Answer to Question #206647 in Abstract Algebra for Yogesh yadav

Question

let R be comutetive ring with unity in which each ideal prime then show that R is a field.

Answer 1

As the zero ideal (0) of R is a proper ideal, it is a prime ideal by assumption.

Hence R=R/{0} is an integral domain.

Let a be an arbitrary nonzero element in R. We prove that a is invertible. Consider the ideal (a²) generated by the element a². If (a²)=R, then there exists b∈R such that 

1=a²b  as 1∈R=(a²). Hence we have 1=a(ab) and a is invertible. Next, if (a²) is a proper ideal, then

(a²) is a prime ideal by assumption. Since the product $a\cdot a=a^2$ is in the prime ideal (a²), it follows that $a\isin (a^2)$ . Thus, there exists $b\isin R$ such that a=a²b.

Equivalently, we have a(ab-1)=0.

We have observed above that R is an integral domain. As $a\ne 0$ , we must have ab-1=0,

and hence ab=1. This implies that a is invertible.

Therefore, every nonzero element of R is invertible.

Hence R is a field.