Question

Let R be a ring with identity and let I be an ideal in R.

1. If ${\mathbf{1}}_{\mathbf{R}}\mathbf{\in }\mathit{I}$, prove that I=R.

Short answer

It is proved I = R.

Step-by-step solution

Ideal definition

By the ideal definition, if $I\subseteq R$, then $R\subseteq I$.

Prove that I = R

Consider that a is the arbitrary element that belongs to R and I;it is ideal, so $1_R\in I$.

As I is ideal,$a{1}_R\in I$ . If I_R is an identity, then $a{1}_R=a$. Thus, $a\in I$.

Also, a is the arbitrary element; then, $R\subseteq I$and $I\subseteq R$.

Therefore, $I=R$.