Why is every ideal in $\mathbb Z[ i ] / (5)$ principal?

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

Prove that any ideal in $R$ is principal.

Any ideas on how to prove this?

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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.

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$\mathbb{Z}[i]$ 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.

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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