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Consider the following (part of a) proof:

A commutative ring $R$ where the only ideals are $(0)$ and $R$ is a field:

Proof:

Suppose that $R$ is not a field. Then there is some non-zero element $x$ $\in$ $R$ which does not have an inverse, so $(x)$ is not $(0)$ and can not contain $1$, so it is not $R$. Hence we have an ideal which is not $(0)$ or $R$.

This is the contrapositive and is a satisfactory proof.

Question

If we are assuming that $R$ is a ring with $1$, then the above proof is trivial, because our ideal $(x)$ doesn't have $1$ in it, and then it definitely isn't $R$.

Do we necessarily need $R$ to be a ring with 1 for the proof to work? Specifically, the step saying that $(x)$ is not R?

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    How do you choose $x$ without knowing what an inverse is ? (What is an inverse if there is no $1$ ?) – 2017-02-06
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    If $1\notin R$, what is $(x)$? Is it elements of the kind $ax, a\in R$ or the minimal ideal containing $x$? – 2017-02-06
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    Wolfram : these two things coincide no matter whether $1\in R$ or not (they don't coincide when $R$ isn't commutative, but here it is) – 2017-02-06

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Consider the ring $2\mathbb{Z}$ with ideal $4\mathbb{Z}$. It is clear that $2\mathbb{Z}$ does not have $1$. The quotient ring $R=2\mathbb{Z}/4\mathbb{Z}$ does not have the identity, it only has trivial ideals, and it is not a field. The only nonzero element is nilpotent.

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    That's efficient as hell ! I was desperately looking for an example of a ring $R$ with $x\in R$ such that for all $y\in R, xy=0$, but couldn't find any. And then you showed up :p – 2017-02-06
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    @max You can take literally any abelian group, define $xy=0$ for all $x,y$ and you have a ring like the one you were looking for. – 2017-02-06
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    rschwieb : Indeed ! Although gobucksmath's answer looks less ad hoc – 2017-02-06