Maximal Ideal: Definition, Examples, Solved Problems

A maximal ideal of a ring R is an ideal that is not contained in any proper ideal of R. For example, 2ℤ is a maximal ideal of ℤ, but 4ℤ is not a maximal of ℤ as 4ℤ ⊂ 2ℤ. In this article, we will study maximal ideals, its definition, examples with some solved problems.

Definition of a Maximal Ideal

An ideal M of a ring R is said to be Maximal if M ≠ R and there is no proper ideal of R strictly containing M.

More precisely, An ideal M is called a maximal ideal in R if it satisfies the property: for any ideal U of R with M ⊂ U ⊂ R then either U=M or U=R.

For example, 3ℤ is a maximal ideal in the ring ℤ.

Examples

A few examples of maximal ideal are listed here.

• The null ideal {0} is a maximal ideal in a field F. This is because there are only two ideal of F, namely {0} and F.
• In a simple ring, the null ideal {0} is always maximal.
• The ideal 2ℤ is a maximal ideal in the ring ℤ.
• If p is a prime, then the ideal pℤ is a maximal ideal in the ring ℤ.

Non-Examples

The ideal 6ℤ is not maximal in the ring ℤ, since 6ℤ is contained in the proper ideal 2ℤ of ℤ. Similarly, 9ℤ, 10ℤ, 12ℤ are not maximal ideals in ℤ.

Consider the ring ℤ × ℤ. Both S = {(a, 0) : a ∈ ℤ} and T = {(a, 2b) : a, b ∈ ℤ} are ideals in ℤ × ℤ. Note that S ⊂ T. So S is not a maximal ideal in ℤ × ℤ.

Properties

1. In a commutative ring R with unity, an ideal M of R is maximal if and only if the quotient ring R/M is a field. But, this fails when the ring has no unity. For example, 4ℤ is a maximal ideal in 2ℤ, but 4ℤ/2ℤ is not a field.
2. In a commutative ring with unity, every maximal ideal is a prime ideal. This is not true for a ring without unity. For example, 4ℤ is a maximal ideal in 2ℤ, but it is a prime ideal in 2ℤ.

Read This: Ring Theory: Definition, Examples, Properties, Theorems

Solved Problems

Question 1: Prove that 2ℤ is a maximal ideal in ℤ.

Answer:

Let U be an ideal of ℤ such that 2ℤ ⊂ U ⊂ ℤ. To prove 2ℤ is a maximal ideal in ℤ, we need to show that U=ℤ.

Note that U is a principal ideal, since ℤ is a principal ideal ring.

So U = <m> for some integer m.

⇒ U = mℤ.

Now, 2ℤ ⊂ U = mℤ implies that 2 ∈ mℤ.

So m divides 2.

This again implies that m=1 or 2.

If m=1 then mℤ=ℤ, and if m=2 them mℤ = 2ℤ.

This shows 2ℤ ⊂ U ⊂ ℤ implies that either U = 2ℤ or U=ℤ. So the ideal 2ℤ is maximal in ℤ.

More Topics:

• Units of a Ring
• Characteristic of a Ring
• Zero Divisors in a Ring
• Integral Domain
• ℤ_n is an Integral Domain iff n is Prime
• Every Finite Integral Domain is a Field

Question 2: Prove that 4ℤ is not a maximal ideal in ℤ.

Answer:

Both 2ℤ and 4ℤ are proper ideals in the ring ℤ. The ideal 4ℤ is strictly contained in 2ℤ, that is, 4ℤ ⊂ 2ℤ. So by definition, 4ℤ is not a maximal ideal in ℤ.

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