# Topics in Algebra, Chapter 3.2

###### 2012-11-11

There are no exercises from 3.1, so this section covers both 3.1 (“Definition and Examples of Rings”) and 3.2 (“Some Special Classes of Rings”). Throughout, $R$ is a ring and $p$ is a prime integer.

### Topics covered: 3.1

• Definition: An associative ring is a non-empty set $R$ with two operations, $+$ and $\cdot$ obeying the usual axioms: $R$ is an abelian group under $+$, $R$ is closed under the associative multiplication $\cdot$, and the distributive law holds in $R$.

• If $R$ contains a multiplicative identity element then it is called unital.

• $R=\mathbb{Z}$ is a ring under the usual operations. $R$ is commutative and unital.

• $R=2\mathbb{Z}$, the even integers, form a commutative ring under the usual operations but $R$ is not unital.

• $R=\mathbb{Q}$ is a commutative, unital ring, but also a field.

• $R=\mathbb{Z}\mathrm{/}7\mathbb{Z}$ is a field under addition and multiplication modulo $7$.

• $R=\mathbb{Z}\mathrm{/}6\mathbb{Z}$ is a ring with zero divisors, e.g. $2\cdot 3=0$.

• $R={\mathrm{M}\mathrm{a}\mathrm{t}}_{2×2}\left(\mathbb{Q}\right)$, the set of $2×2$ matrices with elements taken from $\mathbb{Q}$, under the usual matrix addition and multiplication, is a non-commutative ring. It is unital and has zero-divisors.

• $R=\mathbb{C}$ is a field.

• The real quaternions (${\alpha }_{0}+{\alpha }_{1}i+{\alpha }_{2}j+{\alpha }_{3}k$ with the usual relations on $i,j,k$ and ${\alpha }_{n}\in \mathbb{R}$) form a non-commutative ring, but also a division ring, where every element is invertible but its left and right inverses may be distinct.

### Topics covered: 3.2

• Definition: With $R$ commutative, a non-zero element $a\in R$ is a zero divisor if there is non-zero element $b\in R$ such that $ab=0$.

• Definition: An integral domain is a commutative ring with no zero divisors.

• $\mathbb{Z}$ is an integral domain.

• Definition: A field is a commutative, unital ring where every element has a multiplicative inverse.

• Lemma 3.2.1: For all $a,b\in R$, $a0=0a=0$, $a\left(-b\right)=\left(-a\right)b=-\left(ab\right)$ and $\left(-a\right)\left(-b\right)=ab$. Furthermore, if $R$ is unital, then $\left(-1\right)a=-a$ and $\left(-1\right)\left(-1\right)=1$. In essence, we can treat negatives and zero the same way that we do in ordinary arithmetic.

• Pigeonhole principle

• Lemma 3.2.2: If an integral domain is finite, then it is a field.

• $\mathbb{Z}\mathrm{/}p\mathbb{Z}$ is a field.

• Definition: With $R$ an integral domain, $a\in R$ non-zero, and $m\in \mathbb{Z}$, $R$ is of characteristic zero if $ma=a+a+\cdots +a=0$ implies $m=0$.

• $\mathbb{Z}$, $\mathbb{Q}$, etc. are of characteristic zero

• Definition: With $R$ an integral domain, $R$ is of finite characteristic if there exists a positive integer $m$ such that $ma=0$ for all $a\in R$. In that case, the smallest such $m$ is the characteristic of $R$.

The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.

### Herstein 3.2.1: Let $a,b,c,d\in R$. Evaluate $\left(a+b\right)\left(c+d\right)$.

By the distributive property, $\left(a+b\right)\left(c+d\right)=\left(a+b\right)c+\left(a+b\right)d=ac+bc+ad+bd$. Notably, the multiplication does not have to be commutative, so the order of the four products cannot be ignored.

### Herstein 3.2.2: Let $a,b\in R$. Show that $\left(a+b{\right)}^{2}={a}^{2}+ab+ba+{b}^{2}$ where ${x}^{2}=xx$.

This is an application of 3.2.1, with $c=a$ and $d=b$.

