Correct Answer - Option 1 : M is a maximal ideal of R
Concept:
If R is a commutative ring then,
ab = ba ∀ a,b ∈ R.
M, which is an ideal of R, will be called the maximal ideal of R,
1) if M ⊂ R, M ≠ R (there is at least one element in R that does not belong to M)
2) There should be no ideal 'N', such that M ⊂ N ⊂ R. (there is no ideal between M and R).
Analysis:
R/M is a field [∵ every finite integral domain is a field]
∴ R/M is a ring with unity
∴ 1 + M ≠ M
i.e., 1 ∉ M
Now, one belongs to R, but it does not belong to R.
∴ M ≠ R.
Let I be an ideal of R
Such that M ⊆ I ⊆ R
Let, M ≠ I
∃ a ∈ I, such that a ∉ M
∴ a + M ∉ M
Now, R/M is a field.
∴ Every, non-zero of R/M is revertible
∴ a + M is invertible
∴ ∃ b + M ∈ R/M such that
(a + M) (b + M) = 1 + M
ab + M = 1 + M
ab – 1 ∈ M ⊆ I ---(1)
a ∈ I, b ∈ R
∴ ab ∈ I ---(2) (∵ I is an ideal)
From (1) and (2), we can write
ab – (ab – 1) ∈ I
∴ 1 ∈ I
Now, as unity belongs to ideal, so ideal becomes ring
∴ I = R
∴ M is a maximal ideal of R
If R is a commutative ring with unity then every maximal ideal is a prime ideal.
