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

• MHB
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In summary: This is the definition given in the book.In summary, Bland's proof that I is a maximal ideal of $\mathbb{Z}$ is complete as it stands.
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MHB
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

Last edited:
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.

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.

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

## 1. What is a maximal ideal?

A maximal ideal is a type of ideal in abstract algebra that is not a proper subset of any other ideal in a given ring. It is the largest possible ideal in the ring and cannot be further extended.

## 2. How do you determine if an ideal is maximal?

To determine if an ideal is maximal, one can use the second isomorphism theorem. If the given ideal is the sum of two ideals, and one of those ideals is the entire ring, then the given ideal is maximal.

## 3. What is the significance of maximal ideals?

Maximal ideals are important in ring theory because they allow for the construction of quotient rings, which can reveal the underlying structure of the original ring. They also have applications in algebraic geometry and number theory.

## 4. Can every ring have a maximal ideal?

No, not every ring has a maximal ideal. For example, fields have no proper ideals, so they cannot have maximal ideals. Additionally, some rings may have multiple maximal ideals.

## 5. How are maximal ideals related to prime ideals?

All maximal ideals are prime ideals, but not all prime ideals are maximal. A prime ideal is a proper subset of a ring that does not contain any zero divisors, while a maximal ideal is the largest possible ideal. Every maximal ideal is a prime ideal because it cannot contain any non-zero divisors, but not every prime ideal is maximal because it may be a proper subset of a larger ideal.

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