📂Abstract Algebra

Covariant Ideals

Definition ¹

A prime ideal $P \ne R$ of a commutative ring $R$ is an ideal that, given $a, b \in R$ and $ab \in P$, if $a \in P$ or $b \in P$ then $P$ is said to be a prime ideal in $R$.

Explanation

As the name prime suggests, it originates from the idea of breaking down the product of elements.

For example, considering the ring of integers $\mathbb{Z}$, all elements of $2 \mathbb{Z}$ are represented in the form of $2k$, and since $2 \in 2 \mathbb{Z}$, it becomes a prime ideal. By the same logic, for any prime number $p$, $p \mathbb{Z}$ are all prime ideals in $\mathbb{Z}$. However, considering $6 \mathbb{Z}$, although $2 \cdot 3 \in 6 \mathbb{Z}$ is true, since $2 \notin 6 \mathbb{Z}$ and $3 \notin 6\mathbb{Z}$, $6 \mathbb{Z}$ cannot be a prime ideal in $\mathbb{Z}$.

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

Theorem

Let the commutative ring $R$ have a unit element $1_{R}$.

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

Proof

$( \implies )$

Definition of Integral Domain

1. For a ring $R$, a non-zero $a,b \in R$ satisfying $ab = 0$ is called a Zero Divisor.
2. An integral domain is a ring with a unit element $D$ that does not have any zero divisors.

If $xy + P$ is the identity element $(0 + P)$ of $R/P$, then $xy \in P$ is true. Since $P$ is a prime ideal, it must be that either $x \in P$ or $y \in P$.

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

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

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$( \impliedby )$

Let’s consider $xy \in P$ again.

$$(x + P) ( y + P ) = xy + P = P = 0 + P$$ Since $0 + P$ is the identity element of the integral domain $R/P$, it must be either $(x + P) = (0 + P )$ or $(y + P) = (0 + P )$.

• If we assume $(x + P) = (0 + P ) = P$, then it immediately means $x \in P$.
• If we assume $(y + P) = (0 + P ) = P$, then it immediately means $y \in P$.

Therefore, $P$ becomes a prime ideal.

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Meanwhile, we can derive the following corollary for maximal ideals and prime ideals. It is obvious that a field is first an integral domain. If $I$ is a maximal ideal of $R$, $R / I$ is a field, and since a field is an integral domain, $I$ is a prime ideal.

Corollary

The maximal ideal of a commutative ring $R$ that has a unit element $1_{R}$ is a prime ideal.