Law of the Rings

This ring has got nothing to do the Lord of the Rings.

The previous few posts dealt with abstract objects called groups. Really, we label a mathematical structure $G$ a group if it satisfies several properties. Then whatever is true of groups must be true of $G$ .

The integers $\mathbb Z$ were an exceedingly peculiar group under addition in the following sense:

For any $m,n \in \mathbb Z$ , $m+n \in \mathbb Z$ .
For any $k,m,n \in \mathbb Z$ , $k+(m+n) = (k+m) + n$ .
$0 \in \mathbb Z$ and for any $n \in \mathbb Z$ , $0+n = n = n + 0$ .
For any $n \in \mathbb Z$ , $-n \in \mathbb Z$ and $n+ (-n) = (-n) + n = 0$ .

Furthermore, the integers are furnished with many interesting properties:

It is infinite, since it does not contain a finite number of elements.
It is cyclic, since any integer $n$ can be written as a sum of $1$ or $-1$ terms.
Its subgroups are multiples of itself (i.e. $k \mathbb Z \leq \mathbb Z$ ).
It is Abelian under addition, that is, addition is commutative.
Hence, all of its subgroups are normal, and $\mathbb Z_k := \mathbb Z/ k \mathbb Z$ forms a group.
In fact, $\mathbb Z_k \cong \langle \zeta_k \rangle$ , where $\zeta_k$ is a primitive $k$ -th root of unity.

However, we have not once discussed the multiplicative properties of $\mathbb Z$ , which are just as uncanny, if not more so.

Lemma 1. The following multiplicative properties hold for $\mathbb Z$ .

For any $m,n \in \mathbb Z$ , $m \cdot n \in \mathbb Z$ .
For any $k,m,n \in \mathbb Z$ , $k \cdot (m \cdot n) = (k \cdot m) \cdot n$ .
$1 \in \mathbb Z$ and for any $n \in \mathbb Z$ , $1 \cdot n = n = n \cdot 1$ .
For any $k,m,n \in \mathbb Z$ , $k \cdot (m + n) = (k \cdot m) + (k \cdot n)$ .
For any $k,m,n \in \mathbb Z$ , $(k + m) \cdot n = (k \cdot n) + (m \cdot n)$ .

Proof. A technical proof requires us to use the construction of the integers and, perhaps, a ton of induction.

In this case, we call $\mathbb Z$ a ring.

Definition 1. An Abelian group $(R , + )$ with additive identity $0$ (and additive inverse $-x$ of $x$ ) is a ring if it satisfies the following properties.

For any $r,s \in R$ , $r \cdot s \in R$ .
For any $r,s,t \in R$ , $r \cdot (s \cdot t) = (r \cdot s) \cdot t$ .
For any $r,s,t \in R$ , $r \cdot (s + t) = (r \cdot s) + (r \cdot t)$ .
For any $r,s,t \in R$ , $(r + s) \cdot t = (r \cdot t) + (s \cdot t)$ .

We say that the ring has unity if there exists a multiplicative identity $1 \in R$ such that for any $r \in R$ , $1 \cdot r = r = r \cdot 1$ . Furthermore, we abbreviate $r \cdot s \equiv rs$ for simplicity. Finally, by multiplying first then adding, we suppress the parentheses for brevity:

$(r \cdot s) + (r \cdot t) \equiv rs + rt.$

Henceforth, let $(R, + , \cdot)$ denote a ring. We say that $R$ is commutative if $rs = sr$ .

Example 1. $\mathbb Z$ forms a commutative ring.

When constructing the integers (and even natural numbers), $0 n = 0$ as a definition of multiplication. However, for other kinds of objects, $0 n = 0$ as a theorem.

Theorem 1. For any $r \in R$ , $0 r = 0$ .

Proof. The first claim follows from $0 + 0 = 0$ :

$\begin{aligned} (0 + 0) r &= 0 r \\ 0 r + 0r &= 0r \\ (0r + 0r) + (-0r) &= 0r + (-0r) \\ 0r + (0r + (-0r)) &= 0r + (-0r) \\ 0r + 0 &= 0 \\ 0r &= 0. \end{aligned}$

The second claim follows from the first and that $1r = r$ :

$\begin{aligned} r + (-1)r &= 1r + (-1)r \\ &= (1 +(-1))r \\ &= 0r = 0. \end{aligned}$

Corollary 1. For any $r,s \in R$ ,

$\begin{aligned} (-r) s = -rs, \quad r (-s) = -rs, \quad (-r) (-s) = rs. \end{aligned}$

Proof. The first equality arises from Theorem 1:

$rs + (-r)s = (r+(-r))s = 0 \cdot s = 0.$

Hence, $-rs = (-r)s$ .

