An element $a$ of a commutative ring $R$ is called a zero divisor if there is $b\neq0$ in $R$ such that $ab=0$.
If a ring $R$ contains no nonzero zero divisors, then we call $R$ an integral domain.
Suppose that we have
for $a, b \in R$. To show that $R$ has no nonzero zero divisors, we need to prove that $a$ or $b$ is the zero element.
Since $ab=0\in P$ and $P$ is a prime ideal, either $a\in P$ or $b\in P$.
Without loss of generality, we may assume $a\in P$.
If $a=0$, then we are done.
So assume that $a\neq 0$. Then since $P$ does not contain any nonzero zero divisor, we must have $b=0$, otherwise $ab=0, b\neq 0$ means that $a$ is a nonzero zero divisor in $P$.
Therefore, in any case we have either $a=0$ or $b=0$, and thus the ring $R$ contains no nonzero zero divisors. Hence $R$ is an integral domain.
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 […]
Equivalent Conditions For a Prime Ideal in a Commutative Ring
Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:
(a) The ideal $P$ is a prime ideal.
(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.
Prime Ideal is Irreducible in a Commutative Ring
Let $R$ be a commutative ring. An ideal $I$ of $R$ is said to be irreducible if it cannot be written as an intersection of two ideals of $R$ which are strictly larger than $I$.
Prove that if $\frakp$ is a prime ideal of the commutative ring $R$, then $\frakp$ is […]