If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field.

Prime Ideal Problems and Solution in Ring Theory in Mathematics

Problem 598

Let $R$ be a commutative ring with $1$.

Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.

 
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Proof.

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 $(a^2)$ generated by the element $a^2$.

If $(a^2)=R$, then there exists $b\in R$ such that $1=a^2b$ as $1\in R=(a^2)$.
Hence we have $1=a(ab)$ and $a$ is invertible.

Next, if $(a^2)$ is a proper ideal, then $(a^2)$ is a prime ideal by assumption.
Since the product $a\cdot a=a^2$ is in the prime ideal $(a^2)$, it follows that $a\in (a^2)$.
Thus, there exists $b\in R$ such that $a=a^2b$.
Equivalently, we have $a(ab-1)=0$.


We have observed above that $R$ is an integral domain. As $a\neq 0$, 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.


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