integral domains整域せいいき, fields体たい, and finite fields有限体ゆうげんたい: basic exercises

1Corresponding lectures

2Suggested order

After “Integral domains, zero divisors, and polynomial rings,” complete Problems 1 and 4. After “Field basics,” continue with Problems 2–3 and the proof exercise. After “Introduction to finite fields,” complete Problem 5. Although the proof exercise appears after Problem 5 on this page, it does not depend on Problem 5. The page prerequisites describe what is needed to complete the whole page.

3Related exercises

4Problem 1: find a zero divisor零因子れいいんし

Find one zero divisor零因子れいいんし in $$\mathbb{Z}/8\mathbb{Z}$$\mathbb Z/8\mathbb Z.

4.1Answer

$$\left[2\right]$$[2] is a zero divisor零因子れいいんし. Indeed,

and $$\left[2\right]\ne \left[0\right]$$[2]\ne[0], $$\left[4\right]\ne \left[0\right]$$[4]\ne[0].

4.2Explanation

A zero divisor零因子れいいんし is an element that makes the product of two nonzero elements become 0.

5Problem 2: decide whether it is a field体たい

Is $$\mathbb{Z}/7\mathbb{Z}$$\mathbb Z/7\mathbb Z a field体たい?

5.1Answer

Yes. Since 7 is prime, for every nonzero $$\left[a\right]$$[a] we have $$\text{[PARSE ERROR: Undefined("Command("gcd")")]}(a,7)=1$$\gcd(a,7)=1. By the Bezout identityベズーの等式, there exist integers $$u,v$$u,v such that

Therefore, in residue classes, $$\left[a\right]\left[u\right]=\left[1\right]$$[a][u]=[1], so a multiplicative inverse乗法逆元じょうほうぎゃくげん exists. Moreover, $$\mathbb{Z}/7\mathbb{Z}$$\mathbb Z/7\mathbb Z is a commutative ring under residue-class addition and multiplication, and $$7\nmid 1$$7\nmid1 gives $$\left[0\right]\ne \left[1\right]$$[0]\ne[1]. Thus all conditions in the definition of a field are satisfied.

5.2Explanation

The ring $$\mathbb{Z}/n\mathbb{Z}$$\mathbb Z/n\mathbb Z is a field体たい exactly when $$n$$n is prime.

6Problem 3: explain a non-field体たい example

Explain why $$\mathbb{Z}/9\mathbb{Z}$$\mathbb Z/9\mathbb Z is not a field体たい.

6.1Answer

Although $$\left[3\right]\ne \left[0\right]$$[3]\ne[0],

Thus there is a zero divisor零因子れいいんし, so it is not a field体たい.

6.2Explanation

A field体たい has no zero divisors零因子れいいんし. In general, if a composite number $$n$$n factors as $$n=ab$$n=ab with $$1<a,b<n$$1<a,b<n, then in $$\mathbb{Z}/n\mathbb{Z}$$\mathbb Z/n\mathbb Z,

while $$\left[a\right]\ne \left[0\right]$$[a]\ne[0] and $$\left[b\right]\ne \left[0\right]$$[b]\ne[0]. Thus, in a residue ring modulo a composite number, zero divisors零因子れいいんし arise from factors of the composite number.

7Problem 4: products and degrees of polynomials

Let $$f\left(x\right)=2x^2+1$$f(x)=2x^2+1 and $$g\left(x\right)=3x-4$$g(x)=3x-4 be elements of $$\mathbb{Z}\left[x\right]$$\mathbb Z[x]. Compute $$f\left(x\right)g\left(x\right)$$f(x)g(x), find its leading coefficient and degree, and verify $$\text{[PARSE ERROR: Undefined("Command("deg")")]}\left(fg\right)=\text{[PARSE ERROR: Undefined("Command("deg")")]}f+\text{[PARSE ERROR: Undefined("Command("deg")")]}g$$\deg(fg)=\deg f+\deg g.

7.1Answer

The leading coefficient is 6 and the degree is 3. Since $$\text{[PARSE ERROR: Undefined("Command("deg")")]}f=2$$\deg f=2 and $$\text{[PARSE ERROR: Undefined("Command("deg")")]}g=1$$\deg g=1,

7.2Explanation

Because $$\mathbb{Z}$$\mathbb Z is an integral domain, the product 6 of the nonzero leading coefficients 2 and 3 is nonzero. This prevents the highest-degree term of the product from disappearing.

8Problem 5: the four-element finite field

Let $$F_4=F_2\left[x\right]/(x^2+x+1)$$\mathbb F_4=\mathbb F_2[x]/(x^2+x+1) and write $$\alpha =\left[x\right]$$\alpha=[x]. Using $${\alpha }^2=\alpha +1$$\alpha^2=\alpha+1 and $$1+1=0$$1+1=0, complete the multiplication table for the nonzero elements $$1,\alpha ,\alpha +1$$1,\alpha,\alpha+1. Then read the multiplicative inverse of each element from the table.

8.1Answer

Multiplication by the identity $$1$$1 leaves every element unchanged. The remaining products are

Also,

where we used $$2\alpha =0$$2\alpha=0 and $$1+1=0$$1+1=0 in characteristic 2. Together with commutativity, these calculations give

In each row, the column whose product is $$1$$1 gives the inverse. Therefore

8.2Explanation

In $$\mathbb{Z}/4\mathbb{Z}$$\mathbb Z/4\mathbb Z, $$[2{]}^2=0$$[2]^2=0, whereas every nonzero element $$1,\alpha ,\alpha +1$$1,\alpha,\alpha+1 of $$F_4$$\mathbb F_4 has an inverse. Thus the number of elements alone does not determine whether a ring is a field.

9Proof exercise: every finite integral domain整域せいいき is a field体たい

9.1Problem

Prove that a finite integral domain整域せいいき $$R$$R is a field体たい.

9.2Answer

Take $$a\in R$$a\in R with $$a\ne 0$$a\ne0. Define the map $${\mu }_a:R\to R$$\mu_a:R\to R by $${\mu }_a\left(x\right)=ax$$\mu_a(x)=ax. If $${\mu }_a\left(x\right)={\mu }_a\left(y\right)$$\mu_a(x)=\mu_a(y), then $$ax=ay$$ax=ay. Since $$R$$R is an integral domain整域せいいき, cancellation gives $$x=y$$x=y. Therefore $${\mu }_a$$\mu_a is injective単射たんしゃ.

Because $$R$$R is finite, every injective単射たんしゃ map from $$R$$R to $$R$$R is surjective全射ぜんしゃ. Hence for $$1\in R$$1\in R, there exists $$x\in R$$x\in R such that $$ax=1$$ax=1. Thus $$a$$a has an inverse. Since $$a$$a was an arbitrary nonzero element and the integral domain $$R$$R is a commutative ring with $$0\ne 1$$0\ne1, the ring $$R$$R is a field.

9.3Explanation

Finiteness is used to derive surjectivity from injectivity. In an infinite integral domain整域せいいき, this argument cannot be used as it stands.

For a finite set, injectivity means that the image has the same number of elements as the input. Since the input and target are the same finite set here, the image must be the whole target, so the map is surjective.