13.Subring, Ideal and Quotient ring

Overview

Correspondence with group theory：

Subgroup - Subring

normal subgroup - ideal

quotient group - quotient ring

Subring

Definition

Let $<A,+,\cdot >$ be a ring, and $S$ be a non-empty subset of $A$. If $S$ forms a ring with respect to $+$ and $\cdot$, then $S$ is called a subring of $A$, and $A$ is an extension ring of $S$.

$\{0\}$ and $A$ itself are subrings of $A$, and these two subrings are called trivial subrings.

Criterion

(Subring Criterion) Suppose $S$ is a non-empty subset of a ring. Then the necessary and sufficient condition for $S$ to be a subring of $A$ is that for any $a,b\in S$, both $a-b\in S$ and $ab\in S$.

Proof Necessity is obvious. Now, let’s prove sufficiency:

(1) $<S,+>$ is commutative: $\mathrm{\forall }a,b\in S$, we have $a-b\in S$. It satisfies the subgroup criterion, so $<S,+>$ is a group. Thus, $<S,+>$ is commutative.

(2) $<S,\cdot >$ is a semigroup: $\mathrm{\forall }a,b,c\in S$, we have $ab\in S$. Closure is satisfied, and associativity is natural. Therefore, $<S,\cdot >$ is a semigroup.

(3) The law of distribution naturally holds.

Therefore, $<S,+,\cdot >$ forms a ring, and $S$ is a subring of $A$​.

Ideal

definition

Let $A$ be a ring, and $I$ be its subring. For any $a\in I$ and any $x\in A$,

(1) If $xa\in I$, $I$ is called a left ideal of $A$;

(2) If $ax\in I$, $I$ is called a right ideal of $A$;

(3) If both (1) and (2) are satisfied, $I$ is called an ideal of $A$.

Condition "$\mathrm{\forall }x\in A,xa,ax\in I$" is called absorbent for $A$. Thus, an ideal is an absorbent subring for the ring.

Intuitively understand absorption: any element in the child ring and any element in the parent ring are in the child ring after the result of the operation, just as a sponge is full of water, it can squeeze the water out, but the sponge can also take all the water back through an operation (expansion).

Criterion

(Ideal Criterion) The necessary and sufficient conditions for a non-empty subset $H$ in a ring to be an ideal are:

(1) $\mathrm{\forall }a,b\in H$, we have $a-b\in H$;

(2) $\mathrm{\forall }a\in H$ and $\mathrm{\forall }x\in A$, we have $ax,xa\in H$.

Proof:

Necessity is obvious. Now, let’s prove sufficiency:

(1) $H$ is a subring: According to the subring criterion, we only need to prove that for $\mathrm{\forall }a,b\in H$, $a-b\in H$ and $ab\in H$. From condition 1, $a-b\in H$. Next, prove $ab\in H$. Since $H$ is a subset of $A$, $b\in A$. Substitute into condition 2, we find $ab,ba\in H$, thus $H$ is a subring.

(2) $H$ is absorbent for $A$: Condition 2 already states this.

Therefore, $H$​ is an ideal.

Quotient ring

Suppose $A$ is a ring, and $I$ is an ideal of $A$, then $I$ is the normal subgroup of the additive group $<A,+>$. The additive quotient group of $A$ by $I$ is

$A/I=\{a+I\mid a\in A\}$, denoted as $\bar{a}=a+I$.

Define the "modular addition" on $A/I$ as: $\bar{a}+\bar{b}=\overline{a+b}$.

Define the "modular multiplication" as: $\bar{a}\cdot \bar{b}=\overline{ab}$,

Then the addition and multiplication on $A/I$ can form a ring, which is the quotient ring.

Definition

Let $A$ be a ring, $I$ be an ideal of $A$. The ring formed by the addition and multiplication on the additive quotient group of $A$ by $I$, denoted as $A/I$ or the residue class ring of $A$ modulo $I$, is still written as $A/I$.

Example

$H=\{4k\mid k\in \mathbb{Z}\}$ is an ideal of $\mathbb{Z}$, find the quotient ring of $\mathbb{Z}$ with respect to $H$.

Solution:

$\mathbb{Z}/H=\{m+H\mid m\in \mathbb{Z}\}=\{\bar{0},\bar{1},\bar{2},\bar{3}\}={\mathbb{Z}}_4$.