Cfrgtkky

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 $6cdot29=174=13^2+5$. Thus $(13-sqrt{-5},)(13+sqrt{-5},)in(29)$, so $(29)$ isn’t a prime ideal.

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

$!bmod 11!: left[a^{large 2} equiv -5right]^{large 5}!Rightarrow, a^{large 10}equiv -5(25)^{large 2} equiv -5(3)^{large 2} equiv -1,$ 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!:, {-}5cdot 2^{large 2} &equiv 3^{large 2} \[.3em]
Rightarrow color{#c00}{{-}5} &equiv left(dfrac{3}2right)^{large 2}!!equiv left(dfrac{-26}2right)^{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.