# Prime ideals in certain quadratic ring

Let's consider the quadratic ring $$\mathbb{Z}[\sqrt{-5}]$$ and the principal ideals $$(29)$$ and $$(11)$$. Tell whether or not these ideals are prime.

My approach: In order to solve this theorem I am using the following fact:

Fact: If $$R$$ - commutative ring with $$1_R$$. Ideal $$I$$ in $$R$$ is prime iff factor-ring $$R/I$$ is integral domain.

Using the fact that $$\mathbb{Z}[\sqrt{-5}]\cong \mathbb{Z}[x]/(x^2+5)$$ I have derived the following the following results:

$$\mathbb{Z}[\sqrt{-5}]/(29)\cong \mathbb{Z}_{29}[x]/(x^2+5) \quad \text{and} \quad \mathbb{Z}[\sqrt{-5}]/(11)\cong \mathbb{Z}_{11}[x]/(x^2+5).$$

How to show are these quotient-rings are integral domain or not?

Is there any method except computational one?

Would be very grateful for help!

EDIT: For the second example I know that $$x^2+5$$ is irreducible over $$\mathbb{Z}_{11}$$. Hence this quotient-ring is field. Hence it is an integral domain.

But what about the first one?

It will suffice to show whether or note $$x^2+5$$ is irreducible over $$\Bbb Z_{29}$$. As a quadratic this only factors if there is some $$x$$ such that $$x^2 = -5 \mod 29$$. This can be done by brute force (trying $$x = 0, 1, \ldots, 29$$) as a last resort.

The techniques of number theory can fairly efficiently compute this question too (whether or not $$-5$$ is what is known as a quadratic residue mod $$29$$). However, this requires some familiarity with the Lagrange symbol and quadratic reciprocity.

An alternate approach might be to try to directly find a primitive root mod $$29$$, ie some nonzero value of $$z$$ such that $$z^i \ne 1$$ unless $$i$$ is a multiple of 28. Then the values of $$z^i$$ will consist of all values of $$\Bbb Z_{29}$$. Then you can see if the congruence class of $$-5$$ is an even power of $$z$$ or an odd power ($$x^2+5$$ would be reducible only if $$-5$$ is an even power of $$z$$). I don't think this will save effort compared to just computing the squares mod $$29$$ directly, though.

To go the next step beyond @RolfHeyer’s answer, I noticed that $$6\cdot29=174=13^2+5$$. Thus $$(13-\sqrt{-5}\,)(13+\sqrt{-5}\,)\in(29)$$, so $$(29)$$ isn’t a prime ideal.

• No need to pull $\,\large -5\equiv 13^2\,$ out of a hat since $\,\,\large -5\cdot 2^2\equiv 3^2\pmod{29},\,$ see my answer. Dec 9, 2018 at 3:41

First it is easy to check that $$-5$$ is not a square $$\!\bmod{11}$$ by Euler's criterion, i.e.

$$\!\bmod 11\!:\ \left[a^{\large 2} \equiv -5\right]^{\large 5}\!\Rightarrow\, a^{\large 10}\equiv -5(25)^{\large 2} \equiv -5(3)^{\large 2} \equiv -1\,$$ contra little Fermat.

Therefore $$\ x^{\large 2}\equiv -5\,$$ is unsolvable so $$\,x^2+5\,$$ has no root so is irreducible $$\bmod {11}.\,$$ Otoh

\!\!\begin{align} \bmod 29\!:\,\ {-}5\cdot 2^{\large 2} &\equiv 3^{\large 2}\ \ \\[.3em] \Rightarrow\ \ \ \ \ \ \ \ \color{#c00}{{-}5} &\equiv \left(\dfrac{3}2\right)^{\large 2}\!\!\equiv \left(\dfrac{-26}2\right)^{\large 2}\!\! \equiv \color{#c00}{13^{\large 2}}\\[.3em] \Rightarrow\ \ x^{\large 2}+\color{#c00}{5}&\equiv x^{\large 2}\color{#c00}{-13^2}\equiv (x-13)(x+13) \end{align}

Remark  Here we don't actually need to calculate the value of $$\,3/2\,$$ since we only need to know that $$-5$$ is a square to infer that $$\,x^2+5\,$$ is reducible. But it was easy to do so here so we did it.