Q:

Prove that R is a field ↔ The only ideals in R are R(0) and (0).

Possible Duplicate:
Rings and ideals 

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

A:

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.

