# 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 $$I \triangleleft R$$.

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 $$9\mathbb{Z}$$.

Ben Lynn blynn@cs.stanford.edu 💡