Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R = \mathbb Z[ i ] / (5)$ .

Prove that any ideal in $R$ is principal.

Any ideas on how to prove this?

share|cite|improve this question

Your ring is isomorphic to $\Bbb{Z}[x]/(5,x^2+1) \cong \Bbb{F}_5[x]/(x^2 + 1) $. Now notice that $x^2 + 1 = (x+2)(x+3)$ in here. The ideals $(x+2)$, $(x+3)$ are coprime since $x+3 - x-2 = 1$ and hence by the Chinese Remainder Theorem

$$\Bbb{F}_5[x]/(x^2 + 1) \cong \Bbb{F}_5[x]/(x+2) \oplus \Bbb{F}_5[x](x+3) \cong \Bbb{F}_5 \times \Bbb{F}_5.$$

This is a product of PIRs and hence is a PIR. Alternatively, $\Bbb{Z}[i]$ is a Dedeking domain since it is the ring of integers of $\Bbb{Q}(i)$, and hence by this exercise here is a PIR.

share|cite|improve this answer

$\mathbb{Z}[i]$ is a Euclidean domain hence it is a principal ring. Consider the homomorphism $\phi:\mathbb{Z}[i]\rightarrow \mathbb{Z}[i]/\left<5\right>$ that sends $x$ to $x+\left<5\right>$. This homomorphism is surjective, thus $\mathbb{Z}[i]/\left<5\right>$ is a principal ring.

share|cite|improve this answer
Quotients of PIDs need not be PIDs, and the given $R$ is in fact not even a domain, as $(2+i)(2-i) = 0$. – arkeet Dec 14 '12 at 7:37
Yes I should have said that it is just a principal ring. – Amr Dec 14 '12 at 7:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.