6
$\begingroup$

Is $(11)$ a prime ideal of $\mathbb{Z}[\sqrt{-5}]$? I know that $11$ is an irreducible element in $\mathbb{Z}[\sqrt{-5}]$. Now to determine whether it is prime we can say $\mathbb{Z}[\sqrt{-5}]$ isomorphic to $\mathbb{Z}[x]/(x^2 + 5)$. So we get an isomorphism $$ \mathbb{Z}[\sqrt{-5}]/(11) \;\;\simeq\;\; \mathbb{Z}_{11}[x]/(x^2 + 5) \,.$$

Since $\mathbb{Z}_{11}$ is a field, $\mathbb{Z}_{11}[x]$ is a PID, and since $(x^2 + 5)$ is irreducible over $\mathbb{Z}_{11}[x]$, the ring $\mathbb{Z}_{11}[x]/(x^2 + 5)$ is a field. Hence $(11)$ can be treated as a maximal ideal as well as a prime ideal in the ring $\mathbb{Z}[\sqrt{-5}]$.

  • 1
    Your method is correct and answer too. Refer this [link](http://math.stackexchange.com/questions/1166143/show-that-langle-13-rangle-is-a-prime-ideal-in-mathbbz-sqrt-5?rq=1) for more methods. – 2017-01-30

1 Answers 1

2

As indicated in the comments, yes, you are correct. It might be wise to justify why $(x^2+5)$ is irreducible over $\mathbb{Z}_{11}[x]$ though.

I'm posting this CW answer so that users who confidently concur have something to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue. If however anyone would like to write a more substantial response to the question, please downvote this answer and post your response.