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A subset $I$ of a ring/semiring $R$ is said to be an ideal of $R$ if $x+y\in I$ for all $x,y\in I$ and $x.a\in I$($a.x\in I$) for all $a\in R$. Since $0\in R,$ $x.0\in I$ for all $x\in I$ or $0\in I.$ This shows that the identity $0$ necessarily belongs to $I$. Is this intuition correct?

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  • $\begingroup$ Yes that is correct. Moreover, $(I,+)$ must form a subgroup of $(R,+)$ for a ring $R$ so that we know we must have $0 \in I$. $\endgroup$
    – FofX
    Feb 1, 2019 at 3:31

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That's not just intuition. It's an actual proof that any ideal must contain the additive identity.

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Yes. $0$ has to be in the ideal, since, as you said, $0\in\mathcal R$. So $0=0\cdot x\in\mathscr I$, for any $x\in\mathscr I$.

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