MHB Prime Ideal in a Commutative Ring - Rotman Proposition 7.5

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently studying Section 7.1 Prime Ideals and Maximal Ideals ... ...

I need help with understanding an aspect of the proof of Proposition 7.5

Proposition 7.5 and its proof reads as follows:View attachment 4727In the first part of the proof of the proposition above we read the following:

"Let $$I$$ be a prime ideal. Since $$I$$ is a proper idea, we have $$1 \notin I$$ and so $$1 + I \neq 0 + I$$ in $$R/I$$ ... ... ... "

My question is ... ... why is Rotman taking trouble to show that $$1 + I \neq 0 + I$$ in $$R/I$$?

What is the point Rotman is making ... ... ?

Hope someone can clarify this matter ... ...

Peter

 

[Fallen Angel]

Hi Peter,

He is just avoiding the case when $I$ is not proper and then $R/I\cong \{0\}$.

 

[mathbalarka]

Fallen Angel has answered the question already, but I will elaborate a bit more on this in case it's not clear :

A proper ideal of a ring $R$ is an ideal which is not the whole ring $R$. By definition, prime ideals are proper. If a prime ideal $I \subset R$ contained the identity $1$, then it has to contain every element of $R$ by definition of an ideal, hence forcing it to be not proper - contradiction.

Thus, $1 \notin I$, which in turn implies $0 \mod I \neq 1 \mod I$ (note : I use $a \mod I$ to denote an element of $R/I$ instead of $a + I$).

 

[Math Amateur]

Fallen Angel has answered the question already, but I will elaborate a bit more on this in case it's not clear :

A proper ideal of a ring $R$ is an ideal which is not the whole ring $R$. By definition, prime ideals are proper. If a prime ideal $I \subset R$ contained the identity $1$, then it has to contain every element of $R$ by definition of an ideal, hence forcing it to be not proper - contradiction.

Thus, $1 \notin I$, which in turn implies $0 \mod I \neq 1 \mod I$ (note : I use $a \mod I$ to denote an element of $R/I$ instead of $a + I$).

Fallen Angel, Mathbalarka

Thanks for your help ... appreciate you assistance ...

Peter