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Are there examples of nontrivial subrings of $\mathbb{R}$ that do not contain $1$? If not, how can we prove they don't exist? The definition of "ring" here is really "rng"; rings do not have to contain $1$.


If $m \in \mathbb{Z}$, then $\mathbb{Z}[\sqrt{m}] = \{ a + b \sqrt{m} : a, b \in \mathbb{Z} \}$ is a subring of $\mathbb{R}$, but it contains $1$. Similarly, $\mathbb{Z}$ and $\mathbb{Q}$ are subrings, but they also contain $1$.


EDIT: there are quite a lot of counterexamples ($\mathbb{Z}a$ for $a \in \mathbb{Z}_{\geq 2}$, e.g. the even integers $\mathbb{Z}2$). I am still interested in seeing other counterexamples. Is there a way to classify all the counterexamples?

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  • $\begingroup$ Why was this downvoted? $\endgroup$
    – jskattt797
    Commented Aug 30, 2020 at 16:51

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The easy way to construct more general counterexamples is to take a subring $K<\Bbb R$ containing $1$ and then take its proper ideal $I$. Then $1\not\in I$.

In fact all examples can be obtained this way. Indeed, let $Y$ be a subring of $\mathbb R$ which does not contain $1$. Consider the ring $K=\Bbb{Z}+Y=\{n+y\mid n\in\Bbb Z, y\in Y\}$. Clearly $K$ is a subring containing $1=1+0$ and $Y$ is an ideal in $K$.

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While writing the question, I thought of a family of counterexamples. If $H$ is a nontrivial subgroup of $\mathbb{Z}$, then $H = \mathbb{Z}a = \{ka : k \in \mathbb{Z}\}$, where $a$ is the smallest positive integer in $H$. Such subgroups are also subrings. For $a > 1$, they do not contain $1$.

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