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Is this okay or does it require a different approach? Let $R$ be a commutative ring with unity. Suppose that the only ideals of $R$ are $\{0\}$ and $R$, show that $R$ is a field.

Attempt:Since $\{0\}$ and $R$ are the only ideals, $\{0\}$ is maximal, thus $R/\{0\} \cong R$ so $R$ is a field since $R/\{0\}$ is a field.

mrtaurho
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Scott Frazier
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    How did you prove that $R/M$, where $M$ is a maximal ideal of a commutative ring with unity, is a field? Sometimes one uses this result to prove that, so that would make your argument circular. A direct way is to note that all you need to show $R$ is a field is to show that every nonzero element has a multiplicative inverse. Take $a\neq 0$, and consider $(a)=Ra$. – Arturo Magidin Feb 12 '21 at 06:12

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A general and easy to prove fact:

Fact: Let $R$ be a ring. Then $(r)=R$ if and only if $r\in R^\times$ (i.e. $r$ is a unit in $R$).

Proof. If $r$ is a unit then there is some $s\in R$ such that $s\cdot r=1$. Hence, $1\in(r)$ and so $a=a\cdot1\in(r)$ for all $a\in R$. So $R\subseteq(r)$ and consequently $(r)=R$. Conversely assume that $(r)=R$. Then in particular $1\in(r)$. But then by definition $s\cdot r=1$ for some $s\in R$, that is, $r$ is a unit.$~\square$

Now suppose the only ideals of your ring are $\{0\}$ and $R$. If $r\ne0$, then $(r)\ne\{0\}$ as $r\in(r)$. Thus $(r)=R$ and by the aforegoing fact it follows that $r\in R^\times$. As $r$ was an arbitrary non-zero element of our ring we conclude that every such element is invertible and hence $R^\times=R\setminus\{0\}$. Given that $R$ is commutative and unital we see that then $R$ is in fact a field.

mrtaurho
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