Chapter 13, Problem 1E

To determine

The verification that the given statements are true.

Answer to Problem 1E

The validity of the given statements has been verified.

Explanation of Solution

Given:

The statements: -

1. The ring of integers is an integral domain.
2. The ring of Gaussian integers $\mathbb{Z}\left[i\right]=\left\{a+bi|a,b\in \mathbb{Z}\right\}$ is an integral domain.
3. The ring $\mathbb{Z}\left[x\right]$ of polynomials with integer coefficients is an integral domain.
4. The ring $\mathbb{Z}\left[\sqrt{2}\right]=\left\{a+b\sqrt{2}|a,b\in \mathbb{Z}\right\}$ is an integral domain.
5. The ring ${\mathbb{Z}}_p$ of integers modulo a prime $p$ is an integral domain.
6. The ring ${\mathbb{Z}}_n$ of integers modulo $n$ is not an integral domain when $n$ is not prime.
7. The ring $M_2\left(\mathbb{Z}\right)$ of $2\times 2$ matrices over the integers is not an integral domain.
8. $\mathbb{Z}\oplus \mathbb{Z}$ is not an integral domain.

Concept used:

An integral domain is a commutative ring with unity and no zero-divisors. So, if $ab=0$ where $a$ and $b$ are elements of an integral domain, then either $a=0$ or $b=0$ .

Calculation:

1. The ring of integers is an integral domain.

The multiplication defined in the ring of integers is nothing but integer product. We know that, integer product is commutative. Also, $1\in \mathbb{Z}$ is the multiplicative identity, or unity since $1\cdot a=a\cdot 1=a$ for all $a\in \mathbb{Z}$ . So, the ring of integers is a commutative ring with unity. Now, if $a$ and $b$ are integers, then $ab=0$ implies either $a=0$ or $b=0$ . Equivalently, the product of two non-zero integers is never zero. This implies that the ring of integers has no zero-divisor. Thus, by definition, the ring of integers is an integral domain. This proves the given statement.

2. The ring of Gaussian integers $\mathbb{Z}\left[i\right]=\left\{a+bi|a,b\in \mathbb{Z}\right\}$ is an integral domain.

The multiplication defined in this ring is component-wise integer product.

So, if $a+bi,c+di\in \mathbb{Z}\left[i\right]$ , then $\left(a+bi\right)\cdot \left(c+di\right)=ac-bd+\left(ad+bc\right)i$ . Then, it follows that $\left(c+di\right)\cdot \left(a+bi\right)=ca-db+\left(da+cb\right)i$ . Since, the terms of the results are all integer products, $ac=ca,bd=db,ad=da,bc=cb$ . Thus, $\left(a+bi\right)\cdot \left(c+di\right)=\left(c+di\right)\cdot \left(a+bi\right)$ . Also, $1=1+0i\in \mathbb{Z}\left[i\right]$ is the multiplicative identity or unity as $\left(a+bi\right)\cdot \left(1+0i\right)=\left(1+0i\right)\cdot \left(a+bi\right)=a+bi$ . So, the given ring is a commutative ring with unity.

Now, let $a+bi,c+di\in \mathbb{Z}\left[i\right]$ . Then, $\left(a+bi\right)\cdot \left(c+di\right)=ac-bd+\left(ad+bc\right)i=0$ implies $ac-bd=0,ad+bc=0$ .

Let $c\ne 0$ . Then,

$\begin{array}{l}ac-bd=0\\ ac=bd\\ a=\frac{bd}{c}\end{array}$

Put $a=\frac{bd}{c}$ in $ad+bc=0$ to get,

$\begin{array}{l}\left(\frac{bd}{c}\right)d+bc=0\\ \frac{b}{c}\left(c^2+d^2\right)=0\end{array}$

This implies that either $b=0$ or $c=d=0$ . The latter case is not possible due to our assumption. So, $b=0$ . Put $b=0$ in $a=\frac{bd}{c}$ to get $a=0$ . Then, $a=0$ and $b=0$ implies that $a+bi=0$ . Similarly, if we start by assuming $a\ne 0$ , then we can show that $c+di=0$ . So, $\left(a+bi\right)\cdot \left(c+di\right)=0$ implies that either $a+bi=0$ or $c+di=0$ . Thus, the given ring has no zero-divisor. Then, by definition, it follows that the given ring is an integral domain. This proves the given statement.

