I received an email asking about Zn and proposition 4.14 and 4.15.
In a previous reply I explained Zn , to refresh our memory, it is the set of
all the equivalence classes induced by the equivalence relation ∼n .
For example Z3 = {[0], [1], [2]}. Z4 = {[0], [1], [2], [3]}
You have seen on a problem sheet or in your notes, that you add and multiply
equivalence classes in the way you would expect.
[a] · [b] = [a · b] and also [a] + [b] = [a + b].
For example in Z4 [1] + [2] = [3] , [2] + [2] = [0] , [2][3] = [2] (because 6 gives
remainder 2 when you divide by 4).
In one of the problem sheets you have to write down the multiplication and
addition tables, modulo certain numbers. It will be useful to look at these tables
for the following discussion.
Propositions 4.14 and 4.15 are talking about multiplicative inverses in Zn .
For an equivalence class [a] , a multiplicative inverse is a class [b] so that
[a][b] = [1] and [b][a] = 1.
For example in Z3 the multiplicative inverse of [1] is [1] since [1][1] = [1 · 1] =
[1] .
The multiplicative inverse of [2] is [2] since [2] · [2] = [4] = [1] .
[0] does not have a multiplicative inverse.
Notice in Z3 all of the non zero equivalence classes have a multiplicative
inverse.
Is this always the case? NO
Consider Z4 , [2] does not have a multiplicative inverse since:
[1][2] = [2], [2][2] = [0] , [3][2] = [2]
In light of this discussion have a look at the multiplication tables on the past
problem sheet.
Proposition 4.14 tells us when a multiplicative inverse for [x] exists, if and
only if x and n are coprime.
Does this match our previous ndings? YES. What happens in Z3 ? The
integers 1 and 2 are coprime with 3, and their equivalence classes [1] and [2] have
multiplicative inverses. In Z4 we saw [2] does not have a multiplicative inverse,
which is what we expect by the proposition since 2 and 4 are not coprime.
Remember the Euler ϕ function? It tells us how many numbers are coprime
to n. So, based on what we just said, ϕ(n) tells us how many elements of Zn
have a multiplicative inverse.
Later on you will see that Zp is a eld if and only p is a prime. Again in
light of what we have just said, this seems reasonable, as to be a eld every non
zero element needs to have a multiplicative inverse, and when p is prime all of
the numbers 1, ..., p − 1 are coprime with p.
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