Which elements of the ring #ZZ_14# are invertible?
1 Answer
Explanation:
Ring
A ring is a set
#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)
Invertible elements and zero divisors
An element
#a xx a^(-1) = a^(-1) xx a = 1#
An element
#a xx b = 0#
Note that any zero divisor is not invertible:
Suppose
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
#AA c in R, (c xx a) xx b = c xx (a xx b) = c xx 0 = 0#
Addition and multiplication of integers modulo
- Multiplication is commutative:
#AA a, b in R, a xx b = b xx a" "# (commutativity)
This property makes
However, note that
#2 xx 7 -= 0" "# modulo#14#
So we have seen that
Hence the following numbers are not invertible in
#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.