3. The ring $\mathbb{Z}\left[x\right]$ of polynomials with integer coefficients is an integral domain.

The multiplication defined in a polynomial ring is nothing but the product of two polynomials, which is commutative. Also, the constant polynomial $1\in \mathbb{Z}\left[x\right]$ is the multiplicative identity, or unity as for any polynomial $p\left(x\right)\in \mathbb{Z}\left[x\right]$ , we have, $1\cdot p\left(x\right)=p\left(x\right)\cdot 1=p\left(x\right)$ . So, the ring of polynomials with integer coefficients is a commutative ring with unity. Now, the product of two polynomials is nothing but a term-wise product, where the coefficients are multiplied and the power of the variable is increased accordingly. Since the coefficients of the result are obtained by product of coefficients of the two polynomials, the result can be the constant polynomial $0$ only if either of the polynomials is $0$ . This implies that the ring of polynomials with integer coefficients has no zero-divisor. Thus, by definition, it follows that the given ring is an integral domain. This proves the given statement.

4. The ring $\mathbb{Z}\left[\sqrt{2}\right]=\left\{a+b\sqrt{2}|a,b\in \mathbb{Z}\right\}$ is an integral domain.

The multiplication defined in this ring is component-wise integer product.

If $a+b\sqrt{2},c+d\sqrt{2}\in \mathbb{Z}\left[\sqrt{2}\right]$ , then $\left(a+b\sqrt{2}\right)\cdot \left(c+d\sqrt{2}\right)=ac+2bd+\left(ad+bc\right)\sqrt{2}$ . Then, it follows that $\left(c+d\sqrt{2}\right)\cdot \left(a+b\sqrt{2}\right)=ca+2db+\left(da+cb\right)\sqrt{2}$ . Since, the terms of the results are all integer products, $ac=ca,bd=db,ad=da,bc=cb$ . Thus, $\left(a+b\sqrt{2}\right)\cdot \left(c+d\sqrt{2}\right)=\left(c+d\sqrt{2}\right)\cdot \left(a+b\sqrt{2}\right)$ . Also, $1=1+0\sqrt{2}\in \mathbb{Z}\left[\sqrt{2}\right]$ is the multiplicative identity or unity as $\left(a+b\sqrt{2}\right)\cdot \left(1+0\sqrt{2}\right)=\left(1+0\sqrt{2}\right)\cdot \left(a+b\sqrt{2}\right)=a+b\sqrt{2}$ . So, the given ring is a commutative ring with unity.

Now, let $a+b\sqrt{2},c+d\sqrt{2}\in \mathbb{Z}\left[\sqrt{2}\right]$ . Then, $\left(a+b\sqrt{2}\right)\cdot \left(c+d\sqrt{2}\right)=ac+2bd+\left(ad+bc\right)\sqrt{2}=0$ implies $ac+2bd=0,ad+bc=0$ .

Let $c\ne 0$ . Then,

$\begin{array}{l}ac+2bd=0\\ ac=-2bd\\ a=-2\frac{bd}{c}\end{array}$

Put $a=-2\frac{bd}{c}$ in $ad+bc=0$ to get,

$\begin{array}{l}\left(-2\frac{bd}{c}\right)d+bc=0\\ \frac{b}{c}\left(c^2-2d^2\right)=0\end{array}$

This implies that either $b=0$ or $c^2-2d^2=0$ . Since $c,d\in \mathbb{Z}$ , $c^2-2d^2=0$ implies that $c=d=0$ . The latter case is not possible due to our assumption. So, $b=0$ . Put $b=0$ in $a=-2\frac{bd}{c}$ to get $a=0$ . Then, $a=0$ and $b=0$ implies that $a+b\sqrt{2}=0$ . Similarly, if we start by assuming $a\ne 0$ , then we can show that $c+d\sqrt{2}=0$ . So, $\left(a+b\sqrt{2}\right)\cdot \left(c+d\sqrt{2}\right)=0$ implies that either $a+b\sqrt{2}=0$ or $c+d\sqrt{2}=0$ . Thus, the given ring has no zero-divisor. Then, by definition, it follows that the given ring is an integral domain. This proves the given statement.

