An ideal of a ring $R$ is a nonempty subset $I\subset R$ such that for all $x, y \in I$

  1. $x+y,-x \in I$

  2. $x \cdot y \in I$

The first condition is equivalent to requiring

\[ x-y \in I \]

We write $I \triangleleft R$.

Since $I$ is an additive subgroup of $R$, we can form the quotient group

\[ R/I = \{I+a|a\in R\} \]

which is the group of cosets of $I$ with addition: for $a,b \in R$ we have

\[ (I+a) + (I+b) = I +(a+b) \]

It is easily verified that $R/I$ is in fact a ring by defining multiplication as follows:

\[ (I+a) \cdot (I+b) = I +(a\cdot b) \]

(It needs to be checked that multiplication is well-defined: if $I + a = I + a', I+ b = b'$, then it can be seen that $a b - a' b' = a(b -b') + (a-a')b' \in I$.)

We call $R/I$ a quotient ring.

The mapping $\phi : R \rightarrow R/I$ that takes $x$ to $I+x$ is a surjective ring homomorphism that is called the natural map. We have $ker \phi = I$, so every ideal is the kernel of some ring homomorphism. The converse is easily verified, that is, the kernels of ring homomorphisms with domain $R$ are precisely the ideals of $R$.

The following are easy to verify:

Fundamental Homomorphism Theorem: If $f:R\rightarrow S$ is a ring homomorphism with kernel $I$ and image $C$ then $R/I \cong C$.

Proposition: Let $I\triangleleft R$ and $\phi:R\rightarrow R/I$ be the natural map.

Then the ideals ${\mathcal { I}}$ of $R/I$ have the form $\mathcal{I} = J/I = \{I +j | j\in J\}$ for some $I\subset J\triangleleft R$.

Example: $\mathbb{Z}/9\mathbb{Z} \cong \mathbb{Z}_9$ has ideals $\mathbb{Z}/9\mathbb{Z}, 3\mathbb{Z}/9\mathbb{Z}, 9\mathbb{Z}/9\mathbb{Z}$ that correspond under $\phi^{-1}$ to $\mathbb{Z} \supset 3\mathbb{Z} \supset 9\mathbb{Z}$, which are all the ideals of $\mathbb{Z}$ containing $9\mathbb{Z}$.