A commutative ring $R$ is a principal ideal domain (PID) if $R$ is a domain and any ideal $I$ is generated by a single element $a\in I$, that is $I=(a)$.
Since $R$ is a PID, we can write $P=(a)$, an ideal generated by an element $a\in R$.
Since $P$ is a nonzero ideal, the element $a\neq 0$.
Now suppose that we have
\[P \subset I \subset R\]
for some ideal $I$ of $R$.
We can write $I=(b)$ for some $b \in R$ since $R$ is a PID.
The element $a\in (a) \subset (b)$ and so there is an element $c \in R$ such that $a=bc$.
Since $a=bc$ is in the prime ideal $P$, we have either $b \in P$ or $c \in P$.
If $b\in P$, then it follows that $I=(b)\subset P$, and hence $P=I$.
If $c \in P=(a)$, then we have $d\in R$ such that $c=ad$.
Then we have
and since $R$ is a domain and $a\neq 0$, we have
This yields that $b$ is a unit and hence $I=(b)=R$.
In summary, we observe that whenever we have $P \subset I \subset R$, we have either $I=P$ or $I=R$. Thus $P$ is a maximal ideal.
Characteristic of an Integral Domain is 0 or a Prime Number
Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.
Definition of the characteristic of a ring.
The characteristic of a commutative ring $R$ with $1$ is defined as […]
Every Maximal Ideal of a Commutative Ring is a Prime Ideal
Let $R$ be a commutative ring with unity.
Then show that every maximal ideal of $R$ is a prime ideal.
We give two proofs.
The first proof uses the following facts.
Fact 1. An ideal $I$ of $R$ is a prime ideal if and only if $R/I$ is an integral […]