### Herstein 3.2.3: How does the binomial theorem generalize to an arbitrary ring?

First recall the traditional case. With $a,b$ real numbers and $n$ a non-negative integer, we have $\left(a+b{\right)}^{n}=\sum _{k=0}^{n}\left(\genfrac{}{}{0px}{}{n}{k}\right){a}^{k}{b}^{n-k}\mathrm{.}$

Because integer multiplication is commutative, we were able to collect terms that were equivalent modulo moving factors around. Doing that collection gave us the factors of $\left(\genfrac{}{}{0px}{}{n}{k}\right)$: every term in the expansion has $n$ factors, there are $\left(\genfrac{}{}{0px}{}{n}{k}\right)$ ways of choosing $k$ of those factors to be $a$’s (and the rest are automatically $b$’s). In the non-commutative case, we won’t be able to collect and add up like terms in that way, but there will still be ${2}^{n}$ terms to be added up, each being one of the ways of lining up $n$ copies of the $a$’s and $b$’s.

Now consider a general ring $R$ and let $a,b\in R$. The statement of the above generalization is that $\left(a+b{\right)}^{n}=\sum _{{ϵ}_{1}=0}^{1}\cdots \sum _{{ϵ}_{n}=0}^{1}{a}^{{ϵ}_{1}}{b}^{1-{ϵ}_{1}}{a}^{{ϵ}_{2}}{b}^{1-{ϵ}_{2}}\cdots {a}^{{ϵ}_{n}}{b}^{1-{ϵ}_{n}}\mathrm{.}$

In fact, this is an almost contentless rewriting of $\left(a+b{\right)}^{n}$ because the $n$ sums decouple into a bunch of $\sum _{ϵ=0}^{1}{a}^{ϵ}{b}^{1-ϵ}=a+b\mathrm{.}$

Nevertheless, the insight is that we have to enumerate all ${2}^{n}$ of the length-$n$ strings made up of $a$’s and $b$’s. It is unlikely that there is much more that we can do in a general non-commutative setting.

### Herstein 3.2.4: Show that $R$ is commutative if ${x}^{2}=x$ for every $x\in R$. (“Boolean ring”)

We must show that $xy=yx$ for any $x,y\in R$. The natural thing to consider, in light of what is given, is $\left(x+y{\right)}^{2}={x}^{2}+xy+yx+{y}^{2}=x+y+xy+yx\mathrm{.}$

Of course, we are also given that $\left(x+y{\right)}^{2}=x+y$, so $xy+yx=0.$

In fact, $\left(-r{\right)}^{2}={r}^{2}$ by Lemma 3.2.1, so the Boolean condition also gives us that $x={x}^{2}=\left(-x{\right)}^{2}=-x$, i.e. that every element is its own additive inverse. Therefore $xy=-yx=yx,$

and the claim is proven.

### Herstein 3.2.5: With $a,b\in R$ and $m,n\in \mathbb{N}$, we define $ma$ as the sum of $m$ copies of $a$. Prove that $\left(ma\right)\left(nb\right)=\left(mn\right)ab$.

By the distributive property, \begin{aligned} (ma)(nb)&=(a+\cdots+a)(b+\cdots+b)=a(b+\cdots+b)+\cdots+a(b+\cdots+b)\
=&n(ab)+\cdots+n(ab)=(mn)(ab). \end{aligned}

This illustrates the idea, and the inductive proof is trivial.

### Herstein 3.2.6: Show that if $R$ is an integral domain of finite characteristic, then its characteristic is prime.

Let the characteristic of $R$ be $k$ so that $ka=0$ for all $a\in R$. Furthermore, suppose that $k$ is composite and write $k=mn$ with $m,n$ integers larger than $1$. Let $b\in R$ be an element such that $nb\mathrm{\ne }0$. Such an element exists because otherwise we would say that $n was the characteristic. Also, let $a\in R$ be arbitrary. By problem 3.2.5, we have that $\left(ma\right)\left(nb\right)=\left(mn\right)ab=0$ because $mn=k$ is the characteristic. Now, because $R$ is an integral domain, this forces either $ma=0$ or $nb=0$. The latter is explicitly ruled out by assumption, and therefore $ma=0$ for every $a\in R$, which is the contradiction we seek (the characteristic is defined as the least such positive integer). Thus, $k$ must not admit any non-trivial factorizations, i.e. it must be prime.

