Math 542 Day 3 follow-up
Where is “relatively prime” used in the proof that Znm is isomorphic to Zn x Zm if m
and are relatively prime?
Answer: It is in the proof that these are isomorphic as groups. Your probably thinking
that is a little lame as an answer since that was a long time ago (Section 11). Also that
proof was kind of clever and actually didn’t prove 1-1 or onto directly but rather proved
that Zn x Zm was a cyclic group of order nm generated by (1,1).
So let’s see what happens if we try to directly prove 1-1 right now. Recall that the map
we are using is defined by ϕ(k) = (k, k). Now I said that it seems clear that if ϕ(k) = ϕ(l),
then we have (k, k) = (l, l) and this ought to mean that k = l. Let’s see how bad I abused
the term “clear”.
The first problem is that we know that if we try this from Z4 to Z2 x Z2 we see that this is
a big lie. ϕ(0) = ϕ(2) since (0, 0) is the same as (2, 2) in Z2 x Z2. But of course 0 is not
the same as 2 in Z4. So it is in fact not 1-1. So wouldn’t we all feel better if we saw how
the relatively prime assumption can be used prove 1-1? Of course we would!
So let’s go back and see if we can actually prove that in the relatively prime case (m and
n are relatively prime) (k, k) = (l, l) in Zn x Zm actually does imply that k = l in Znm.
We do know that (k, k) = (l, l) in Zn x Zm implies that k ≡ l mod m and k = l mod n. We
want do deduce from this that k = l mod mn. So we have that k = l + pm for some p in Z
and k = l + qn for some q in Z.
I seem to remember that one can always write the GCD of two integers as a linear
combination of the two integers. This means that there exists integers r and s such that rm
+ sn = 1. Perhaps this will be useful?
Well this tells me that k = krm + ksn. (*)
[I just figured I should work k into this equation and hoped it would help.]
Then if I use my two equations k = l + pm and k = l + qn with my (*) equation, I can get l
involved as well and hope for a miracle. But how do I use these equations?
[Really when I tried this I was just hoping it would work out, knowing that the rm + sn =
1 equation is usually the trick for this sort of thing. But I knew I needed to get mn in there
somewhere, so I decided to substitute l + qn for the first k on the right hand side of the
(*) equation and l + pm for the second k.]
The aforementioned substitutions give me k = (l + qn)rm + (l + pm)sn. And behold when
I multiply the desired miracle occurs:
k = (l + qn)rm + (l + pm)sn = lrm + qnrm + lsn + pmsn = l(rm + sn) + mn(qr + ps).
Now since rm + sn = 1 we have k = l + mn(qr + ps). Which means that k and l are
equivalent mod mn.
So if m and n are relatively prime, it is “clear” that the defined mapping is 1-1. So it
appears that the abuse of the term “clear” is pretty severe in this case!
The proof (that the map from Znm to Zn x Zm defined by ϕ(k) = (k, k) is 1-1) without the
color commentary and with some calculations suppressed:
Suppose that ϕ(k) = ϕ(l), then (k, k) = (l, l) in Zn x Zm. Thus we have that k ≡ l mod m and
k = l mod n. This means that there exists integers p and q in Z such that k = l + pm and k
= l + qn.
Now since m and n are relatively prime their GCD is 1 so there are integers r and s
such that rm + sn = 1. Then we have k = krm + ksn.
Substituting we get k = (l + qn)rm + (l + pm)sn. Multiplying and substituting 1 for rm +
sn = 1 we get k = l + mn(qr + ps). Which means that k and l equal in Znm.
So the map is 1-1.
So now we could use the pigeonhole principle and conclude that the map was onto (since
the two rings are the same size). But this is still somewhat unsatisfying because I really
want to know why the map is onto in the case where m and n are relatively prime.
Basically I want to know why everything in Zn x Zm can be expressed as something that
looks like (k, k).
So for my next trick I will directly prove that the map is onto. Here goes.
Let (a, b) be an element of Zn x Zm. We want to show that there exists an element in Znm
that maps to (a, b). Since the map is defined by ϕ(k) = (k, k), we are looking for an
integer k that is equivalent to a mod n and equivalent to b mod m.
Consider the integers a, a + n, a + 2n, … , a + (m-1)n.
The next one on the list would be a + mn, which would be equivalent to a mod m.
However, I claim all the numbers in this list are actually non-equivalent mod m. Note that
the are all equivalent to a mod n.
Proof of my claim: Suppose that a + pn ≡ a + qn mod m, where n and m are different and
both less than m and non-negative (in other words they are two different guys from the
list). Well then a + pn – (a + qn) = dm for some integer d. Simplifying we se that (p-q)n =
dm. This means that m divides (p-q)n. Now since m and n are relatively prime, none of
the factors of m are factors of n, this means that all of the factors of m must be factors of
p-q. This means that m must divide p-q. Of course that is just plain silly since p-q is less
than m!
So our list a, a + n, a + 2n, … , a + (m-1)n contains m different integers that belong in m
different congruence classes mod m. Of course there are only m congruence classes mod
m, so this list has a representative from every congruence class mod m.
Ha! So something in this list must be equivalent to b mod m. So there exists an integer j
(between 0 and m-1) such that a + jn ≡ b mod m. But of course we also have a + jn ≡ a
mod n. So the element we are looking for is a + jn!
So there you have it. We know where the relatively prime condition is needed to give us
onto and where it is needed to give us 1-1. And that’s all I have to say about that.
Oh, and note that I also proved #52 from Section 18 directly.