# Thread: Units of integers mod m

1. ## Units of integers mod m

Show that U(I_m) = {[k] in I_m : (k,m) = 1}.

I_m is integers mod m.

I know that if m is prime then the result is obvious. But do i need to break it into cases.???

2. ## Re: Units of integers mod m

How is the case (m prime) obvious?

Why not use set inclusion, and that (m, k) = 1 iff am + bk = 1 for some a, b...

3. ## Re: Units of integers mod m

Should I assume (m,k)= 1 to try to show that [k] is in I_m?

4. ## Re: Units of integers mod m

I think I got it. Say (k,m)= 1 then there exist a, b that ak+bm =1 I can put bm on the other side to get

ak-1=bm. Now we know that m| ak -1 and so ak is congruent to 1mod m. That proves [k] is in I_m. further more ak is congruent to 1 mod m therefore there exist a c such that kc =1 and that is the definition of a unit.

Did I miss something?

5. ## Re: Units of integers mod m

a couple of things:

[k] is, by definition, in I_m. what you are trying to prove is that [k] is in U(I_m). so you don't want a c such that kc = 1 (unless k =1, no such integer exists), you want a [c] such that [c][k] = [1].

now you have shown that [ak] = [1], so you're almost there. the [c] that you're looking for is clearly [a]. what you haven't done, is show if 0 < c < m such that [c] = [a], that (c,m) = 1.

6. ## Re: Units of integers mod m

I have this piace of a proof from Rotman advanced algebra book. I dont understand how or what it means by solve of [x].

If (a,m) = 1, then [a][x] = [1] can be solved for
[x] in Im
. Now(x,m) = 1, for rx + sm = 1 for some integer s, and so Proposition 1.13
on page 5 gives (x,m) = 1; therefore, [x] ∈ U(Im),

7. ## Re: Units of integers mod m

we are given that (k,m) = 1 by the definition of U(I_m).

so (by the defintion of gcd) we have a,b in Z such that ak + bm = 1.

if two integers are equal, then they remain equal mod m, so

[ak + bm] = [1]
[a][k] + [b][m] = [1]
[a][k] + [b][0] = [1]
[a][k] = [1].

now we just take the integer c, where 0 < c < m, so that [c] = [a] (in other words, reduce the integer a used in the gcd formula mod m).

to give a concrete example, suppose m = 6 and k = 5. (5,6) = 1, so [5] is in U(6). note that 5(-7) + 6(6) = 1.

so we can take a = -7. this means that [5][-7] = [1] (mod 6). but -1 isn't in the right range, so we add multiples of 6 until it is:

-7 + 2(6) = 5. so [-1] = [5], so [5] is its own inverse in U(6). and indeed, [5][5] = [25] = [1], since 25 = 1 mod 6.

well, here, we obviously have (5,6) = 1. ok, back to the general case:

we know that [a] ( = [c]) comes from ak + bm = 1, so it's trivial that (a,m) = 1 (just switch the role of a and k).

what's not so trivial, and what your proof from Rotman doesn't state explicitly, is that , in fact, (c,m) = 1.

but this isn't that much extra difficulty: [c] = [a], so c = a + tm, for some integer t.

therefore ak + bm = 1 --> (c - tm)k + bm = 1, and re-arranging slightly:

kc + (b - tk)m = 1, so since there exists integers k, b - tk such that kc + (b - tk)m = 1, (c,m) = 1.