# Which elements of the ring ZZ_14 are invertible?

Nov 12, 2016

$1 , 3 , 5 , 9 , 11 , 13$

#### Explanation:

${\mathbb{Z}}_{14}$, otherwise known as $\mathbb{Z} \text{/} 14$ or $\mathbb{Z} \text{/} 14 \mathbb{Z}$ is the ring of integers modulo $14$.

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Ring

A ring is a set $R$ equipped with addition $+$ and multiplication $\times$ satisfying the following properties:

• $R$ is an abelian group under addition:

$\forall a , b , c \in R , \left(a + b\right) + c = a + \left(b + c\right) \text{ }$ (associativity)

$\forall a , b \in R , a + b = b + a \text{ }$ (commutativity)

$\exists 0 \in R : \forall a \in R , a + 0 = 0 + a = a \text{ }$ (identity)

$\forall a \in R , \exists \left(- a\right) \in R : a + \left(- a\right) = \left(- a\right) + a = 0 \text{ }$ (inverse)

• The non-zero elements of $R$ are a monoid under multiplication:

$\forall a , b , c \in R \text{\" {0}, (axxb)xxc = axx(bxxc)" }$ (associativity)

$\exists 1 \in R : \forall a \in R \text{\" {0}, a xx 1 = 1 xx a = a" }$ (identity)

• Multiplication is (left and right) distributive over addition:

$\forall a , b , c \in R , a \times \left(b + c\right) = \left(a \times b\right) + \left(a \times c\right) \text{ }$ (left distributivity)

$\forall a , b , c \in R , \left(a + b\right) \times c = \left(a \times c\right) + \left(b \times c\right) \text{ }$ (right distributivity)

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Invertible elements and zero divisors

An element $a \in R$ is invertible if there is some element ${a}^{- 1} \in R$ such that:

$a \times {a}^{- 1} = {a}^{- 1} \times a = 1$

An element $a \in R$ is a zero divisor if there is some element $b \ne 0$ in $R$ such that:

$a \times b = 0$

Note that any zero divisor is not invertible:

Suppose $a$ has a multiplicative inverse ${a}^{- 1}$ and $a \times b = 0$.

Then:

$b = 1 \times b = \left({a}^{- 1} \times a\right) \times b = {a}^{- 1} \times \left(a \times b\right) = {a}^{- 1} \times 0 = 0$

Any multiple of a zero divisor is also a zero divisor:

If $a \times b = 0$ then:

$\forall c \in R , \left(c \times a\right) \times b = c \times \left(a \times b\right) = c \times 0 = 0$

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$\boldsymbol{{\mathbb{Z}}_{14}}$ as a Commutative Ring

Addition and multiplication of integers modulo $14$ satisfy these properties. In addition it satisfies:

• Multiplication is commutative:

$\forall a , b \in R , a \times b = b \times a \text{ }$ (commutativity)

This property makes ${\mathbb{Z}}_{14}$ a commutative ring

However, note that ${\mathbb{Z}}_{14} \text{\} \left\{0\right\}$ is not a group under multiplication. It has zero divisors: pairs of non-zero elements which multiply to give $0$. For example:

$2 \times 7 \equiv 0 \text{ }$ modulo $14$

So we have seen that $2$ and $7$ are zero divisors. So any multiples will also be zero divisors and therefore not be invertible.

Hence the following numbers are not invertible in ${\mathbb{Z}}_{14}$:

$0 , 2 , 4 , 6 , 7 , 8 , 10 , 12$

That leaves:

$1 , 3 , 5 , 9 , 11 , 13$

To confirm, note that:

$3 \times 5 = 15 \equiv 1 \text{ }$ modulo $14$

$9 \times 11 = 99 \equiv 1 \text{ }$ modulo $14$

$13 \times 13 = 169 \equiv 1 \text{ }$ modulo $14$

So these elements are all invertible.