Which elements of the ring #ZZ_14# are invertible?

1 Answer
Nov 12, 2016

Answer:

#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.