Commutative ring/Ideal/Introduction/Section

Definition

A subset $𝔞$ of a commutative ring $R$ is called an ideal, if the following conditions are fulfilled:

$0 \in 𝔞$ .
For all $a, b \in 𝔞$ , we have $a + b \in 𝔞$ .
For all $a \in 𝔞$ and $r \in R$ , we have $r a \in 𝔞$ .

The property $0 \in 𝔞$ can be replaced by the condition that $𝔞$ is not empty. An ideal is a subgroup of the additive group of $R$ , which, moreover, is also closed under scalar multiplication.

For a family of element $a_{1}, a_{2}, \dots, a_{n} \in R$ in a commutative ring $R$ , we denote by $(a_{1}, a_{2}, \dots, a_{n})$ the ideal generated by these elements. It consists of all linear combinations

r_{1} a_{1} + r_{2} a_{2} + \dots + r_{n} a_{n},

where

r_{1}, r_{2}, \dots, r_{n} \in R

.

Definition

An ideal $𝔞$ in a commutative ring $R$ of the form

𝔞 = (a) = R a = {r a : r \in R}

is called a principal ideal.

The zero element forms, in every ring, the so-called zero ideal. We write this simply as $0 = (0) = {0}$ . write. The $1$ and, moreover, every unit, generates as an ideal the total ring.

Definition

The unit ideal in a commutative ring

R

is the ring itself.

In a field, there exist exactly two ideals.

Lemma

Let $R$ be a commutative ring. Then the following statements are equivalent.

$R$ is a field.
There exist exactly two ideals in $R$ .

Proof

If $R$ is a field, then there exists the zero ideal and the unit ideal, and these are different ideals. Let $I$ be an ideal in $R$ different from $0$ . Then $I$ contains some element $x \neq 0$ , which is a unit. Therefore, $1 = x x^{- 1} \in I$ and thus $I = R$ .

Suppose now that $R$ is a commutative ring with exactly two ideals. Then $R$ is not the zero ring. Let now $x$ be an element in $R$ different from $0$ . The principal ideal generated by $x$ , that is $R x$ , is $\neq 0$ , and therefore it must be the other ideal, which is the unit ideal. In particular, this means $1 \in R x$ . Hence, $1 = x r$ for some $r \in R$ , so that $x$ is a unit.

◻