91: Quotient Ring of Ring by Ideal

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of quotient ring of ring by ideal

---

Topics

About: ring

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note

---

Starting Context

• The reader knows a definition of ring.
• The reader knows a definition of ideal of ring.
• The reader knows a definition of quotient group of group by normal subgroup.

---

Target Context

• The reader will have a definition of quotient ring of ring by ideal.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the rings }\}$
$I$: $\in \{\text{ the both-sided ideals of }R\}$
$\ast R/I$: $=\text{ the additive quotient group }$ as the ring with the multiplication specified below
//

Conditions:
$\mathrm{\forall }[p_1],[p_2]\in R/I([p_1][p_2]=[p_1{p}_2])$
//

As $I$ is an additive normal subgroup, the additive quotient group makes sense. The addition of $R/I$ is of course the addition of the additive quotient group.

The multiplication is well-defined: for each $[p_1]=[p_1^{\prime }]$ and $[p_2]=[p_2^{\prime }]$, there are some $i_1,i_2\in I$ such that $p_1^{\prime }=p_1+i_1$ and $p_2^{\prime }=p_2+i_2$, and $[p_1^{\prime }{p}_2^{\prime }]=[(p_1+i_1)(p_2+i_2)]=[p_1{p}_2+p_1{i}_2+i_1{p}_2+i_1{i}_2]=[p_1{p}_2]$, because $p_1{i}_2+i_1{p}_2+i_1{i}_2\in I$.

---

2: Natural Language Description

For any ring, $R$, and its any both-sided ideal, $I$, the additive quotient group, $R/I$, as the ring with the multiplication, $[p_1][p_2]=[p_1{p}_2]$

---

3: Note

$R/I$ is indeed a ring: 1) it is an Abelian group under addition: it is the additive quotient group of the Abelian group; 2) it is a monoid under multiplication: the associativity holds, because $([p_1][p_2])[p_3]=[p_1{p}_2][p_3]=[(p_1{p}_2)p_3]=[p_1(p_2{p}_3)]=[p_1][p_2{p}_3]=[p_1]([p_2][p_3])$; $[1]$ is the identity, because $[1][p]=[1p]=[p]=[p1]=[p][1]$; 3) multiplication is distributive with respect to addition: $[p_1]([p_2]+[p_3])=[p_1][p_2+p_3]=[p_1(p_2+p_3)]=[p_1{p}_2+p_1{p}_3]=[p_1{p}_2]+[p_1{p}_3]=[p_1][p_2]+[p_1][p_3]$ and $([p_1]+[p_2])[p_3]=[p_1+p_2][p_3]=[(p_1+p_2)p_3]=[p_1{p}_3+p_2{p}_3]=[p_1{p}_3]+[p_2{p}_3]=[p_1][p_3]+[p_2][p_3]$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>