Question

Let$\mathit{R}$ be a commutative ring with identity${\mathbf{1}}_{\mathbf{R}}\mathbf{\ne }\mathbf{0}$ whose only ideals are$\left(0_R\right)$ and $\mathit{R}$. Prove that$\mathit{R}$ is a field. [Hint: If $\mathit{a}\mathbf{\ne }{\mathbf{0}}_{\mathbf{R}}$, use the ideal$\left(a\right)$ to find a multiplicative inverse for $\mathit{a}$.]

Short answer

It is proved that R is a field.

Step-by-step solution

Write the conditions for the ring R

Consider$a\in R$be any non-zero element.

As$\left(a\right)\ne 0$ then by hypothesis,

$\left(a\right)=R$so,$1\in \left(a\right)$

Show that R is a field 

By using the definition of rings we get,$b\in R$ whereas, $ab=ba=1$.

Now, use the definition of the multiplicative inverse $b=a^{-1}$.

As$a$ is an arbitrary non-zero element$R$ is a field.

Therefore, for any non-zero element we have,$1\in \left(a\right)$

Hence, the statement is proved.