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I have read the theorem which states that "Every Euclidean domain is a Principal ideal domain." I found that every non-zero ideal $I$ of a Euclidean domain $R$ is generated by a non-zero element $a \in I$ for which $\delta (a)$ is the least, where $\delta$ is a Euclidean valuation on $R$ defined by $\delta (a + bi) = a^{2} + b^{2}$, for all $a + bi \in \mathbb Z {[i]}$ .I have proved that $\mathbb Z {[i]}$ is a Euclidean domain.Now by the above mentioned theorem we find that any non-zero ideal is generated by one of $1, -1, i, -i$.But all those elements generates the whole of $\mathbb Z {[i]}$.Which in turn implies $\mathbb Z {[i]}$ is simple.

I think I have done mistakes in my above consideration.Because we know that "If $R$ is a commutative ring with identity then $R$ is simple if and only if $R$ is a field." Now if $\mathbb Z{[i]}$ were simple then by this theorem it becomes a field.Which is not true.

Please tell me where have I done mistakes.

Thank you in advance.

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$I$ is generated by an $a$ where $\delta(a)$ is minimal among elements of $I$, not elements of $R$. So , for instance, $\{a+bi\mid a,b\text{ are even}\}\subseteq \Bbb Z[i]$ is generated by $2$, and cannot be generated by $1,-1,i,-i$ because none of those are in $I$.

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    Yes you are right.2017-01-10