idealsイデアル and quotient rings商環しょうかん

To construct a quotient group商群しょうぐん from a group, a normal subgroup正規部分群せいきぶぶんぐん was needed. To construct a quotient ring商環しょうかん from a ring環かん, we need a subset that plays the corresponding role. That subset is an idealイデアル.

1Definition of an idealイデアル

A subset $$I$$I of a ring $$R$$R is an idealイデアル if $$0\in I$$0\in I and it satisfies the following conditions.

The condition $$0\in I$$0\in I makes $$I$$I nonempty. The first condition says that $$I$$I is a subgroup部分群ぶぶんぐん with respect to addition. The second says that multiplying by any element of the ring still leaves the result inside $$I$$I.

In a commutative ring, $$ra$$ra and $$ar$$ar are the same, so it is enough to check one side.

2Why idealsイデアル are necessary

In a quotient ring商環しょうかん, elements of $$I$$I are treated as 0. In other words, $$a$$a and $$b$$b are treated as the same element when their difference belongs to $$I$$I.

Because $$I$$I is an additive subgroup, this is an equivalence relation. Reflexivity follows from $$a-a=0\in I$$a-a=0\in I. If $$a-b\in I$$a-b\in I, then $$b-a=-(a-b)\in I$$b-a=-(a-b)\in I, which gives symmetry. If $$a-b\in I$$a-b\in I and $$b-c\in I$$b-c\in I, then

which gives transitivity.

Using the equivalence classes built from this equivalence relation同値関係どうちかんけい, we want to define

Assuming that $$I$$I is an additive subgroup, absorption from both sides—the remaining ideal condition—is necessary for this multiplication to be independent of representatives.

Here the equivalence class of $$a$$a can be written as $$a+I=\{a+i\mid i\in I\}$$a+I=\{a+i\mid i\in I\}. In a quotient ring, these classes are treated as the elements.

3Example from the integers

$$n\mathbb{Z}=\{nk\mid k\in \mathbb{Z}\}$$n\mathbb Z=\{nk\mid k\in\mathbb Z\} is an idealイデアル of $$\mathbb{Z}$$\mathbb Z.

The quotient ring商環しょうかん

groups integers while ignoring differences that are multiples of $$n$$n. This is exactly the world of congruences合同式ごうどうしき.

4Correspondence with quotient groups商群しょうぐん

In both settings, the part collapsed as a kernel核かく is identified with the relevant identity element to form a quotient. In multiplicative notation, a group kernel collapses to the group identity $$e$$e, whereas a ring kernel collapses to the additive identity 0.

Ring homomorphisms and the first isomorphism theorem are treated later. This table is a roadmap showing that ideals will later appear as kernels.

5What changes and what is preserved

In a quotient ring商環しょうかん, differences inside the idealイデアル are collapsed to 0. What changes is the granularity of elements. On the other hand, addition, multiplication, and distributive laws are preserved on the quotient in a well-defined way.

6Proof supplement: why quotient-ring operations do not depend on representatives

Let $$I$$I be an idealイデアル of a ring $$R$$R. Define the sum and product of residue classes by

We prove that this definition定義ていぎ does not depend on representatives.

Suppose $$a+I=a^{\prime }+I$$a+I=a'+I and $$b+I=b^{\prime }+I$$b+I=b'+I. This means $$a^{\prime }-a\in I$$a'-a\in I and $$b^{\prime }-b\in I$$b'-b\in I. For addition,

so $$(a^{\prime }+b^{\prime })+I=(a+b)+I$$(a'+b')+I=(a+b)+I.

For multiplication,

Since $$b^{\prime }-b\in I$$b'-b\in I and $$I$$I remains inside itself when multiplied by elements of the ring, $$a^{\prime }(b^{\prime }-b)\in I$$a'(b'-b)\in I. Similarly, $$(a^{\prime }-a)b\in I$$(a'-a)b\in I. Therefore $$a^{\prime }{b}^{\prime }-ab\in I$$a'b'-ab\in I, so $$a^{\prime }{b}^{\prime }+I=ab+I$$a'b'+I=ab+I.

This proof証明しょうめい shows that the idealイデアル condition is exactly the condition needed so that multiplying classes of remainders does not break the quotient.

Conversely, suppose that $$I$$I is an additive subgroup and that $$(a+I)(b+I)=ab+I$$(a+I)(b+I)=ab+I is independent of representatives. If $$i\in I$$i\in I, then $$i+I=0+I$$i+I=0+I. Hence, for every $$r\in R$$r\in R,

Thus $$ri+I=0+I$$ri+I=0+I and $$ir+I=0+I$$ir+I=0+I, so $$ri,ir\in I$$ri,ir\in I. Therefore well-definedness of multiplication forces absorption from both sides. Together with the additive-subgroup condition, the ideal condition is not only sufficient but also necessary.

Now that the operations are well-defined, the ring axioms can also be checked. The additive identity is $$0+I$$0+I, the additive inverse of $$a+I$$a+I is $$(-a)+I$$(-a)+I, and the multiplicative identity is $$1+I$$1+I. For example, distributivity follows from

Associativity and commutativity of addition, associativity of multiplication, and the other distributive law follow by applying the corresponding axiom in $$R$$R to representatives. Therefore the set of residue classes $$R/I$$R/I with these operations is a ring. When $$I=R$$I=R, we have $$0+I=1+I$$0+I=1+I and obtain the zero ring, which this material includes among rings.

7Exercise link

8Summary

An idealイデアル is the kind of subset needed to construct a quotient ring商環しょうかん. In a quotient ring商環しょうかん, elements of the idealイデアル are treated as 0, and addition and multiplication are placed on the remaining residue classes. The ring $$\mathbb{Z}/n\mathbb{Z}$$\mathbb Z/n\mathbb Z is the most basic example of a quotient ring商環しょうかん.