Question

Which elements of the ring `ZZ_14` are invertible?

Answer 1

`1, 3, 5, 9, 11, 13`

Explanation:

`ZZ_14`, otherwise known as `ZZ"/"14` or `ZZ"/"14ZZ` is the ring of integers modulo `14`.

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Ring

A ring is a set `R` equipped with addition `+` and multiplication `xx` satisfying the following properties:

• `R` is an abelian group under addition:

`AA a, b, c in R, (a+b)+c = a+(b+c)" "` (associativity)

`AA a, b in R, a+b = b+a" "` (commutativity)

`EE 0 in R : AA a in R, a + 0 = 0 + a = a" "` (identity)

`AA a in R, EE (-a) in R : a + (-a) = (-a) + a = 0" "` (inverse)

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

`AA a, b, c in R "\" {0}, (axxb)xxc = axx(bxxc)" "` (associativity)

`EE 1 in R : AA a in R "\" {0}, a xx 1 = 1 xx a = a" "` (identity)

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

`AA a, b, c in R, a xx (b+c) = (a xx b) + (a xx c)" "` (left distributivity)

`AA a, b, c in R, (a+b) xx c = (a xx c) + (b xx c)" "` (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 xx a^(-1) = a^(-1) xx a = 1`

An element `a in R` is a zero divisor if there is some element `b != 0` in `R` such that:

`a xx b = 0`

Note that any zero divisor is not invertible:

Suppose `a` has a multiplicative inverse `a^(-1)` and `a xx b = 0`.

Then:

`b = 1 xx b = (a^(-1) xx a) xx b = a^(-1) xx (a xx b) = a^(-1) xx 0 = 0`

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

If `a xx b = 0` then:

`AA c in R, (c xx a) xx b = c xx (a xx b) = c xx 0 = 0`

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`bb(ZZ_14)` as a Commutative Ring

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

• Multiplication is commutative:

`AA a, b in R, a xx b = b xx a" "` (commutativity)

This property makes `ZZ_14` a commutative ring

However, note that `ZZ_14 "\" { 0 }` is not a group under multiplication. It has zero divisors: pairs of non-zero elements which multiply to give `0`. For example:

`2 xx 7 -= 0" "` 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 `ZZ_14`:

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

That leaves:

`1, 3, 5, 9, 11, 13`

To confirm, note that:

`3 xx 5 = 15 -= 1" "` modulo `14`

`9 xx 11 = 99 -= 1" "` modulo `14`

`13 xx 13 = 169 -= 1" "` modulo `14`

So these elements are all invertible.