How Does the Correspondence Theorem for Rings Prove Maximal Ideals?

[Math Amateur]

I am reading Joseph J. Rotman's book: Advanced Modern Algebra (AMA) and I am currently focused on Section 5.1 Prime Ideals and Maximal Ideals ...

I need some help with understanding the proof of Proposition 5.9 ... ...Proposition 5.9 reads as follows:
View attachment 5934 In the proof of Proposition 5.9, Rotman writes:

" ... ... The Correspondence Theorem for Rings shows that \(\displaystyle I\) is a maximal ideal if and only if \(\displaystyle R/I\) has no ideals other than \(\displaystyle (0)\) and \(\displaystyle R/I\) itself ... ... "

My question is: how exactly (in clear and simple terms) does Rotman's statement of the Correspondence Theorem for Rings lead to the statement that "\(\displaystyle I\) is a maximal ideal if and only if \(\displaystyle R/I\) has no ideals other than \(\displaystyle (0)\) and \(\displaystyle R/I\) itself" ... ...

Hope that someone can help ...

Peter

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The above post refers to Rotman's statement of the Correspondence Theorem for Rings, so I am providing a statement of that theorem and its proof, as follows:View attachment 5936

 

[Math Amateur]

I am reading Joseph J. Rotman's book: Advanced Modern Algebra (AMA) and I am currently focused on Section 5.1 Prime Ideals and Maximal Ideals ...

I need some help with understanding the proof of Proposition 5.9 ... ...Proposition 5.9 reads as follows:
In the proof of Proposition 5.9, Rotman writes:

" ... ... The Correspondence Theorem for Rings shows that \(\displaystyle I\) is a maximal ideal if and only if \(\displaystyle R/I\) has no ideals other than \(\displaystyle (0)\) and \(\displaystyle R/I\) itself ... ... "

My question is: how exactly (in clear and simple terms) does Rotman's statement of the Correspondence Theorem for Rings lead to the statement that "\(\displaystyle I\) is a maximal ideal if and only if \(\displaystyle R/I\) has no ideals other than \(\displaystyle (0)\) and \(\displaystyle R/I\) itself" ... ...

Hope that someone can help ...

Peter

============================================================

The above post refers to Rotman's statement of the Correspondence Theorem for Rings, so I am providing a statement of that theorem and its proof, as follows:

Maybe I should not be responding to my own post but I have been reflecting on the question in the above post and now suspect that the answer is quite simple and goes along the lines ... ... as follows:

\(\displaystyle I\) maximal

\(\displaystyle \Longrightarrow\) there are no ideals in \(\displaystyle R\) that contain \(\displaystyle I \) except \(\displaystyle R\) itself ...

\(\displaystyle \Longrightarrow\) there are no ideals in \(\displaystyle R/I\) (except \(\displaystyle R/I\) itself) since there exists a bijection between the set of ideals of \(\displaystyle R/I\) and the ideals of \(\displaystyle R\) containing \(\displaystyle I\) ... ...

BUT ... it seems that the only ideal in \(\displaystyle R/I\) is \(\displaystyle R/I\) itself ... but how do we explain the existence of \(\displaystyle (0)\) ...?

Seems that I still need some help ... ...

Peter

 

[steenis]

Given $I\lhd R$ (notation: $I$ is ideal in $R$), $I$ is proper, i.e., $I\neq (0)$ and $I\neq R$.
If $J\lhd R$, define $\overline J = \{a+I \mid a\in J \}$, you can prove that $\overline J = J/I$.

Define $A= \{ J\lhd R \mid I\subset J \}$ and $B=\{ K \lhd R/I \}$.

The Correspondence Theorem says that there is a bijection $\phi : A\to B$ given by $J\mapsto \overline J=J/I$.

What does this say?
a) If we have an ideal $K\lhd R/I$ then there exists an ideal $J\lhd R$ with $I\subset J$ and $J/I=K$

b) If we have an ideal $J\lhd R$ such that $I\subset J$ then $J/I \lhd R/I$

Let $I\lhd R$ be maximal, then $I$ is proper and there are no ideals between $I$ and $R$.
This means that $A$ consists of two elements: $A= \{ I, R \}$.
Therefore, $B$ consists of two elements: $B=\{ \phi (I), \phi (R) \}$.
We have $\phi (I) = (0)$ and $\phi (R) = R/I$. Thus $B=\{ (0), R/I \}$.
$B$ is the set of ideals in $R/I$, so $R/I$ has no other ideals than $(0)$ and $R/I$.

Conversely, $R/I$ has no other ideals than $(0)$ and $R/I$, i.e., $B=\{ (0), R/I \}$.
Then $A= \{ \phi ^{-1} ((0)), \phi ^{-1} (R/I) \} = \{ I, R \}$

Can you fill in the the details and apply example 5.8, now? I am going to have a break. If necessary, we continue later.

 

[Math Amateur]

Given $I\lhd R$ (notation: $I$ is ideal in $R$), $I$ is proper, i.e., $I\neq (0)$ and $I\neq R$.
If $J\lhd R$, define $\overline J = \{a+I \mid a\in J \}$, you can prove that $\overline J = J/I$.

Define $A= \{ J\lhd R \mid I\subset J \}$ and $B=\{ K \lhd R/I \}$.

The Correspondence Theorem says that there is a bijection $\phi : A\to B$ given by $J\mapsto \overline J=J/I$.

What does this say?
a) If we have an ideal $K\lhd R/I$ then there exists an ideal $J\lhd R$ with $I\subset J$ and $J/I=K$

b) If we have an ideal $J\lhd R$ such that $I\subset J$ then $J/I \lhd R/I$

Let $I\lhd R$ be maximal, then $I$ is proper and there are no ideals between $I$ and $R$.
This means that $A$ consists of two elements: $A= \{ I, R \}$.
Therefore, $B$ consists of two elements: $B=\{ \phi (I), \phi (R) \}$.
We have $\phi (I) = (0)$ and $\phi (R) = R/I$. Thus $B=\{ (0), R/I \}$.
$B$ is the set of ideals in $R/I$, so $R/I$ has no other ideals than $(0)$ and $R/I$.

Conversely, $R/I$ has no other ideals than $(0)$ and $R/I$, i.e., $B=\{ (0), R/I \}$.
Then $A= \{ \phi ^{-1} ((0)), \phi ^{-1} (R/I) \} = \{ I, R \}$

Can you fill in the the details and apply example 5.8, now? I am going to have a break. If necessary, we continue later.

Thanks for for your assistance, Steenis ... most helpful ...

Reflecting on what you have said ...

Thanks again,

Peter