📂Abstract Algebra

Prime Ideal

Definition¹

Let an ideal $P \ne R$ of a commutative ring $R$ be such that for $a, b \in R$, $ab \in P$ implies $a \in P$ or $b \in P$. Then $P$ is called a prime Ideal in $R$.

Explanation

As the name prime suggests, the idea starts from splitting apart a multiplied element.

For example, consider the ring of integers $\mathbb{Z}$. Every element of $2 \mathbb{Z}$ has the form $2k$, and since $2 \in 2 \mathbb{Z}$, it is a prime ideal. By the same logic, for any prime number $p$, every $p \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$. However, considering $6 \mathbb{Z}$, we have $2 \cdot 3 \in 6 \mathbb{Z}$ but $2 \notin 6 \mathbb{Z}$ and $3 \notin 6\mathbb{Z}$, so $6 \mathbb{Z}$ cannot be a prime ideal in $\mathbb{Z}$.

Meanwhile, prime ideals have the following property related to integral domains. This is analogous to the relationship between maximal ideals and fields.

Theorem

Let a commutative ring $R$ have a unity $1_{R}$.

• $P$ is a prime ideal of $R$ $\iff$ $R / P$ is an integral domain

Proof

$( \implies )$

Definition of an integral domain

1. For a ring $R$, nonzero $a,b \in R$ satisfying $ab = 0$ are called zero divisors.
2. If $D$ with unity $1 \ne 0$ has no zero divisors, it is called an integral domain.

Suppose $xy + P$ is the identity $(0 + P)$ of $R/P$; then $xy \in P$. Since $P$ is a prime ideal, we must have $x \in P$ or $y \in P$.

• If $x \in P$, then $x + P = P = 0 + P$, so $x = 0$
• If $y \in P$, then $y + P = P = 0 + P$, so $y = 0$

Therefore $R/P$ cannot have zero divisors, and $R / P$ is an integral domain.

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$( \impliedby )$

Suppose $xy \in P$.

$$ (x + P) ( y + P ) = xy + P = P = 0 + P $$ Since $0 + P$ is the identity of the integral domain $R/P$, we must have $(x + P) = (0 + P )$ or $(y + P) = (0 + P )$.

• If $(x + P) = (0 + P ) = P$, then immediately $x \in P$.
• Even if $(y + P) = (0 + P ) = P$, then immediately $y \in P$.

Therefore $P$ is a prime ideal.

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Meanwhile, for maximal ideals and prime ideals we can obtain the corollary below. It is obvious from the fact that a field is, first of all, an integral domain. If $I$ is a maximal ideal of $R$, then $R / I$ is a field, and since a field is an integral domain, $I$ is a prime ideal.

Corollary

A maximal ideal of a commutative ring $R$ with a unity $1_{R}$ is a prime ideal.