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

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In summary: 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)There are the following cases: If $rI\neq R$, then it must be $rI\subseteq I$, since $I$ is the unique maximal right ideal of $R$. If $r
  • #1
mathmari
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Hey! :eek:

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)
 
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  • #2
mathmari said:
Hey! :eek:

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 $rI$?
 
  • #3
Opalg said:
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)
 
  • #4
mathmari said:
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 ...
 
  • #5
Opalg said:
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)
 
  • #6
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)
 

Related to Is $I$ an Ideal in a Ring $R$?

1. What is an ideal?

An ideal is a mathematical concept used in abstract algebra that represents a subset of a ring with special properties. It is a generalization of the concept of a multiple of an integer.

2. How is an ideal denoted?

An ideal is denoted by the symbol I, usually in italics. It is written as I ⊂ R, where R is the ring that the ideal is a subset of.

3. What are the requirements for I to be an ideal?

In order for I to be an ideal, it must satisfy two conditions: closure under addition and closure under multiplication by elements of the ring R. This means that for any x, y ∈ I and r ∈ R, x + y ∈ I and rx ∈ I.

4. How is an ideal different from a subring?

An ideal is a subset of a ring that is closed under multiplication by elements of the ring, whereas a subring is a subset of a ring that is closed under addition, subtraction, and multiplication. Additionally, every ideal is a subring, but not every subring is an ideal.

5. What are the applications of ideals?

Ideals are used in many areas of mathematics, including number theory, algebraic geometry, and coding theory. They also have applications in physics and engineering, such as in the study of symmetries and conservation laws.

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