The second equality follows similarly:

$rs + r(-s) = r(s + (-s)) = r \cdot 0 = 0,$

so that $r(-s) = -rs$ . Finally, the last equality follows from

$(-r)(-s) = -r(-s) = -(-(rs)) = rs,$

where the last result is group-theoretic since

$-x + x = 0 \quad \Rightarrow \quad -(-x) = x.$

There are many other arithmetic properties that we usually take for granted that hold for rings in general.

Before diving into ring theory, we should ask the golden question: why bother? To solve equations. More specifically, congruence equations.

Lemma 2. Fix $k, m, n \in \mathbb Z$ with $k > 0$ . Denote elements in $\mathbb Z_k$ by $[n]_k$ , i.e.

$x \in [n]_k \quad \iff \quad k \mid (x-n).$

Recall that $[m]_k + [n]_k = [m+n]_k$ by the group structure of $\mathbb Z$ under $+$ .

Then $x \in [m]_k$ and $y \in [n]_k$ implies that $xy \in [mn]_k$ .

Proof. We remark that $x \in [m]_k$ if and only if there exists an integer $x_0$ such that $x = m + kx_0$ . Therefore,

$xy =(m + k x_0) (n + k y_0) = mn + k(nx_0 + my_0 + x_0 y_0)$

implies that $xy \in [mn]_k$ .

Henceforth, define $[m]_k \cdot [n]_k := [m \cdot n]_k$ .

Theorem 2. $\mathbb Z_k$ forms a commutative ring.

Proof. Most ring properties are consequences of the ring properties in $\mathbb Z$ . Associativity, for example follows from

$\begin{aligned} [n_1]_k \cdot ([n_2]_k \cdot [n_3]_k) &= [n_1]_k \cdot [n_2 \cdot n_3]_k \\ &= [n_1 \cdot (n_2 \cdot n_3)]_k \\ &= [ (n_1 \cdot n_2) \cdot n_3 ]_k \\ &= [n_1 \cdot n_2]_k \cdot [n_3]_k \\ &= ([n_1]_k \cdot [n_2]_k) \cdot [n_3]_k. \end{aligned}$

Example 1. Evaluate $[3]_5 \cdot [2]_5$ . Hence, determine the value of $x$ such that $[2]_5 \cdot [x]_5 = [3]_5$ , where $0 \leq x < 5$ .

Proof. By a direct calculation,

$[3]_5 \cdot [2]_5 = [3 \cdot 2]_5 = [6]_5 = [1]_5.$

Therefore, we can left-multiply $[3]_5$ on both sides:

$\begin{aligned} [3]_5 \cdot ([2]_5 \cdot [x]_5) &= [3]_5 \cdot [3]_5 \\ ([3]_5 \cdot [2]_5) \cdot [x]_5 &= [3 \cdot 3]_5 \\ [1]_5 \cdot [x]_5 &= [9]_5 \\ [x]_5 &= [4]_5. \end{aligned}$

where $[n]_5 = [n + 5t]_5$ for any integer $t$ . Since the map $\{0,1,2,3,4\} \to \mathbb Z_5$ defined by $n \mapsto [n]_5$ is injective, $x = 4$ is the required solution.

Remark 1. In the language of congruence equations and modular arithmetic, we seek the set of values of $x$ such that

$5 \mid (2x-3) \quad \iff \quad 2x \equiv 3 \mod 5 \quad \iff \quad 2x \equiv_5 3.$

However, since we like to keep equality as equality, we will work in the language of congruence classes $[x]_k$ . In particular, solving congruence equations modulo $k$ amounts to solving equations in the ring $\mathbb Z_k$ .

Example 2. Show that there are no values of $x$ such that $[2]_4 \cdot [x]_4 = [3]_4$ .

Proof. Suppose for a contradiction that there exists some $0 \leq x< 4$ such that $[2]_4 \cdot [x]_4 = [2x]_4$ . Now

$\begin{aligned} [2x]_4 = [3]_4 \quad &\Rightarrow \quad 4 \mid (2x-3) \\ &\Rightarrow \quad 2 \mid (2x-3), \end{aligned}$

a blatant contradiction. Therefore, for any $x$ , $[2]_4 \cdot [x]_4 \neq [3]_4$ .

When then, can we solve congruence equations? Furthermore, can we solve systems of congruence equations?

Lemma 3. Fix integers $k, n$ with $k > 0$ . Then there exists a unique integer $0 \leq m < k$ such that

$[n]_k \cdot [m]_k = [m]_k \cdot [n]_k = [1]_k$

if and only if $\gcd(n, k) = 1$ . In this case, we denote $[m]_k =: [n]_k^{-1}$ , and we have $([n]_k^{-1})^{-1} = [n]_k$ .

Proof. By multiplication of congruence classes,

$[m \cdot n]_k = [m]_k \cdot [n]_k = [1]_k \quad \iff \quad k \mid (mn - 1).$

The latter holds if and only if there exists an integer $\ell$ such that

$mn - 1 = k\ell.$

$(\Rightarrow)$ Write

$mn - 1 = k\ell \quad \iff \quad n \cdot m + k \cdot (-\ell) = 1.$

By Bézout’s lemma, $\gcd(n, k) = 1$ .

