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Can an improper ideal ($\varnothing$ or $R$) be a principal one in the ring $R$?

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closed as off-topic by GNUSupporter 8964民主女神 地下教會, callculus, José Carlos Santos, Leucippus, YuiTo Cheng May 5 at 1:09

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Proper or not, an ideal is an additive subgroup of $R$, therefore it isn't empty. $\{0\}$ is a principal ideal, its generator being $0$. $R$ is principal in rings with $1$, and $1$ itself is its most notable generator; it may not be principal in rings without unity.

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  • $\begingroup$ Thank you! Now it is clear to me. $\endgroup$ – Bonrey May 4 at 15:31
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The ideal $\langle\{1\}\rangle = R$ is principle, where $1$ is the unit element in $R$, and the ideal $\langle\emptyset\rangle = \{0\}$ is principle, where $0$ is the zero element in $R$.

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