Rings

Section 2.1 Rings

In a sense the motivation for the mathematical definition of a ring in mathematics is more immediate than that for a group. From very early on in our mathematical training we are introduced to quite a rich variety of different number systems including the integers $\Z\text{,}$ the rational numbers $\Q\text{,}$ the real numbers $\R\text{,}$ and the complex numbers $\C\text{.}$ Roughly understood, a number system is a set together with two distinct operations, addition and multiplication, which are each reasonably well-behaved and moreover play nicely together. Definition 2.1.1 makes precise exactly what me mean by this.

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Definition 2.1.1. Ring.

A ring is a triple $(R,+,\cdot)\text{,}$ where $R$ is a nonempty set, and $+$ and $\cdot$ are binary operations

\begin{align*} R\times R \amp \xrightarrow{\phantom{XX}+\phantom{XX}} R \amp R\times R\amp \xrightarrow{\phantom{XX}\bullet\phantom{XX}} R\text{,} \end{align*}

called ring addition and ring multiplication, respectively, that satisfy the following axioms.

$(R,+)$ is an abelian group.
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Ring multiplication is associative: i.e., $a\cdot(b\cdot c)=(a\cdot b)\cdot c$ for all $a,b,c\in R\text{.}$
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Ring multiplication distributes over ring addition: i.e., we have

\begin{align} a\cdot (b+c) \amp =a\cdot b+a\cdot c\tag{2.1.1}\\ (b+c)\cdot d \amp =b\cdot d+c\cdot d\tag{2.1.2} \end{align}

for all $a,b,c,d\in R\text{.}$

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There is a multiplicative identity element $1\in R$ satisfying

\begin{equation} 1\cdot a=a\cdot 1=a\tag{2.1.3} \end{equation}

for all $a\in R\text{.}$

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A ring $R$ is commutative if its ring multiplication is commutative and noncommutative otherwise.

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Warning 2.1.2.

The Dummit and Foote text does not include (iv) as one of the axioms a ring needs to satisfy. Instead it distinguishes between rings with and without identity. We will not follow suit and will conscientiously point out resulting discrepancies between our exposition and the book’s.

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Notation 2.1.3.

Addition in a ring will always be denoted by $+\text{,}$ and this notation will never be suppressed. In contrast, we are free to denote the ring multiplication operation in any manner we please, but will more often than not omit the notation in ring expressions.

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Additionally we will use the following integer multiple notation:

\begin{align*} n a \amp = \underset{n \text{ times}}{\underbrace{a+a+\cdots +a}} \amp a\in R, n\in \Z_{\geq 1}\\ (-n)a \amp = n(-a) \amp a\in R, n\in \Z_{\geq 1}\text{.} \end{align*}

Additionally, in a ring $R$ we will denote $n=n\cdot 1_R$ for all $n\in \Z\text{.}$

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As mentioned at the top, our definition of a ring is chosen to generalize the familiar number systems ($\Z\text{,}$ $\Q\text{,}$ $\R\text{,}$ $\C$) of our mathematical youth. That these sets are rings (indeed, commutative rings) with respect to their usual arithmetic operations is a result of arithmetic properties you learned long ago. We add to this collection the slightly more exotic modular rings $\Z/n\Z$ for $n$ a positive integer; the ring axiom identities for modular addition and multiplication follow directly from the corresponding identities for the integer operations.

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Specimen 17. Elementary commutative rings.

The sets $\Z\text{,}$ $\Q\text{,}$ $\R\text{,}$ and $\C$ are all commutative rings with respect to their usual addition and multiplication operations. Additionally, for any positive integer $n\text{,}$ the set $\Z/n\Z$ is a commutative ring with respect to its modular addition and multiplication operations.

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Example 2.1.4. Trivial ring.

Show that a singleton $\{a\}$ can be given a unique ring structure. We call such rings trivial.

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Show that a ring $R$ is trivial if and only if the equality $0=1$ holds in $R\text{.}$

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Solution.

Left to the reader.

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We need not travel too far to find an examples of noncommutative rings. Matrix arithmetic furnishes us with a familiar enough example.

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Specimen 18. Matrix rings.