### Herstein 3.2.7: Give an example of an integral domain with infinitely many elements but finite characteristic.

Let $R$ be the set of polynomials, in a variable $x$, with coefficients taken from $\mathbb{Z}\mathrm{/}2\mathbb{Z}$. This set is infinite (we can enumerate a countably infinite subset by considering $\left\{{x}^{k}\mid k=0,1,2,\dots \right\}$) and forms a ring under the familiar addition and multiplication (i.e. powers of $x$ are increased by multiplication). Furthermore, it is an integral domain because multiplying non-zero polynomials produces a polynomial of equal or larger degree. More concretely, if $p,q\in R$ have leading terms ${x}^{n}$ and ${x}^{m}$, respectively (the coefficients must be one to be relevant as leading terms), then $pq\in R$ has leading term ${x}^{n+m}$, and hence $pq\mathrm{\ne }0$.

Finally, $R$ has characteristic $2$ because twice any coefficient is zero in $\mathbb{Z}\mathrm{/}2\mathbb{Z}$.

### Herstein 3.2.8: Let $R$ be an integral domain and suppose there exists $n\in \mathbb{N}$ and non-zero $a\in R$ such that $na=0$. Prove that $R$ has finite characteristic.

Let $b\in R$ and consider $0=\left(na\right)b=a\left(nb\right)$. The last manipulation is justified by problem 3.2.5. As $R$ is an integral domain and $a\mathrm{\ne }0$, we have that $nb=0$ for any $b\in R$. This implies that the characteristic of $R$ is finite, less than or equal to $n$.

### Herstein 3.2.9: If $R$ is a set satisfying all the ring axioms except possibly commutativity of addition, and $R$ has a multiplicative identity $1$, then prove that addition must be commutative in $R$, and so $R$ is a ring.

Following the hint in Herstein, we let $a,b\in R$ and compute $\left(a+b\right)\left(1+1\right)$ in two different ways. First, $\left(a+b\right)\left(1+1\right)=\left(a+b\right)+\left(a+b\right)=a+b+a+b\mathrm{.}$

Second, $\left(a+b\right)\left(1+1\right)=a\left(1+1\right)+b\left(1+1\right)=a+a+b+b\mathrm{.}$

These must be equal, $a+b+a+b=a+a+b+b$, and subtracting $a$ from the left and $b$ from the right on both sides, we find $b+a=a+b$, the desired result.

### Herstein 3.2.10: Let $R$ be a commutative ring and let $a,b,c\in R$. Show that $R$ is an integral domain if and only if $ab=ac$ with $a\mathrm{\ne }0$ implies that $b=c$. In other words, $R$ is an integral domain if and only if one can cancel out a common non-zero factor in an equation.

If $R$ is an integral domain and $a\mathrm{\ne }0$, then $0=ab-ac=a\left(b-c\right)$ forces $b=c$.

On the other hand, suppose that we can cancel out common non-zero factors in an equation. Let $a,b\in R$ be such that $ab=0$ and assume that $a\mathrm{\ne }0$. Then $ab=a0$ implies $b=0$. Alternately, suppose $b\mathrm{\ne }0$ and we have no information about $a$. Then $ab=0b$ implies $a=0$. Therefore, for the product to be zero, one of $a$ or $b$ must be zero, which is the statement that $R$ is an integral domain.

### Herstein 3.2.11: Prove that an infinite integral domain need not be a field (compare with Lemma 3.2.2).

$\mathbb{Z}$ is an infinite integral domain which is not a field.

### Herstein 3.2.12: Prove that a field is an integral domain.

A field is a commutative ring where the non-zero elements form a group under multiplication, so a product of non-zero elements will never give zero. This is the definition of an integral domain.

### Herstein 3.2.13: Let $a,b,m,n\in \mathbb{Z}$ with $\left(m,n\right)=1$. Use the pigeonhole principle to show that there exists an integer $x$ such that $x\equiv a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0.6666666666666666em}{0ex}}\mathrm{m}\mathrm{o}\mathrm{d}\text{\hspace{0.17em}}\text{\hspace{0.17em}}m$ and $x\equiv b\phantom{\rule{0ex}{0ex}}\phantom{\rule{0.6666666666666666em}{0ex}}\mathrm{m}\mathrm{o}\mathrm{d}\text{\hspace{0.17em}}\text{\hspace{0.17em}}m$.

Note that this problem is identical to problem 1.3.15, except for the new suggestion to use the pigeonhole principle. Also note that this is a special case of the chinese remainder theorem.

As suggested by Herstein, consider the set $S=\left\{a,\text{\hspace{0.17em}}a+m,\text{\hspace{0.17em}}a+2m,\text{\hspace{0.17em}}\dots ,\text{\hspace{0.17em}}a+\left(n-1\right)m\right\}$

which contains $n$ integers. Clearly every element of $S$ is congruent to $a$ modulo $m$. If we could also show that one element of $S$ was congruent to $b$ modulo $n$, then that would be the desired $x$. If it were the case that no element of $S$ is congruent to $b$ modulo $n$, then the pigeonhole principle says that, of the $n$ elements of $S$ computed modulo $n$, some output occurs at least twice, say $a+im\equiv a+jm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0.6666666666666666em}{0ex}}\mathrm{m}\mathrm{o}\mathrm{d}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n$ for some $0\le i,j and $i\mathrm{\ne }j$. Then we have $\left(a+im\right)-\left(a+jm\right)=\left(i-j\right)m\equiv 0\phantom{\rule{0ex}{0ex}}\phantom{\rule{1em}{0ex}}\mathrm{m}\mathrm{o}\mathrm{d}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\mathrm{.}$

Because $m$ is coprime to $n$, it can be canceled, leaving $i-j\equiv 0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0.6666666666666666em}{0ex}}\mathrm{m}\mathrm{o}\mathrm{d}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n$ which is impossible under the conditions imposed on $i$ and $j$. Therefore, there is an element of $S$ congruent to $b$ modulo $n$, and it is the desired $x$.

### Herstein 3.2.14: Use the pigeonhole principle to prove that the decimal expansion of a rational number must eventually become repeating.

Consider a rational number $m\mathrm{/}n$, with $m$ and $n$ coprime integers and $n\mathrm{\ne }0$. We can compute the decimal expansion by the Euclidean algorithm (i.e. long division). We proceed as $m={a}_{0}n+{r}_{0},\phantom{\rule{2em}{0ex}}10{r}_{0}={a}_{1}n+{r}_{1},\phantom{\rule{2em}{0ex}}10{r}_{1}={a}_{2}n+{r}_{2},\dots ,$

where each remainder ${r}_{k}\in \left\{0,\dots ,n-1\right\}$, and then we find the decimal expansion to be the string of digits ${a}_{0}{a}_{1}{a}_{2}\cdots$ with a decimal point in the appropriate position. Now, once we find the $k$th remainder, ${r}_{k}$, we find the next digit in a straightforward way, ${a}_{k+1}=⌊\frac{10{r}_{k}}{n}⌋\mathrm{.}$

The subsequent remainder ${r}_{k+1}$ is similarly found, and the sequence proceeds from there entirely deterministically. Therefore, if a remainder is ever repeated, it signals that the decimal has begun to repeat and will continue indefinitely. There are $n$ values allowed for the remainder, and infinitely many digits to compute, so the pigeonhole principle says that this repetition must happen, but also that the first period will terminate before the $\left(n+1\right)$-st digit.

As an example, consider the fraction $5\mathrm{/}7$, $m=5$ and $n=7$. The process described above manifests as $\begin{gathered} 5=0\cdot7+5\qquad\longrightarrow&\qquad a_0=0,\qquad r_0=5\$
$\cdots$