5. The ring ${\mathbb{Z}}_p$ of integers modulo a prime $p$ is an integral domain.

The multiplication defined in this ring is nothing but integer product modulo a prime $p$ . Since integer product is commutative, the multiplication defined in this ring is also commutative. Also, $\overline{1}\in {\mathbb{Z}}_p$ is the multiplicative identity, or unity as for any $\overline{a}\in {\mathbb{Z}}_p$ , $\overline{1}\cdot \overline{a}=\overline{a}\cdot \overline{1}=\overline{a}$ . So, ${\mathbb{Z}}_p$ is a commutative ring with unity.

Let $\overline{a},\overline{b}\in {\mathbb{Z}}_p$ . Then, $\overline{a}\cdot \overline{b}=\overline{0}$ implies either $ab=0$ or $ab=p$ . However, $ab=p$ implies that $a$ and $b$ are factors of $p$ , which is not possible as it is given that $p$ is prime. Then, the only possibility is $ab=0$ which gives either $a=0$ or $b=0$ . That is, either $\overline{a}=\overline{0}$ or $\overline{b}=\overline{0}$ . So, $\overline{a}\cdot \overline{b}=\overline{0}$ implies either $\overline{a}=\overline{0}$ or $\overline{b}=\overline{0}$ . Thus, ${\mathbb{Z}}_p$ has no zero-divisor. Then, by definition, ${\mathbb{Z}}_p$ is an integral domain.

6. The ring ${\mathbb{Z}}_n$ of integers modulo $n$ is not an integral domain when $n$ is not prime.

The multiplication defined in this ring is nothing but integer product modulo a prime $n$ . Since integer product is commutative, the multiplication defined in this ring is also commutative. Also, $\overline{1}\in {\mathbb{Z}}_n$ is the multiplicative identity, or unity as for any $\overline{a}\in {\mathbb{Z}}_n$ , $\overline{1}\cdot \overline{a}=\overline{a}\cdot \overline{1}=\overline{a}$ . So, ${\mathbb{Z}}_n$ is a commutative ring with unity.

Now, since $n$ is not a prime, it has non-trivial factors. Let $a\left(\ne 0\right)$ be a factor of $n$ such that $ab=n$ . Then, for $\overline{a},\overline{b}\in {\mathbb{Z}}_n$ , $\overline{a}\cdot \overline{b}=\overline{ab}=\overline{n}=\overline{0}$ . This implies that ${\mathbb{Z}}_n$ has zero-divisors. Thus, it follows that the ring ${\mathbb{Z}}_n$ is not an integral domain when $n$ is not prime. This proves the given statement.

7. The ring $M_2\left(\mathbb{Z}\right)$ of $2\times 2$ matrices over the integers is not an integral domain.

The multiplication defined in this ring is nothing but matrix multiplication, which is not commutative. So, the given ring is not commutative at all. Thus, by definition, the given ring is not an integral domain. This proves the given statement.

8. $\mathbb{Z}\oplus \mathbb{Z}$ is not an integral domain.

Let $\left(a,b\right),\left(c,d\right)\in \mathbb{Z}\oplus \mathbb{Z}$ . Then, the multiplication is defined as $\left(a,b\right)\cdot \left(c,d\right)=\left(ac,bd\right)$ . Clearly, $a,b,c,d\in \mathbb{Z}$ . Then, $ac=ca,bd=db$ as integer product is commutative. Then, $\left(c,d\right)\cdot \left(a,b\right)=\left(ca,db\right)=\left(ac,bd\right)$ . So, $\left(a,b\right)\cdot \left(c,d\right)=\left(c,d\right)\cdot \left(a,b\right)$ . Also, $\left(1,1\right)\in \mathbb{Z}\oplus \mathbb{Z}$ is the multiplicative identity, or unity as $\left(a,b\right)\cdot \left(1,1\right)=\left(1,1\right)\cdot \left(a,b\right)=\left(a,b\right)$ . So, $\mathbb{Z}\oplus \mathbb{Z}$ is a commutative ring with unity.

Now, let $\left(1,0\right),\left(0,1\right)\in \mathbb{Z}\oplus \mathbb{Z}$ . Then, $\left(1,0\right)\cdot \left(0,1\right)=\left(0,0\right)$ . Clearly, the product of two non-zero elements is the zero element of the ring. Thus, the ring $\mathbb{Z}\oplus \mathbb{Z}$ has zero-divisors. Then, by definition, the given ring is not an integral domain. This proves the given statement.

Conclusion:

The validity of the given statements has been verified.