Prime and Maximal Ideals

In the rings $\mathbb{Z}$ and $F[x]$, we have seen when $(n)$ and $(p(x))$ has some “primeness” property, then we would have that $\mathbb{Z}/(n)$ and $F[x]/(p(x))$ are fields. We will define prime ideals this section, and we will see in general quotient by prime ideals gives us only integral domains. To get fields, we will need to look at maximal ideals.

Prime Ideals

Definition

A prime ideal $P$ of a commutative unitary ring $R$ is an ideal such that: for any $a,b\in R$, if $ab\in P$, then $a\in P$ or $b\in P$.

This definition is a generalization of the key property of prime numbers. We will shown it is equivalent to the following:

Theorem

Let $I$ be an ideal of a commutative unitary ring $R$. The following are equivalent:

1. $I$ is a prime ideal.
2. $R/I$ is an integral domain.

Proof

1. $I$ is a prime ideal $⟹R/I$ is an integral domain:
   • Let $a,b\in R$ such that $(a+I)(b+I)=0+I$. Then we have $ab\in I$, since $I$ is prime we have $a\in I$ or $b\in I$. Thus $a+I=0+I$ or $b+I=0+I$. Therefore $R/I$ is an integral domain.
2. $R/I$ is an integral domain $⟹I$ is a prime ideal:
   • Let $a,b\in R$ such that $ab\in I$. Then we have $(a+I)(b+I)=0+I$. Since $R/I$ is an integral domain, we have $a+I=0+I$ or $b+I=0+I$. Thus $a\in I$ or $b\in I$. Therefore $I$ is a prime ideal. $■$

Warning

Quotient by prime ideals always gives us integral domains, but not always fields. For example $(2)\subset \mathbb{Z}[x]$ is a prime ideal, but $\mathbb{Z}[x]/(2)$ is not a field. It is a special property of prime ideals in $\mathbb{Z}$ and $F[x]$ that they automatically are maximal ideals.

Maximal Ideals

Definition

A maximal ideal $M$ of a commutative unitary ring $R$ is an ideal such that:

1. $M$ is a proper ideal of $R$, i.e. $M\ne R$.
2. There are no proper ideals $I$ such that $M\subset I\subset R$.

It turns out that maximal ideals are the ones that give us fields when we take the quotient. We will show this in the following theorem:

Theorem

Let $M$ be a ideal of a commutative unitary ring $R$. The following are equivalent:

1. $M$ is a maximal ideal.
2. $R/M$ is a field.

Proof

1. $M$ is a maximal ideal $⟹R/M$ is a field:
   • Let $a+M\in R/M$ be a nonzero element. Then $a\notin M$. Since $M$ is maximal, we have $M+(a)=R$. Thus there exists $b\in R$ and $m\in M$ such that $m+ab\in M$. Therefore $(a+M)(b+M)=ab+M=0+M$. Thus $b+M$ is the multiplicative inverse of $a+M$. Therefore $R/M$ is a field.
2. $R/M$ is a field $⟹M$ is a maximal ideal:
   • Let $I$ be a proper ideal such that $M\subset I\subset R$. Then $I/M$ is a proper ideal of $R/M$. Since $R/M$ is a field, we have $I/M=0+M$. Thus $I=M$. Therefore $M$ is a maximal ideal. $■$