$(\Leftarrow)$ Conversely, use Bézout’s lemma to furnish integers $r,s$ such that

$r \cdot n + s \cdot k = 1 \quad \iff \quad rn - 1 = k(-s).$

Use the division algorithm to furnish integers $q, m$ with $0 \leq m < k$ such that

$r = qk + m\quad \Rightarrow \quad mn - 1 = k(-s-qn).$

Set $\ell := -s - qn$ to yield the desired result.

Example 3. $[2]_5^{-1} = [3]_5$ and $[3]_5^{-1} = ([2]_5^{-1})^{-1} = [2]_5$ .

Proof. Since $2, 5$ are distinct prime numbers, $\gcd(2, 5) = 1$ . Write

$5 = 2 \cdot 2 + 1 \quad \Rightarrow \quad (-2) \cdot 2 - 1 = 5 \cdot (-1).$

By the division algorithm, $-2 = (-1) \cdot 5 + 3$ , so that $m = 3$ as per Lemma 3, and

$[2]_k^{-1} = [m]_k = [3]_k.$

Theorem 3. Let $a,b,k$ be integers with $k > 0$ . If $\gcd(a,k) = 1$ , then there exists a unique $0 \leq x < k$ such that

$[a]_k \cdot [x]_k = [b]_k.$

Proof. By Lemma 3, $[a]_k^{-1}$ exists, so that we left-multiply by $[a]_k^{-1}$ :

$\begin{aligned} [a]_k^{-1} \cdot ([a]_k \cdot [x]_k) &= [a]_k^{-1} \cdot [b]_k. \end{aligned}$

The left-hand side simplifies to

$\begin{aligned} [a]_k^{-1} \cdot ([a]_k \cdot [x]_k) &= ([a]_k^{-1} \cdot [a]_k) \cdot [x]_k \\ &= [1]_k \cdot [x]_k \\ &= [x]_k. \end{aligned}$

Therefore,

$\begin{aligned} [x]_k &= [a]_k^{-1} \cdot ([a]_k \cdot [x]_k) = [a]_k^{-1} \cdot [b]_k. \end{aligned}$

Denote $\mathbb Z_k^* := \mathbb Z_k \backslash \{ [0]_k\}$ . Let $p$ be a prime.

Lemma 4. $\mathbb Z_p^*$ forms a group of order $p-1$ under multiplication.

Proof. Write $\mathbb Z_p^* = \{ [1]_p, [2]_p, \dots, [p-1]_p \}$ . For each $1 \leq n \leq p-1$ , by Lemma 3,

$\begin{aligned} p \nmid n \quad &\Rightarrow \quad \gcd(n, p) = 1 \\ &\Rightarrow \quad[n]_p^{-1} \in \mathbb Z_p^*. \end{aligned}$

Theorem 4 (Fermat’s Little Theorem). For any $1 \leq a \leq p-1$ ,

$[a]_p^{p-1} = [1]_p.$

More generally, for any integer $a$ , $[a]_p^p = [a]_p$ .

Proof. By Lemma 4, $\mathbb Z_p^*$ forms a group of order $p-1$ under multiplication. Furthermore, $\langle [a]_p \rangle \leq \mathbb Z_p^*$ . By Lagrange’s theorem

$m:= | \langle [a]_p \rangle | \mid | \mathbb Z_p^* | = p-1.$

Find an integer $k$ such that $p-1 = k m$ . Then

$\begin{aligned} [a]_p^{p-1} &= [a]_p^{k m} \\ &= ([a]_p^{ m })^k \\ &= [1]_p^k = [1]_p. \end{aligned}$

The result holds trivially for $a = 0$ . For the general case, use the division algorithm to obtain unique integers $q, r$ such that

$a = qp + r,\quad 0 \leq r < p.$

Then the special case yields

$[a]_p^{p-1} = [r]_p^{p-1} = [1]_p.$

Definition 2. A ring $R$ is a field if $R^* := R\backslash \{0\}$ forms an Abelian group under multiplication.

Example 4. For prime numbers $p$ , $\mathbb Z_p$ forms a finite field with $p$ elements, and is often denoted $\mathbb F_p \equiv \mathbb Z_p$ to emphasise its field structure.

Example 5. The commonly used infinite sets $\mathbb Q, \mathbb R, \mathbb C$ form fields.

Now we could go down the number-theoretic rabbit-hole a bit further, but I hope I’ve given sufficient examples in number theory that motivate a study of ring theory. Next time, we will look at the various sub-structures, and in particular, ideals.

—Joel Kindiak, 20 Jan 26, 1950H

KindiakMath

Law of the Rings

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Law of the Rings

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