MHB Is $I$ an Ideal in a Ring $R$?

[mathmari]

Hey!

Let $R$ be a ring and let $I\subseteq R$ the unique maximal right ideal of $R$.
I want to show the following:

1. $I$ is an ideal
2. each element $a\in R-I$ is invertible
3. $I$ is the unique maximal left ideal of $R$

For 1. we have to show that $I$ is also a left ideal, right? (Wondering)

Could you give me some hints how we could show that? (Wondering)

 

[Opalg]

Hey!

Let $R$ be a ring and let $I\subseteq R$ the unique maximal right ideal of $R$.
I want to show the following:

1. $I$ is an ideal
2. each element $a\in R-I$ is invertible
3. $I$ is the unique maximal left ideal of $R$

For 1. we have to show that $I$ is also a left ideal, right? (Wondering)

Could you give me some hints how we could show that? (Wondering)

For 1., let $r\in R$. What can you say about the set $rI$?

 

[mathmari]

For 1., let $r\in R$. What can you say about the set $rI$?

We have that $I$ is a right ideal, so it is an additive subgroup of $R$, so when $i,j\in I$ then $-i, i+j\in I$.

Let $a,b\in rI$, then $a=ri_1, b=ri_2, \ i_1, i_2\in I$.
Then we have the following:
$$-a=-(ri_1)=r(-i_1)\in rI \\
a+b=ri_1+ri_2=r(i_1+i_2)\in rI$$

So, $rI$ is an additive subgroup of $R$. For $x\in R$, $a'=ra\in rI$ we have that $a'x=rax\in rI$, since $I$ is a right ideal and so $ax\in I$.

That means that $rI$ is also a right ideal, right? (Wondering)

 

[Opalg]

We have that $I$ is a right ideal, so it is an additive subgroup of $R$, so when $i,j\in I$ then $-i, i+j\in I$.

Let $a,b\in rI$, then $a=ri_1, b=ri_2, \ i_1, i_2\in I$.
Then we have the following:
$$-a=-(ri_1)=r(-i_1)\in rI \\
a+b=ri_1+ri_2=r(i_1+i_2)\in rI$$

So, $rI$ is an additive subgroup of $R$. For $x\in R$, $a'=ra\in rI$ we have that $a'x=rax\in rI$, since $I$ is a right ideal and so $ax\in I$.

That means that $rI$ is also a right ideal, right? (Wondering)

That is correct. For the next step, what I'm hoping is true is that every right ideal should be contained in a maximal right ideal ...

 

[mathmari]

That is correct. For the next step, what I'm hoping is true is that every right ideal should be contained in a maximal right ideal ...

Since $I$ is a maximal right ideal, $rI$ is either a subset of $I$ or it is the entire ring $R$.

If $r$ has not a right inverse in $I$, then $rI$ doesn't contain $1$, so it must be $rI\subset I$.

Is this correct? (Wondering)

What if $r$ has a right inverse? (Wondering)

 

[mathmari]

There are the following cases:

1. If $rI\neq R$, then it must be $rI\subseteq I$, since $I$ is the unique maximal right ideal of $R$.
   If $rI\subseteq I$, then $ri\in I, \forall i\in I, \forall r\in R$.
   So, $I$ is a left ideal.
   Since $I$ is a right and a left ideal, it follows that $I$ is an ideal.
   
   $$$$
   
2. If $rI=R$, then $ri=1$ for some $i\in I$.
   We have that $ir\in I$ since $I$ is a right ideal, so $ir\neq 1$.
   Then I thought to use post #8 of http://mathhelpboards.com/linear-abstract-algebra-14/statement-true-18331.html#post84301 to say that since $1-ir$ has no right inverse, it follows that $(1-ir)R\neq R$.
   Then $(1-ir)R$ is contained in the maximal right ideal, i.e., $(1-ir)R\subseteq I$. Then $1-ir\in I$.
   We have that $1=ir+(1-ir)\in I$, a contradiction.
   So, $rI\neq R$.
   
   But at the other post they told me that the argument of post #8 is not true... Do you maybe have an other idea how we can reject the case $rI=R$ ? (Wondering)