Show that $\langle 13 \rangle$ is a prime ideal in $\mathbb{Z[\sqrt{-5}]}$ 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.
 A: 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 $\Bbb Q(\sqrt{-5}\,)$.
A: 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.
A: 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.
A: I think this is one of those questions that is better answered by looking at the numbers rather than the ideals.
If $\langle 13 \rangle$ is a prime ideal, then $13$ is a prime number. Which means that whenever $13 \mid ab$, then $13 \mid a$ or $13 \mid b$. If there were numbers $a$ and $b$ that satisfied the first condition but neither the second nor the third condition, they would be of the form $x \pm \sqrt{-5}$, where $x \in \mathbb Z$ is a solution to $x^2 \equiv -5 \pmod p$ with $p = 13$.
The famous example of $6 = (1 - \sqrt{-5})(1 + \sqrt{-5})$ gives an idea of how this can happen. If we didn't discover it by simply looking at numbers with small norms, we'd discover it by finding that $1^2 \equiv -5 \pmod 3$. This shows that $3$ is irreducible but not prime.
It's also true of $7$, since $3^2 \equiv -5 \pmod 7$, leading us to find that $(3 - \sqrt{-5})(3 + \sqrt{-5}) = 14$, showing that $7$ is also irreducible but not prime. In terms of ideals, we have $\langle 7 \rangle \subset \langle 7, 3 - \sqrt{-5} \rangle$ and $\langle 7 \rangle \subset \langle 7, 3 + \sqrt{-5} \rangle$.
And we only need to look at $p$ squares to make this determination. The squares modulo $13$ are $1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1, 0$. As we don't find $13 - 5 = 8$ in this list, this means no number of the form $x \pm \sqrt{-5}$ has a norm that is a multiple of $13$.
Therefore, if $\langle 13 \rangle \subseteq I \subseteq \mathbb Z[\sqrt{-5}]$, either $I = \langle 13 \rangle$ or $I = \mathbb Z[\sqrt{-5}]$, proving $\langle 13 \rangle$ is a prime ideal.
A: $$\left (\sqrt{-5}\right )^2\equiv_{13}8$$
Since $8$ is a quadratic non-residue $\mathbb{Z}[\sqrt{-5}]/(13)$ is a finite integral domain, hence a field. Therefore $(13)$ is a maximal ideal.
