Maximal Ideal .... Bland - AA - Example 2, Section 3.2.12 .... ....

[Math Amateur]

I am reading The Basics of Abstract Algebra by Paul E. Bland ...

I am focused on Section 3.2 Subrings, Ideals and Factor Rings ... ...

I need help with the proof of Example 2, Section 3.2.12, pages 147 to 148 ... ... Example 2, Section 3.2.12 reads as follows:
View attachment 8262
https://www.physicsforums.com/attachments/8263
In the above example Bland shows that if \(\displaystyle I\) is an ideal of \(\displaystyle \mathbb{Z}\) such that \(\displaystyle 5 \mathbb{Z} \subset I \subseteq \mathbb{Z}\) then \(\displaystyle I = \mathbb{Z}\) ... Bland then claims that \(\displaystyle I\) is a maximal ideal of \(\displaystyle \mathbb{Z}\) ...... BUT ...... doesn't Bland also have to show that if \(\displaystyle I\) is an ideal of \(\displaystyle \mathbb{Z}\) such that \(\displaystyle 5 \mathbb{Z} \subseteq I \subset \mathbb{Z}\) then \(\displaystyle I = 5 \mathbb{Z}\) ... ?Can someone explain why Bland's proof is complete as it stands ...

Peter============================================================================***NOTE***

It may help readers to have access to Bland's definition of a maximal ideal ... so I am providing the same as follows:https://www.physicsforums.com/attachments/8264
https://www.physicsforums.com/attachments/8265Sorry about the legibility ... but Bland shades his definitions ...Peter

 

[castor28]

Hi Peter,

Actually, Bland claims that $5\mathbb{Z}$ (not $I$ as you wrote) is a maximal ideal of $\mathbb{Z}$.

The two statements are equivalent to ‘‘if $5\mathbb{Z}\subseteq I \subseteq\mathbb{Z}$, then $I=5\mathbb{Z}$ or $I=\mathbb{Z}$’’ (this is the definition in the book).

To see this, let us assume that $5\mathbb{Z}\subseteq I\subset\mathbb{Z}$. If the first inclusion is proper, then, by what has been proved in the example, we must have $I=\mathbb{Z}$, and this contradicts the assumption that the second inclusion is proper.

 

[Olinguito]

In general, to show that an ideal $M$ of a ring $R$ (i.e. commutative ring with multiplicative identity) is maximal, you can show that for any ideal $I$, either$$M\subset I\subseteq\mathbb Z\ \implies\ I=\mathbb Z$$or$$M\subseteq I\subset\mathbb Z\ \implies\ I=M.$$You don’t have to do both.

 

[steenis]

A similar argument shows that $p \mathbb{Z}$ is a maximal ideal of $\mathbb{Z}$ for each prime $p$.