rings環かん, idealsイデアル, and quotient rings商環しょうかん: basic exercises

1Corresponding lectures

2Suggested order

After “Ring basics,” complete Problems 1–2. After “Ideals and quotient rings,” continue with Problems 3–4 and the proof exercise. The page prerequisites describe what is needed to complete the whole page.

3Related exercises

4Problem 1: give an example of a ring

Is $$\mathbb{Z}$$\mathbb Z a ring環かん under ordinary addition and multiplication?

4.1Answer

Yes. It is an abelian group under addition, multiplication satisfies associativity, and the distributive law分配法則ぶんぱいほうそくs hold. The multiplicative identity is 1.

4.2Explanation

Division is generally not closed in $$\mathbb{Z}$$\mathbb Z, but a ring does not require every element to have a multiplicative inverse.

5Problem 2: basic ring identities

For arbitrary $$a,b\in R$$a,b\in R in a ring $$R$$R, prove $$a0=0a=0$$a0=0a=0 and $$(-a)b=-\left(ab\right)$$(-a)b=-(ab) from the ring axioms.

5.1Answer

By distributivity,

Adding $$-\left(a0\right)$$-(a0) to both sides in the additive group gives $$a0=0$$a0=0. Similarly, $$0b=(0+0)b=0b+0b$$0b=(0+0)b=0b+0b gives $$0b=0$$0b=0. Also,

Thus $$(-a)b$$(-a)b is the additive inverse of $$ab$$ab, so $$(-a)b=-\left(ab\right)$$(-a)b=-(ab).

5.2Explanation

The rules $$a0=0$$a0=0 and the sign laws are not special properties of integers. They necessarily follow from distributivity and the additive-group structure of every ring.

6Problem 3: check an idealイデアル

Is $$3\mathbb{Z}$$3\mathbb Z an idealイデアル of $$\mathbb{Z}$$\mathbb Z?

6.1Answer

Yes. First, $$0=3·0\in 3\mathbb{Z}$$0=3\cdot0\in3\mathbb Z, so $$3\mathbb{Z}$$3\mathbb Z is nonempty. For any $$3a,3b\in 3\mathbb{Z}$$3a,3b\in3\mathbb Z,

Also, for any $$r\in \mathbb{Z}$$r\in\mathbb Z,

Since $$\mathbb{Z}$$\mathbb Z is commutative, we also have $$\left(3a\right)r=r\left(3a\right)\in 3\mathbb{Z}$$(3a)r=r(3a)\in3\mathbb Z. Thus $$3\mathbb{Z}$$3\mathbb Z absorbs multiplication from both sides and is an ideal of $$\mathbb{Z}$$\mathbb Z.

6.2Explanation

For an idealイデアル, we check closure閉包性へいほうせい under subtraction and that multiplying by any element of the ring leaves the result inside.

7Problem 4: read a quotient ring商環しょうかん

In $$\mathbb{Z}/3\mathbb{Z}$$\mathbb Z/3\mathbb Z, what are $$\left[2\right]+\left[2\right]$$[2]+[2] and $$\left[2\right]\left[2\right]$$[2][2]?

7.1Answer

7.2Explanation

In a quotient ring商環しょうかん, the result of a calculation is represented again by the same residue class. A representative is not fixed uniquely.

8Proof exercise: quotient-ring operations are well-defined

8.1Problem

Let $$I$$I be an idealイデアル of a ring $$R$$R. Prove that if $$a+I=a^{\prime }+I$$a+I=a'+I and $$b+I=b^{\prime }+I$$b+I=b'+I, then $$(a+b)+I=(a^{\prime }+b^{\prime })+I$$(a+b)+I=(a'+b')+I and $$ab+I=a^{\prime }{b}^{\prime }+I$$ab+I=a'b'+I.

8.2Answer

We have $$a^{\prime }-a\in I$$a'-a\in I and $$b^{\prime }-b\in I$$b'-b\in I. For addition,

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

For multiplication,

Since $$I$$I is an idealイデアル, $$a^{\prime }(b^{\prime }-b)\in I$$a'(b'-b)\in I and $$(a^{\prime }-a)b\in I$$(a'-a)b\in I. Therefore $$a^{\prime }{b}^{\prime }-ab\in I$$a'b'-ab\in I, so $$ab+I=a^{\prime }{b}^{\prime }+I$$ab+I=a'b'+I.

8.3Explanation

The absorption property of an idealイデアル guarantees that changing representatives does not change the class of the product.