$I$ is the maximal left ideal Let $R$ be a ring and $I\subseteq R$ the unique maximal right ideal of $R$. 
I have shown that $I$ is an ideal and that each element $a\in R-I$ is invertible. 
I want to show that $I$ is the unique maximal left ideal of $R$. 
I found the following solution: 
If $R\neq T$ is a left ideal then $x\in T$, $x$ is not invertible. So
It must be $x\in I$. 
Therefore, each right ideal $T\subseteq I$. 
Could you explain to me that solution?.I haven't understood that... 
 A: Let $I$ be the unique maximal right ideal of $R$. This means that


*

*the only right ideal properly containing $I$ is $R$ (maximality);

*every proper right ideal of $R$ is contained in $I$.


Condition 2 follows from the fact that every proper right ideal is contained in a maximal right ideal.
If you have correctly proved that each element of $R\setminus I$ is invertible, then you have also proved that $I$ consists of all non-invertible elements, because a proper ideal cannot contain an invertible element. To be more explicit, let $U(R)$ be the set of invertible elements. Then you have proved that $R\setminus I\subseteq U(R)$, which implies $I\supseteq R\setminus U(R)$; but you also know that $I\cap U(R)=\emptyset$, so $I\subseteq R\setminus U(R)$. Therefore $I=R\setminus U(R)$.
Now, it's easy to prove that $I$ is also a two-sided ideal: if $x\in I$ and $r\in R$, then $rx$ cannot be invertible, so it belongs to $I$.
Let $M$ be a maximal left ideal of $R$. Then no element of $M$ is invertible, which means $M\subseteq I$, because $I$ consists precisely of all non-invertible elements. By maximality of $M$, we conclude $M=I$.
