Show that $\langle 13 \rangle$ is a prime ideal in $\mathbb{Z[\sqrt{-5}]}$

Question

To show that $\langle 13 \rangle$ is a prime ideal in $D= \mathbb{Z[\sqrt{-5}]}$, I could show that $13$ is an irreducible element of $D$ but as $D$ is not a U.F.D, it is not of much use I guess. How can I prove this.

Answer 1

You should compute $\mathbb Z[\sqrt{-5}]/(13) \cong \mathbb Z[X]/(13,X^2+5) \cong (\mathbb Z/13\mathbb Z)[X]/(X^2+5)$.

It is easy to see whether this is an integral domain or not.

Answer 2

This is a hint too long for a comment.

To show that $\langle 13 \rangle$ is a prime ideal, you need to show that it is a maximal ideal, that the only ideal that properly contains it is the whole ring itself. But if it is not a prime ideal, then you can find it properly contained in another ideal that is not the whole ring.

It's a lot like in a principal ideal domain (PID), e.g., $\mathbb{Z}[\sqrt{-2}]$, but unique factorization provides a neat shortcut: if $x$, not 0 nor unit, is irreducible, then $\langle x \rangle$ is a prime ideal. Primes and irreducibles are the same when there is unique factorization, so if $p$ is irreducible and $p|ab$, then at least one of these must hold true: $p|a$ or $p|b$.

That shortcut is not available in $\mathbb{Z}[\sqrt{-5}]$. For example, 3 is an irreducible number in this domain, but $\langle 3 \rangle$ is not a prime ideal. Its norm is 9, and we see that $9 = 3^2 = (2 - \sqrt{-5})(2 + \sqrt{-5})$, but $3 \nmid (2 \pm \sqrt{-5})$. So $\langle 3 \rangle$ must be contained within an ideal that also contains $\langle 2 - \sqrt{-5} \rangle$ and $\langle 2 + \sqrt{-5} \rangle$. I don't know what that ideal is, but I'm pretty sure it's not the whole ring, and furthermore I suspect this might be an exercise in your book. There's also the much more famous $6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5})$.

If $\langle 13 \rangle$ is not a prime ideal, then there exist ideals $\mathfrak{a}$ and $\mathfrak{b}$ such that $\mathfrak{a} \mathfrak{b} = \langle 13 \rangle$. Since $N(\mathfrak{a}) N(\mathfrak{b}) = 169$, we can deduce $N(\mathfrak{a}) = 13$. Yada, yada, yada, draw a contradiction about $N(\mathfrak{a})$.

EDIT: It seems I jumped to a wrong conclusion about what ideals contain $\langle 3 \rangle$. But my overall point stands: the generator of the ideal being irreducible does not automatically mean the ideal is prime.

Answer 3

The question can be answered with quadratic reciprocity.

Generally, in a quadratic field $\Bbb Q(\sqrt n\,)$, a rational prime $p$ can behave in three different ways: (1) $(p)=\mathfrak p$. i.e. $p$ remains prime; (2) $(p)=\mathfrak p_1\mathfrak p_2$, i.e. $p$ splits into two conjugate primes; or (3) $(p)=\mathfrak p^2$, $p$ ramifies. In our example, (3) doesn’t happen ’cause $2$ and $5$ are the only ramified primes of $\Bbb Q(\sqrt{-5}\,)$. So it boils down to a decision between (1)and (2).

You see that case (2) happens exactly when $n$, $-5$ in our case, is a square modulo $p$, i.e. when the quadratic symbol $\left(\frac n p\right)=1$, that is, if $\left(\frac{-5}{13}\right)=1$ in our case.

But $\left(\frac{-5}{13}\right)=\left(\frac8{13}\right)=\left(\frac2{13}\right)$, which is equal to $-1$ because $\left(\frac2p\right)=1$ if and only if $p\equiv\pm1\pmod8$.

The upshot is that $-5$ is not a square modulo $13$, so $13$ doesn’t split, i.e. remains prime in $\Bbb Q(\sqrt{-5}\,)$.