Let $R$ be a commutative ring. Given a positive integer $n\text{,}$ we define $\ML_n(R)$ to be the set of all $n\times n$ matrices with coefficients in $R\text{.}$

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Given matrices $A=[a_{ij}],B=[b_{ij}]\in \ML_n(R)\text{,}$ we define their sum $A+B$ and product $AB$ as follows:

\begin{align} A+B \amp =[a_{ij}+b_{ij}]\tag{2.1.4}\\ AB \amp = [c_{ij}] \amp c_{ij}=\sum_{k=1}^na_{ik}b_{kj}\text{.}\tag{2.1.5} \end{align}

The triple $(\ML_n(R), +,\cdot)\text{,}$ where $+$ and $\cdot$ are the matrix addition and multiplication operations defined above, is a ring called the ring of $n\times n$ matrices over $R$ (or with coefficients in $R$).

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If $n\geq 2\text{,}$ then $\ML_n(R)$ is a noncommutative ring, as an argument very similar to the one in Example 1.1.4 shows.

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Remark 2.1.5. Matrix rings.

We can remove the restriction that $R$ be commutative in Specimen 18. In other words, the matrix operations (2.1.4) satisfy the ring axioms whether or not $R$ is commutative. We will restrict our attention to the case where $R$ is commutative, however, as here the algebraic properties of $\ML_n(R)$ are somewhat simpler. In particular, when $R$ is commutative, we can define a determinant function $\det\colon \ML_n(R)\rightarrow R$ in the usual fashion, and $\det$ enjoys essentially all of the properties you are familiar with in the case where $R=\R\text{.}$ See Theorem 2.2.2 below.

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We now dispense with the usual “general properties” result for our newly defined mathematical friends.

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Proposition 2.1.6. Ring properties.

Let $R$ be a ring.

The multiplicative identity element of $R$ is unique: i.e., there is exactly one element of $R$ satisfying the identity (2.1.3).
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$0a=a0=0$ for all $a\in R\text{.}$
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$-a=(-1)a=a(-1)$ for all $a\in R\text{.}$
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$(-a)b=a(-b)=-ab$ for all $a,b\in R\text{.}$
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Proof.

We prove (1) and (2), leaving (3) and (4) to the reader.

If an element $b\in R$ satisfies $ba=ab=a$ for all $a\in R\text{,}$ then in particular, it satisfies $b1=1\text{.}$ But by definition of the the multiplicative identity, we have $b1=b\text{.}$ Thus $b=1\text{,}$ as desired.
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We have

\begin{align*} 0a \amp =(0+0)a\\ \amp = 0a+0a\text{.} \end{align*}

Using the additive inverse $-0a\text{,}$ we see that

\begin{align*} 0a=0a+0a \amp \implies -0a+0a=-0a+0a+0a\\ \amp \implies 0=0+0a\\ \amp \implies 0=0a\text{,} \end{align*}

as desired. A similar argument shows $a0=0$.

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Our matrix ring example is in fact an example of a ring construction: a method of building new rings from existing ones. The product ring construction is another such example.

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Specimen 19. Product rings.

Let $I$ be a nonempty set, and let $(R_i)_{i\in I}$ be a family of rings indexed by $I\text{.}$ We define an addition and multiplication operation on $R=\prod_{i\in I}R_i$ as follows:

\begin{align} R\times R \amp \xrightarrow{\phantom{XX}+\phantom{XX}} R \amp R\times R\amp \xrightarrow{\phantom{XX}\bullet\phantom{XX}} R \tag{2.1.6}\\ (a_i)_{i\in I}, (b_i)_{i\in I} \amp \mapsto (a_i+b_i)_{i\in I} \amp (a_i)_{i\in I},(b_i)_{i\in I}\amp\mapsto (a_ib_i)_{i\in I}\text{.}\notag \end{align}

In other words, addition and multiplication of tuples is defined component-wise.

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The triple $(R,+,\cdot)$ is ring called the product of the rings $R_i\text{.}$ Note that we have already seen that $(R,+)$ is a group. (See Proposition 1.1.7.) The identity element of $R$ is the tuple $1=(1_{R_i})_{i\in I}\text{,}$ where for all $i\in I\text{,}$ $1_{R_i}$ is the identity element of $R_i\text{.}$

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Technically speaking, the ring of functions construction introduced next is just a special case of the product ring construction: that is, by definition we have

\begin{equation*} \mathcal{F}(X,R)=R^X=\prod_{x\in X}R\text{.} \end{equation*}

We include it nonetheless, as this construction (in this particular guise) is particular important in analysis contexts.

