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Rings and ideals

The question is the title, any help would be greatly appreciated!

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    PS: The statement is false as given: the ring of $2\times 2$ matrices over a field are simple (so the only ideals are $(0)$ and the entire ring) but they are not even a division ring. You should (i) put the question in the body, not just the title; and (ii) put all the hypothesis in the question. This problem almost certainly included the words "commutative" and "with unity" somewhere along the line. – Arturo Magidin Jan 21 '12 at 23:04
  • I would think the definition of a ring does not [almost certainly not require unity]. $;$ –  Jan 21 '12 at 23:12

1 Answers1

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Below is a proof that a commutative ring $R$ is a field iff its only ideals are $R$ and $(0)$:

Suppose first that $R$ is a field. Then any nonzero ideal $I$ must contain $1$, as for any $0\neq x\in I$ it must contain $x^{-1}x=1$, hence contains $r\cdot1=r$ for any $r\in R$ so $R\subseteq I$. By definition $I\subseteq R$, so $I=R$. Suppose instead that $R$ is not a field. Then we have some nonzero element $x\in R$ which has no inverse, so $(x)=\{rx | r\in R\}$ is not $(0)$ and cannot contain $1$ (as this would mean we have some $r\in R$ such that $rx=1$, and so since $R$ is commutative $rx=xr=1$ hence $x^{-1}=r$) so is not $R$. Thus $R$ is a field iff its only ideals are $R$ and $(0)$.

Note that $R$ may or may not be considered an ideal of $R$, depending on convention. Some authors reserve the term "ideal" for proper ideals, that is ideals which are not the whole ring. This convention is typical among authors who require rings to have $1\neq 0$, as this means that the trivial ring $R/R$ is not a ring under their definition.

Alex Becker
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  • Your answer implicitly assumes that R contains identity when you claim that (x) doesn't contain 1 hence is not R. This is not clear at all as to why not containing 1 means (x) is not R unless R is known to contain 1. – ImBatman Apr 24 '23 at 15:49