Question

Prove: 2Z is a maximal ideal of Z.

Answer 1

The statement is incorrect. 2Z is not a maximal ideal of Z.

An ideal 4Z in Z, comprising integers divisible by 4, properly contains 2Z, disproving the claim that 2Z is a maximal ideal.

Step-by-step explanation:

In order to determine whether 2Z is a maximal ideal of Z, let's first recall the definitions. An ideal I of a ring R is said to be maximal if there are no ideals J in R such that I is a proper subset of J and J is a proper subset of R.

Now, consider the ring Z (integers) and the ideal 2Z, which consists of all even integers. The claim that 2Z is a maximal ideal is not true. To prove this, we need to show the existence of an ideal J in Z such that 2Z is a proper subset of J and J is a proper subset of Z.

Consider the ideal 4Z, which consists of all integers divisible by 4. It is evident that 2Z is a proper subset of 4Z (as 4 is not in 2Z), and 4Z is a proper subset of Z (as it does not contain odd integers). Thus, 2Z is not maximal, as we have found an ideal (4Z) that properly contains it.

In conclusion, 2Z is not a maximal ideal of Z, as there exists a larger ideal (4Z) within Z, proving the statement to be false.

Answer 2

2Z is a maximal ideal of Z proved.

How the proof was done

To prove that 2Z (the set of all even integers) is a maximal ideal of Z (the set of all integers),

Let's show that

1. 2Z is an ideal of Z.

2.2Z is maximal among all proper ideals of Z.

Closure under addition: For any (a, b ∈ 2Z, (a + b) ∈Z because the sum of two even numbers is even.

Closure under multiplication by integers: For any (a ∈ 2Z) and (n ∈ Z), na ∈ 2Z because the product of an even number and any integer is even.

Assume I is an ideal of Z: 2Z ⊂ I ⊂ Z).

We know that (I) must be of the form I = nZ for some integer n.

If n is odd, then 2 is not in I because 2 is even. But 2 is ∈ 2Z.

Thus, I ⊄ 2Z

Therefore, 2Z is a maximal ideal.

In conclusion, 2Z is a maximal ideal of Z.