# Commutative ring/Residue class ring/Group known/Introduction/Section

Due to fact, the kernel of a ring homomorphism is an ideal. We will construct, for any given ideal $I\subseteq R$ in a (commutative) ring, a commutative ring $R/I$ and a surjective ring homomorphism
    $R\longrightarrow R/I,$
dessen kernel is the ideal $I$. Therefore, ideals and kernels of ring homomorphisms are essentially equivalent objects, in the same way as, in group theory, normal subgroups and kernels of group homomorphisms are the same. In fact, the same homomorphism theorems hold again, and follow easily from the group situation.

## Definition

Let $R$ be a commutative ring, and let $I\subseteq R$ be an ideal in $R$. For $a\in R$, the subset
    $a+I={\left\{a+f\mid f\in I\right\}}\,$
$a$
$I$
$I$
These cosets are just the cosets of the (additive) subgroup $I\subseteq R$, which is a normal subgroup, as $(R,+,0)$ is a commutative group. Two elements $a,b\in R$ define the same coset, that is, $a+I=b+I$, if and only if their difference $a-b$ belongs to the ideal. We also say that $a$ and $b$ represent the same coset.

## Definition

Let $R$ be a commutative ring, and let $I\subseteq R$ be an ideal in $R$. Then the residue class ring $R/I$ (R modulo I) is a commutative ring that is determined by the following data.
* As a set, $R/I$ is the set of all cosets to $I$.
* An addition of cosets is defined by
    $(a+I)+(b+I):=(a+b+I)\,.$
* A multiplication of cosets is defined by
    $(a+I)\cdot (b+I):=(a\cdot b+I)\,.$
* ${\bar {0}}=0+I=I$ is the neutral element of the addition (the zero class).
* ${\bar {1}}=1+I$ is the neutral element of the multiplication (the unit class).
We have to show that these mappings (that is, addition and multiplication) are well-defined, that is, independent of the choice of representatives, and that the ring axioms are fulfilled. As $I$ is, in particular, a subgroup of the commutative group $(R,+,0)$, it is a normal subgroup, so that $R/I$ is a group and that the residue class mapping
    $R\longrightarrow R/I,a\longmapsto a+I=:{\bar {a}},$
is a group homomorphism. The only new feature compared with the group situation is that we now have also a multiplication. That the multiplication is well-defined can been seen as follows: Let two residue classes be given with different representatives, that is, ${\overline {a}}\,={\overline {a'}}\,$ and ${\overline {b}}\,={\overline {b'}}\,$. Then $a-a'\in I$ and $b-b'\in I$, and $a'=a+x$ and $b'=b+y$ with $x,y\in I$. This implies
    $a'b'=(a+x)(b+y)=ab+ay+xb+xy\,.$
The three summands on the right belong to the ideal, so that the difference $a'b'-ab\in I$. From the well-definedness, the other properties follows, in particular, that we have a ring homomorphism to the residue class ring. Again, this morphism is called the residue class mapping, or the residue class homomorphism. The image of $a\in R$ in $R/I$ is often denoted by $[a]$, ${\bar {a}}$, or just by $a$ again, and it is called the residue class of $a$. Under this mapping, exactly the elements from the ideal are sent to $0$, that is, the kernel of the residue class mapping is the given ideal.
The simplest example of this process is the mapping that sends an integer number $a$ to its remainder after division by a fixed number $d$. Every remainder is represented by one of the numbers $0,1,2,\ldots ,d-1$. In general, there does not always exist such a nice system of representatives.