Residue classes
Modular arithmetics as a way of calculating modulo some number $n$ is a powerful tool. However, it can be made simpler, and thus even more effective.
First, recall that numbers having the same remainder when divided by $n$ behave in exactly the same way in such a calculation and yield exactly the same results.
A quick example: $8 * 4 + 11 \equiv 3*(-1) + 6 \equiv 3 (\mod 5)$.
One might say that there are only $n$ essentially different objects for this calculation - the numbers that give distinct remainders modulo $n$. The numbers having the same remainder can be considered to be essentially the same object.
The next step would be to define addition and multiplication between such objects in a suitable way compatible with the original addition and multiplication.
This approach leads us to the notion of residue classes.
A residue class modulo $n$ is a set of all integers that give the same remainder when divided by $n$.
Formally, the residue class of $a \in \mathbb Z$ modulo $n \in \mathbb N$ is $\overline a = \{b \in \mathbb Z \mid a \equiv b \ (\mod n)\} = \{a + kn \mid k \in \mathbb Z\}$. In this notation, the modulus $n$ is implicit.
As the following exercises show, all residue classes modulo $n$ partition the set $\mathbb Z$ into $n$ parts.
Show that for all $a, b \in \mathbb Z$, the following three claims are equivalent:
- $\overline a = \overline b$,
- $\overline a \cap \overline b = \emptyset$,
- $a \equiv b \ (\mod n)$,
- exists $k \in \mathbb Z$, such that $a = b + kn$.
Show that $\overline 0, \overline 1, \dots , \overline{n − 1}$ are different residue classes.
The first of those two exercises shows that there are at most $n$ different residue classes modulo $n$, because there are $n$ different remainders when dividing by $n$. The second exercise shows that the number of residue classes is at least $n$.
Denote $\mathbb Z_n = \{\overline 0, \overline 1, \dots , \overline{n − 1}\}$.
We can define addition and multiplication on $\mathbb Z_n$ as follows:
- $\overline a+\overline b=\overline{a+b}$,
- $\overline a \cdot \overline b=\overline{a\cdot b}$.
Thanks to the properties of $\equiv$, shown in the previous section, these operations are well-defined. Namely, when we write $\overline{a + b}$, the integer $a$ is only defined as an arbitrary element of the residue class $\overline a$. When we take an arbitrary element $a$ of $\overline a$, and an arbitrary element $b$ of $\overline b$, is it the case that the residue class $\overline{a + b}$ is always the same? Yes, because $a_1 + b_1 \equiv a_2 + b_2 \ (\mod n)$ whenever $a_1 \equiv a_2 \ (\mod n)$ and $b_1 \equiv b_2 \ (\mod n)$. Similarly, the multiplication of residue classes is well-defined.
Having defined $\mathbb Z_n$, we can rewrite our quick example from the beginning of the lesson as $\overline{8} \cdot \overline 4 + \overline{11}= \overline 3 \cdot \overline{-1} +\overline 6 = \overline{3}$ in $\mathbb Z_5$, where the first equality holds trivially simply because the corresponding objects are equal.
Write down the addition and multiplication tables for $\mathbb Z_7$ and $\mathbb Z_8$.
Similarities to integers
It turns out that both operations of addition and multiplication on $\mathbb Z_n$ have similar properties as the same operations on integers:
- They are both associative, meaning $(\overline a + \overline b) + \overline c = \overline a + (\overline b + \overline c)$ and $(\overline a \cdot \overline b) \cdot \overline c = \overline a \cdot (\overline b \cdot \overline c)$.
This property allows the formula $\overline a + \overline b + \overline c$ to be correctly understood in a single way. The same is true for $a \cdot b \cdot c$.
- They are both commutative, meaning $\overline a + \overline b = \overline b + \overline a$ and $\overline a \cdot \overline b = \overline b \cdot \overline a$.
- They both have units. For addition, $\overline a + \overline 0 = \overline 0 + \overline a = \overline a$. For multiplication, $\overline a \cdot \overline 1 = \overline 1 \cdot \overline a = \overline a$.
