]> Commutative Algebra - Ideals

## Ideals

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 $I◃R$.

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

$R/I=\left\{I+a\mid a\in R\right\}$

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

$\left(I+a\right)+\left(I+b\right)=I+\left(a+b\right)$

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

$\left(I+a\right)\cdot \left(I+b\right)=I+\left(a\cdot b\right)$

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

We call $R/I$ a quotient ring.

The mapping $\varphi :R\to R/I$ that takes $x$ to $I+x$ is a surjective ring homomorphism that is called the natural map. We have $\mathrm{ker}\varphi =I$, 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 $f:R\to S$ is a ring homomorphism with kernel $I$ and image $C$ then $R/I\cong C$.

Proposition: Let $I◃R$ and $\varphi :R\to R/I$ be the natural map.

Then the ideals $ℐ$ of $R/I$ have the form $ℐ=J/I=\left\{I+j\mid j\in J\right\}$ for some $I\subset J◃R$.

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