show that a commutative Ring with unity is a field if and only if it has no proper non trivial ideal

R is a commutative Ring with Unity
assume that are R is a field
there for R has no proper non trivial ideal

(wacth this video for these steps N=R)

N is an ideal of R ⇒ N is a subring of R
N ⊆ R ⇒(1)
N is an ideal of R containing a unit u
u ∈ N
(unit:An element which has its inverse)

u ∈ N ⇒ u ∈ R
(since N is a subring of R)
u ∈ R ⇒ u' ∈ R
(since u is a element in a field, every non zero element in field has inverse)

Nb ⊆ N ∀ b ∈ R ( equation of ideal)
Nu' ⊆ N ∀ u' ∈ R
uu' ∈ N , since u ∈ N
uu'=1
1 ∈ N

let x∈ R
Nb ⊆N ,∀ b ∈R
Nr ⊆ N , ∀ r ∈ R
1.x⊆ N , since 1 ∈N
x∈ N

R ⊆ N ⇒(2)

from (1) & (2)we get R=N

conversely asume that R has no proper non trivial ideal so that maximal ideal of R is {0}
we know that R is commutative Ring with Unity then M is maximal ideal iff R/M is a field

here {0} is maximal ideal
R/{0} is a field
R is a field