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.