- There is always an opposite element for addition, meaning that for any $\overline a$ there exists $-\overline a = \overline{-a}$ such that $\overline a + -\overline{a} = \overline{0}$.
- Multiplication is distributive over addition:
\begin{align*} (\overline a+\overline b)\cdot \overline c &= \overline a\cdot \overline c + \overline b\cdot \overline c\enspace,\\ \overline a\cdot (\overline b + \overline c) &= \overline a\cdot \overline c + \overline a\cdot \overline c\enspace. \end{align*}
These properties actually mean that both $\mathbb Z$ and $\mathbb Z_n$ are commutative rings with respect to addition and multiplication. For an introduction to rings, see ?.
Because of the latter fact $\mathbb Z_n$ is also called the residue class ring modulo $n$.
Show that $\mathbb Z_n$, together with the addition and multiplication operations, is a commutative ring.
Differences
Not everything in $\mathbb Z_n$ is the same way as in $\mathbb Z$.
Take the equation $4a = 4b$ with integers $a$ and $b$. We know that it yields $a = b$. More generally, for any integer $c \neq 0$ the equality $ca = c b$ implies $a = b$ because we can divide both sides by $c$. On the other hand, for example, in $\mathbb Z_{12}$ the equality $\overline{4}\cdot \overline{a} = \overline{4}\cdot \overline{b}$ does not necessarily mean that $\overline{a} = \overline{b}$, simply because $\overline{4}\cdot \overline{1} = \overline{4}\cdot \overline{4}$.
The numbers as the number $c$ above are called cancellative.
Thus we made an important observation that every nonzero integer is cancellative while for some numbers $n$ there are nonzero nonconcellative elements of $\mathbb Z_n$.
Find noncancellative elements of $\mathbb Z_{12}$, $\mathbb Z_7$, and $\mathbb Z_4$.
Similarly, consider the equation $ab = 0$ with integers $a$ and $b$. It means that at least one of the numbers $a$ and $b$ is equal to $0$. Again, this is not the case in $\mathbb Z_{12}$ because we can find nonzero $\overline{a}$ and $\overline b$ such that $\overline{a} \cdot \overline{b} = \overline{0}$, e.g., $\overline{4} \cdot \overline{3} = \overline{0}$.
Nonzero $a$ such that $ab = 0$ for some nonzero $b$ is called a zero divisor.
Thus we observed that there are no zero divisors in $\mathbb Z$ but for some numbers $n$ there are zero divisors in $\mathbb Z_n$.
It is relatively easy to see that a zero divisor is always noncancellative.
Show that a zero divisor is never cancellative. Solution
It turns out that in $\mathbb Z_n$ the notions of zero divisors and noncancellative nonzero elements actually coincide. In fact, a nonzero element $\overline a$ in $\mathbb Z_n$ is a zero divisor and noncancellative if and only if $\gcd(a,n) > 1$.
As we already observed in the proof, the condition $\gcd(a,n) = 1$ actually describes those elements in $\mathbb Z_n$, which have multiplicative inverse. That is, those elements $\overline a$ for which there is $x \in \mathbb Z_n$ such that $\overline a \cdot \overline x = \overline x \cdot \overline a = \overline 1$. For example, in $\mathbb Z_7$ the multiplicative inverse of $\overline 3$ is $\overline 5$, because $\overline 3 \cdot \overline 5 = \overline 1$ in $\mathbb Z_7$. Then $\overline x$ is denoted by $\overline a^{-1}$. If $\overline a$ has a multiplicative inverse, then $\overline a$ is called a unit in $\mathbb Z_n$. In our example, $\overline 3$ is a unit and $\overline 3^{-1} = \overline 5$ in $\mathbb Z_7$.
Find multiplicative inverses of units in $\mathbb Z_{12}$, $\mathbb Z_7$, $\mathbb Z_{20}$. Remember that units in $\mathbb Z_n$ are exactly those elements $a$ for which $\gcd(a,n) = 1$.
As you may have noticed…