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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?

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3 Answers 3

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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ Dec 9, 2018 at 3:41
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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.

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