \not\equiv 0 and 1+\sqrt{-5} \not\equiv 1 mod B. Likewise, if 1+\sqrt{-5} \in B, we have a contradiction, so that 1+\sqrt{-5} \not\equiv 0 and \sqrt{-5} \not\equiv 1 mod B.

We claim that \mathcal{O}/B \cong \mathbb{F}_2[x]/(x^2). To see this, define \varphi : \mathbb{F}_2[x] \rightarrow \mathcal{O}/B by x \mapsto 1+\sqrt{-5}. Certainly \varphi is surjective. Now \varphi(x^2) = 0, so that (x^2) \subseteq \mathsf{ker}\ \varphi. Now if p(x) = a_0 + a_1x + p^\prime(x)x^2 \in \mathsf{ker}\ \varphi, then \varphi(p(x)) = (a_0+a_1) + a_1\sqrt{-5} \equiv 0. Thus a_1 \equiv a_0 \equiv 0 mod 2. By the first isomorphism theorem for rings, \mathcal{O}/B \cong \mathbb{F}_2(x)/(x^2).

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: