Rings

Maggie Schildt

26 Rings

In the previous sections we looked at sets with one binary operation, in this section we will be looking at sets with two binary operations.

Definition.

A ring $\left\langle R, +, \cdot \right\rangle$ is a set $R$ together with two binary operations + and $\cdot$ , defined on $R$ such that the following axioms are satisfied:

$\left\langle R, + \right\rangle$ is an abelian group.
Multiplication is associative
For all $a,b,c\in R$ , the left distributive law, $a \cdot (b+c)=(a\cdot b) + (a \cdot c)$ and the right distributive law $(a+b) \cdot c = (a\cdot c)+(b\cdot c)$ hold.

Here we are introduced to two new terms, the left and right distributive laws. These laws are the structural requirements for how two binary operations, addition and multiplication, are applied to the set $R$ .

Being Abstract Algebra there is always additional properties to categorize a structure further. We can categorize rings in the following ways:

The Commutative Ring: A ring in which the multiplication is commutative is known as a commutative ring.

Ring with Unity: A ring with the multiplicative identity element 1, which we call “unity” in abstract algebra, is known as a ring with unity [12].

Division Ring: Let element $u$ in ring $R$ , with unity, be a unit of $R$ if $u$ has a multiplicative inverse in $R$ . If every nonzero element of $R$ is a unit, then $R$ is a division ring.

Field: A field is a commutative division ring.

There is little further explanation needed for these different categorizations of rings since we are already familiar with the terms and the algebraic operations that define each. Now we can observe how we might go about finding a ring as well as categorizing it using the above terms.

Example:

Decide whether $\left\lbrace a + b\sqrt{2}| a,b\in \mathbb{Z} \right\rbrace$ with the usual addition and multiplication are closed on the set, give a ring structure. If a ring is formed, state whether the ring is commutative, has unity, and is a field [12].

Let $a,b,c,d\in \mathbb{Q}$ then we have $a+b\sqrt{2}, c+d\sqrt{2}\in R$ such that

$\begin{equation*} (a+b\sqrt{2}) + (c+d\sqrt{2}=(a+c)+(b+d)\sqrt{2}\in R \end{equation*}$

and thus we can conclude that $R$ is closed under addition.

Let $a,b,c,d\in \mathbb{Q}$ then we have $a+b\sqrt{2}, c+d\sqrt{2}\in R$ such that

$\begin{equation*} \begin{split} (a+b\sqrt{2})(c+d\sqrt{2})&=ac+ad\sqrt{2}+cb\sqrt{2}+bd2\\ &= (ac+2bd)+(ad+cb)\sqrt{2} \end{split} \end{equation*}$

Where $(ac+2bd)+(ad+cb)\sqrt{2}\in R$ , thus we can conclude that $R$ is closed under multiplication. Thus we have proven $R$ is closed for addition and multiplication.

Thus we have proven $R$ is closed for addition and multiplication. Now consider that $\left\langle R, +, \cdot \right\rangle$ is a ring, we need to satisfy all three axioms for a ring.

$\mathbb{R}$ is an abelian group under addition and it has been proven that $R\subseteq \mathbb{R}$ such that for all $a+b\sqrt{2}, c+d\sqrt{2}\in R$ it follows that

$\begin{equation*} \begin{split} (a+b\sqrt{2})+(c+d\sqrt{2})&=(a+c)+(b+d)\sqrt{2}\\ &=(c+a)+(d+b)\sqrt{2}\\ &=(c+d\sqrt{2})+(a+b\sqrt{2}) \end{split} \end{equation*}$

thus we conclude that $R$ is an abelian group.

Given $R\subseteq \mathbb{R}$ then elements of $\mathbb{R}$ are also elements of $R$ and since $\mathbb{R}$ is associative then we can conclude that multiplication on ring $R$ is associative.
Given $R\subseteq \mathbb{R}$ then since the right and left distribution laws hold for $\mathbb{R}$ then they hold for $R$ .

Thus we can conclude that $R$ is a ring.

Now since $\mathbb{R}$ is commutative then the multiplication on $R$ is commutative, thus $\left\langle R, +, \cdot \right\rangle$ is a commutative ring.

Let 1 be the identity element in $R$ under multiplication such that $1=1+0\sqrt{2}\in R$ and thus $1(a+b\sqrt{2}) = a+b\sqrt{2}$ for all $a+b\sqrt{2}\in R$ . We can thus conclude that $R$ is a ring with unity 1.

Let $a+b\sqrt{2}\in R$ then $a+b\sqrt{2}$ is a non-zero element of the field $\mathbb{R}$ so the inverse $\frac{1}{a+b\sqrt{2}}\in \mathbb{R}$ . We can rationalize this as

$\begin{equation*} \frac{1}{a+b\sqrt{2}}=\frac{1}{a+b\sqrt{2}}\times \frac{a-b\sqrt{2}}{a-b\sqrt{2}}=\frac{a-b\sqrt{2}}{a^{2}-2b^{2}}=\frac{a}{a^{2}-2b^{2}}+\frac{-b}{a^{2}-2b^{2}} \sqrt{2}. \end{equation*}$

Since $\frac{a}{a^{2}-2b^{2}},-\frac{b}{a^{2}-2b^{2}}\in \mathbb{Q}$ then $\frac{a}{a^{2}-2b^{2}}+\frac{-b}{a^{2}-2b^{2}}\sqrt{2}\in \mathbb{R}$ and so we can write $a+b\sqrt{2}, \frac{1}{a+b\sqrt{2}}\in R$ for which

$\begin{equation*} a+b\sqrt{2}\times \frac{1}{a+b\sqrt{2}}=\frac{a+b\sqrt{2}}{a+b\sqrt{2}}=1. \end{equation*}$

Thus $\left\langle R, +, \cdot \right\rangle$ is a field.

We should now have a good understanding from the above example, about how to find rings as well as how to prove their commutative, unity, division, and field property. There is so much more to discover about rings in Abstract Algebra, how they apply to polynomials, the concept of sub-rings and how they apply to our next topic, fields. In the next section we will look at fields a bit more and look at some more examples with a stronger focus on fields.

22.1 Fields

We have already given the brief description that a field is a ring which satisfies both the commutative property and the property for division ring. But fields deserve their own formal definition and axioms [12].

Definition.

Let $F$ be a group with two binary operations, + and $\cdot$ , with respect to the following axioms:

For all k $a,b\in F$ it follows that $a+b\in F$ and $a\cdot b\in F$ such that $F$ is closed under addition and multiplication.
For all $a,b,c\in F$ it follows that $a+(b+c) = (a+b)+c$ and $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ such that $F$ is associative.
For all $a,b\in F$ it follows that $a+b=b+a$ and $a \cdot b = b \cdot a$ such that both addition and multiplication under $F$ is commutative.

We can see from this definition that a field is a set that satisfies the axioms for a group, as well as satisfying every commutative property so that a field $F$ is a commutative ring. Let us now introduce a few structural classifications for fields [12].

Theorem: Field of Quotients of $D$

Any integral domain $D$ can be enlarged to a field $F$ such that every element of $F$ can be expressed as a quotient of two elements of $D$ .

It should be noted from this theorem that every field $F$ containing an integral domain $D$ contains a field of quotients of $D$ . This is a similar rule to cyclic subgroups where every group has within it a cyclic subgroup. This comparison tells us there are so many structural similarities and axioms that carry over to different categories that a set may qualify under. We truly can see that Abstract Algebra is all about the study of set structure and the properties that help to classify them. Being able to classify sets helps mathematicians be able to instantly understand the properties of a set depending on whether they are a ring, group, field, subgroup, etc. Just as in rings and groups, fields have underlying structures [12].

Theorem:

For $[(a,b)]$ and $[(c,d)]$ in $F$ , the equations $[(a,b)]+[(c,d)]=[(ad+bc, bd)]$ and $[(a,b)][(c,d)]=[(ac,bd)]$ give well-defined operations of addition and multiplication on $F$ .

Here, we can see the structure for elements of a set under field multiplication and addition. If we observe the above theorem we see that when we sum equivalence classes $(a,b)$ and $(c,d)$ , which are arbitrary classes in $F$ the axiom states that the sum is equivalent to $[(ad+bc), bd)]$ . Under each operation never can $bd=0$ or for that matter $b=0$ or $d=0$ . Just as we have seen before throughout our study of mathematics, there are always axioms for binary operations. Fields, as we can see are no different. We can see an example of how we might apply these axioms to a problem, below, to prove that a field $F$ also contains identity elements and inverse elements under addition and multiplication.

Example:

Now $[(a,b)]+[(c,d)]$ is by definition $[(ad+bc, bd)]$ . Also $[(c,d)]+[(a,b)]$ is by definition $[(cb+da, ab0]$ . Prove that $[(0,1)]$ is an identity element for addition in $F$ . Prove $[(-a, b)]$ is an additive inverse for $[(a,b)]$ in $F$ .

Consider $[(0,1)]\in F$ such that

$\begin{equation*} [(0,1)]+[(a,b)]=[(0b+1a), 1b]=[(a,b)] \end{equation*}$

thus we can conclude that $[(0,1)]$ is an identity element for addition in $F$ . Now consider $[(-a, b)]\in F$ such that

$\begin{equation*} [(-a, b)]+[(a,b)]=[(-ab+ba, b^{2})]=[(0, b^{2})] \end{equation*}$

and since $0(1)=0(b^{2})=0$ then $[(0,b^{2})]=[(0,1)]$ and if

$\begin{equation*} \begin{split} [(a,b)]+[(-a,b)]&=[(ab+b(-a), b^{2})]\\ &=[(ab-ba, b^{2})]\\ &=[(0,b^{2})]=[(0,1)] \end{split} \end{equation*}$

thus we can conclude that $[(-a,b)]$ is an additive inverse for $[(a,b)]\in F$ .

Here we can make out similarities between groups, rings, and fields. Although the structure for addition and multiplication, as it pertains to each category, is formatted specifically for a group, a ring, or a field we have some commonalities. Each group, ring, and field must contain an inverse element and an identity element under the one or two binary operations applied to it. From this last example we saw that elements of fields can be expressed as $[(a,b)]$ . These are called equivalence classes and $F$ is considered the set of all equivalence classes for two elements $a,b\in S$ where $S$ is a set. We can further explain that an equivalence class $[(a,b)]$ is for $(a,b)$ in $S$ under the relation $\sim$ . There are three properties for an equivalence relation [12]:

Reflexive Let $S=\left\lbrace (a,b)|a,b\in D, b\neq 0 \right\rbrace$ then $(a,b)\sim (a,b)$ since $ab=ba$ for multiplication in $D$ , where $D$ is the integral domain.
Symmetric If $(a,b)\sim (c,d)$ , then $ad=bc$ . Since multiplication in $D$ is commutative then $cb=da$ thus resulting in $(c,d)\sim (a,b)$ .
Transitive If $(a,b)\sim (c,d)$ , and $(c,d)\sim (r,s)$ , then $ad=bc$ and $cs=dr$ . Since $D$ is commutative, we have $asd=sad=sbc=bcs=bdr=brd \Rightarrow asd=brd \Rightarrow as=br$ and thus $(a,b)\sim (r,s)$

We should already be familiar with the concepts of reflexive, symmetric, and transitive from our discussion of set theory so the notion that these same properties can apply for an equivalence class of a set should seem straight forward.

Before we move on to the next topic, observe the following example that helps to relate sub-domains to domains using the properties for a field of quotients [12].

Example:

Show by an example that a field $F'$ of quotients of a proper sub-domain $D'$ of an integral domain $D$ may also be a field of quotients for $D$ .

Let $D=\mathbb{Z}[\frac{1}{2}]=\left\lbrace \frac{a}{b}\in \mathbb{Q}|b=2^{q}, q\in \mathbb{Z}^{+} \right\rbrace$ be an integral domain, then $D'=\mathbb{Z}$ such that $F'=\mathbb{Q}$ since a rational number formed by two integers is a quotient. Since $\mathbb{Q}$ is a field then $D\subseteq \mathbb{Q}$ . If $F$ is the field of quotients for $D$ then we also say $F\subseteq \mathbb{Q}$ and since $\mathbb{Q}$ is the field of quotients for $D'$ then we can also say that $Q\subseteq F$ thus we can conclude that $F=\mathbb{Q}$ . Thus a field of quotients for the sub-domain $D'$ is also the field of quotients for integral domain $D$ .

With such a focus on integral domains in this example, beyond assisting in the definition of a field of quotients, it makes sense to have a separate section to better explain integral domains and their use.

22.2 Integral Domains

We gave a brief introduction to integral domains as the pertain to the definitions for fields but as always there is a formal definition which we will cover here as well as a diagram of how all these terms relate to one another.

Definition. Zero Divisor

If $a$ and $b$ are two ring elements with $a\neq 0 \neq b$ but $ab=0$ then $a$ and $b$ are zero-divisors.

This might not make any sense at first, how could two non zero elements multiplied together, produce the quantity 0? This is easiest to explain by looking at an example.

Example:

Given $x^{2}+2x+4=0$ in $\mathbb{Z}_{6}$ find all the solutions.

We address each element in $\mathbb{Z}_{6}$ ; $0,1,2,3,4,5$ .

$\begin{equation*} \begin{split} 0^{2}+2(0)+4 &=4\\ 1^{2}+2(1)+4 &=7=1\\ 2^{2}+2(2)+4 &=12=0\\ 3^{2}+2(3)+4 &=19=1\\ 4^{2}+2(4)+4 &=28=4\\ 5^{2}+2(5)+4 &=39=3 \end{split} \end{equation*}$

Thus we can conclude that the only solution in $\mathbb{Z}_{6}$ is when $x=2$ .

Here we are given a function in the ring $\mathbb{Z}_{6}$ thus the we can put every result from the function as they relate to 6 since 6 divides 12 evenly we let $12=0$ where as since 6 divides 7 with a remainder of 1 we let $7=1$ . With this rough description in mind we introduce a theorem for finding the zero-divisors of a ring.

Theorem:

In the ring $\mathbb{Z}_{n}$ , the divisors of 0 are precisely those nonzero elements that are not relatively prime to $n$ .

The counter view to this which supports this theorem is that if $n$ for $\mathbb{Z}_{n}$ is prime then $\mathbb{Z}_{n}$ has no zero divisor because the only two integers $a$ and $b$ whose product $ab$ is a prime number is when $a=1$ and $b=n$ where $n$ is the prime number such that $ab=1n=n$ and since $n$ cannot be an element in $\mathbb{Z}_{n}$ then we say $\mathbb{Z}_{n}$ has no divisor of 0.

Now that we have introduced the concept of divisors we can introduce integral domains, formally.

Definition.

An integral domain $D$ is a commutative ring with unity $1\neq 0$ and containing no divisors of 0.

This will make more sense when we look at a few examples later on but for now we can see how the definition of an integral domain applies to rings and fields. Now we are familiar with fields by now and we know that fields are commutative rings so it follows that all fields are integral domains and thus every finite integral domain is a field.

Theorem:

If $n$ is a prime, then $\mathbb{Z}_{n}$ is a field.

This theorem holds because $\mathbb{Z}_{n}$ is a finite integral domain since $n$ is prime and thus $\mathbb{Z}_{n}$ is also a field. Continuing on our relations between integral domains and other algebraic sets we introduce rings in relation to integral domains [12].

Definition. The characteristic of the ring $R$

If for a ring $R$ a positive integer $n$ exists such that $n\cdot a=0$ for all $a\in R$ , then the least positive integer is the characteristic of the ring $R$ . If no such positive integer exists, then $R$ is of characteristic 0.

Theorem:

Let $R$ be a ring with unity. If $n\cdot 1\neq 0$ for all $n\in \mathbb{Z}^{+}$ , then $R$ has characteristic 0. If $n\cdot 1=0$ for some $n\in \mathbb{Z}^{+}$ , then the smallest such integer $n$ is the characteristic of $R$ .

Example:

Show that the characteristic of an integral domain $D$ must be either 0 or a prime $p$ .

Let $mn$ be the characteristic of $D$ for $m,n\in \mathbb{Z}$ such that $(m\cdot 1)(n\cdot 1)=mn=1\in D$ .

Consider the characteristic of $D$ is not prime. Then there exists $m,n\in \mathbb{N}$ where $m>1$ and $n>1$ such that the characteristic of $D$ is $mn$ . If we let $mn$ be the smallest positive integer such that $(mn)\cdot 1=0$ for $n\in \mathbb{Z}$ . Then we let

$\begin{equation*} (mn)\cdot 1=n\cdot 1+n\cdot 1+n\cdot 1+...+n\cdot 1 \end{equation*}$

with $m$ sums of $n\cdot 1$ with 1 as the unit of ring $R$ . Next

$\begin{equation*} \begin{split} (mn)\cdot 1&=n\cdot 1+n\cdot 1+n\cdot 1+...+n\cdot 1 \\ &= n\cdot 1(1+1+1+...+1)\\ &=(n\cdot 1)(m\cdot 1) \\ &= (m\cdot 1)(n\cdot 1) = 0 \end{split} \end{equation*}$

Since $m<mn$ and $n<mn$ then by definition for the characteristic of a ring $mn$ is not the least positive integer, this is a contradiction to the claim that $mn$ is the characteristic of the integral domain $D$ . Thus we conclude that the integral domain $D$ has characteristic 0 or prime $p$ .

This proof supports the idea that an integral domain, which is itself a ring, has a characteristic of 0 which supports the definition for the characteristic of the ring. As well, this supports the theorem that if a ring or the integral domain has unity then either the characteristic is 0 or it is the smallest positive integer. The characteristic of a ring helps us understand what kind of set we are dealing with if all we are given is the characteristic. The common sets $\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$ can all be classified as having a characteristic of 0 because there is no least positive integer in each set because they are infinite.

We will look at one more example, this time with a defined characteristic that is not 0.

Example:

Let $R$ be a commutative ring with unity of characteristic 4. Compute and simplify $(a+b)^{4}$ for $a,b\in R$ [12].

Let $R$ be a commutative ring with unity of characteristic 4. By the binomial theorem for expansion, for $a,b\in R$ we have

$\begin{equation*} \begin{split} (a+b)^{4} &=(a+b)(a+b)(a+b)(a+b)\\ &=(a+b)(a+b)(a^{2}+2ab+b^{2})\\ &=(a+b)(a^{3}+3a^{2}b+3ab^{2}+b^{3})\\ &=a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4} \end{split} \end{equation*}$

Since we were given that $4n=0$ we can simplify this formula to the following

$\begin{equation*} \begin{split} a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4} &= a^{4}+(0\cdot a^{3}b)+(0\cdot a^{2}b^{2})+(0\cdot ab^{3})+b^{4}\\ &=a^{4}+b^{4} \end{split} \end{equation*}$

Therefore we can conclude that $(a+b)^{4}=a^{4}+b^{4}$ for all $a,b\in R$ .

In this last example we can see how a characteristic of a ring effects a ring under certain binary operations.

We have now covered groups, various families of rings, integral domains, and fields. In the following diagram we will look at how all these terms relate to one another and how they are all involved with one another.

This diagram gives us the best idea for the families under groups and how they all relate to one another. We can also relate the axioms pertaining to each specified ring and how they carry over. This diagram begins with the largest circle which is groups, this means that any inner circle must qualify as a group. If you are asked to prove a set is a field if we can not prove it is first a group then we do not need to go any further. This also means that the axioms that justify a group carry over to every inner family. This diagram shows us that groups are the largest family and with in groups we either have rings or not. With in rings we may have specified commutative rings or rings with unity which sometimes overlap. Within the overlap we have a special family of rings with qualify as integral domains and with in this family some qualify as fields.

Now that we have a strong understanding of how each specified family relates to one another we can look at rings and fields again, this time when determining polynomials.

26 Rings

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