characterization of fieldCharacterizationOfField2003-09-09 12:57:262004-10-05 13:02:04Theoremfield% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Rats}{\mathbb{Q}}\begin{prop}
Let $\mathcal{R}\neq 0$ be a commutative ring with identity. The ring $\mathcal{R}$ (as above) is a field if and only if
$\mathcal{R}$ has exactly two ideals: $(0),\mathcal{R}$.
\end{prop}
\begin{proof}
{\bf ($\Rightarrow$)} Suppose $\mathcal{R}$ is a field and let
$\mathcal{A}$ be a non-zero ideal of $\mathcal{R}$. Then there
exists $r\in \mathcal{A}\subseteq \mathcal{R}$ with $r\neq 0$.
Since $\mathcal{R}$ is a field and $r$ is a non-zero element,
there exists $s\in \mathcal{R}$ such that
$$s\cdot r =1 \in \mathcal{R}$$
Moreover, $\mathcal{A}$ is an ideal, $r\in \mathcal{A}, s\in
\mathcal{S}$, so $s\cdot r =1 \in \mathcal{A}$. Hence
$\mathcal{A}=\mathcal{R}$. We have proved that the only ideals of
$\mathcal{R}$ are $(0)$ and $\mathcal{R}$ as desired.
{\bf ($\Leftarrow$)} Suppose the ring $\mathcal{R}$ has only two
ideals, namely $(0),\mathcal{R}$. Let $a\in \mathcal{R}$ be a
non-zero element; we would like to prove the existence of a
multiplicative inverse for $a$ in $\mathcal{R}$. Define the
following set:
$$\mathcal{A}=(a)=\{r\in\mathcal{R} \mid r=s\cdot a, \text{ for
some } s\in\mathcal{R}\}$$ This is clearly an ideal, the ideal
generated by the element $a$. Moreover, this ideal is not the zero
ideal because $a\in \mathcal{A}$ and $a$ was assumed to be
non-zero. Thus, since there are only two ideals, we conclude
$\mathcal{A}=\mathcal{R}$. Therefore $1\in
\mathcal{A}=\mathcal{R}$ so there exists an element $s\in
\mathcal{R}$ such that
$$s\cdot a=1 \in \mathcal{R}$$
Hence for all non-zero $a\in \mathcal{R}$, $a$ has a multiplicative inverse in $\mathcal{R}$, so $\mathcal{R}$ is, in fact, a field.
\end{proof}