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Specimen 20. Ring of functions.

Let $R$ be a ring, let $X$ be a nonempty set, and let $\mathcal{F}(X,R)$ be the set of all functions from $X$ to $R\text{.}$ Given functions $f,g\in F(X,R)$ we define their sum $f+g$ and product $fg$ as follows:

\begin{align*} (f+g)(x) \amp = f(x)+g(x) \text{ for all } x\in X\\ (fg)(x) \amp =f(x)g(x) \text{ for all } x\in X\text{.} \end{align*}

The triple $(\mathcal{F}(X,R),+,\cdot)\text{,}$ where $+$ and $\cdot$ are the function addition and multiplication operations defined above, is a ring called the ring of functions from $X$ to $R\text{.}$

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The additive identity of $\mathcal{F}(X,R)$ is the zero function $0_X\colon X\rightarrow R$ defined as $0_X(x)=0$ for all $x\in X\text{;}$ the multiplicative identity of $\mathcal{F}(X,R)$ is the constant function $1_X\colon X\rightarrow R$ defined as $1_X(x)=1$ for all $x\in X\text{.}$

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Another method of building a new ring from an existing one is to identify a subring.

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Definition 2.1.7. Subring.

A subset $S$ of the ring $R$ is a subring if it satisfies the following conditions.

$S$ is a subgroup of $(R,+)\text{.}$
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$1\in S\text{.}$
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$S$ is closed under multiplication: i.e., for all $a,b\in S\text{,}$ we have $ab\in S\text{.}$
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It follows from the ring axioms that a subring $S$ of $R$ is itself a ring with respect to ring operations of $R\text{,}$ restricted to $S\text{.}$

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Specimen 21. Quadratic extensions of $\Q$.

Let $q\in \Q$ be a rational number that does not have a rational square root, and let $\sqrt{q}$ denote one of the two square roots of $q$ lying in $\C\text{.}$ We denote by $\Q[\sqrt{q}]$ the subset of $\C$ consisting of all rational linear combinations of $1$ and $\sqrt{q}\text{:}$ i.e.,

\begin{equation} \Q[\sqrt{q}]=\{z\in \C\mid z=a+b\sqrt{q} \text{ for some } a,b\in \Q\}\text{.}\tag{2.1.7} \end{equation}

The set $\Q[\sqrt{q}]$ is a subring of $\C\text{,}$ and hence a ring in its own right.

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Definition 2.1.8. Center of ring.

The center of a ring $R\text{,}$ denoted $Z(R)\text{,}$ is the set of elements of $R$ that commute with all elements of $R\text{:}$ i.e.,

\begin{equation} Z(R)=\{a\in R\mid ab=ba \text{ for all } b\in R\}\text{.}\tag{2.1.8} \end{equation}

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Example 2.1.9. Center of a ring.

Let $R$ be a ring. Show that $Z(R)$ is a subring of $R\text{.}$

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Solution.

First we show that $Z(R)$ is a subring of $(R,+)\text{.}$ From Proposition 2.1.6 we know that $0\in Z(R)\text{.}$ Next, assume $a,b\in Z(R)\text{.}$ Given any $c\in R\text{,}$ we have

\begin{align*} (a+b)c \amp =ac+bc=ca+cb\\ \amp =ca+cb\\ \amp = c(a+b)\\ (-a)c \amp =a(-c)\\ \amp =(-c)a\\ \amp =c(-a)\text{,} \end{align*}

and thus $a+b, -a\in Z(R)\text{.}$

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Next, we have $1\in Z(R)$ by Proposition 2.1.6. Finally, we show that $Z(R)$ is closed under multiplication. Assume $a,b\in Z(R)\text{.}$ Given $c\in R\text{,}$ we have

\begin{align*} (ab)c \amp = a(bc) \\ \amp =a(cb)\\ \amp =(ac)b\\ \amp =(ca)b\\ \amp =c(ab)\text{,} \end{align*}

showing $ab\in Z(R)\text{,}$ as desired.

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