1. ## Non-isomorphic integer subrings

Is the reason that $n\mathbb{Z} \not\cong m\mathbb{Z}$ in general because if we define a map of the type $x \mapsto ax$, the inverse $ax \mapsto \frac{1}{a}ax$ doesn't exist, since $\dfrac{1}{a}$ isn't in $n\mathbb{Z}$ or $m\mathbb{Z}$? (the text doesn't require a ring to have unity)

I have a follow-up question depending on the answer. Thanks.

(by the way, I am new to this forum and read the rules, but if there's anything I should be aware of, please let me know)

2. suppose φ:nZ-->mZ is a ring isomorphism. then φ(n) = mk, for some integer k.

since φ is a ring homomorphism, φ(n^2) = φ(n+n+...+n) (n times)

= φ(n)+φ(n)+...+φ(n) (n times)

= n(φ(n)) = n(mk). but φ(n^2) = [φ(n)]^2 = (mk)^2 as well, leading to n = mk. in general, this is not true.

in fact, you can even show that mZ and (km)Z are not isomorphic (unless k = ±1), in much the same way:

φ(m) = r(km) = (rk)m for some integer r. so φ(m^2) = (rkm)^2 = r^2k^2m^2, but also φ(m^2) = m(φ(m)) = (rk)m^2.

thus rk = 1, which means that k = ±1.

3. Understood, thanks. I asked that question because I'm trying to figure out how to go about finding all homomorphisms $\mathbb{Z} \to \mathbb{Z}$. I wanted to know if I can rule out any map of the form $x \mapsto ax$. I believe the only homomorphism is the identity...is this correct?

EDIT: I wasn't thinking of what I had already decided when I posted that. My reasoning is that if $\phi$ is of the form $\phi (x) = x+a$ then 0 isn't preserved, and if $\phi$ is of the form $\phi (x) = ax, a \ne 1$ then 1 isn't preserved, so the only homomorphism of these types is $\phi (x) = x$.

4. if you stop and think about it, any ring homomorphism φ:Z-->Z is completely determined by its value at 1: φ(k) = kφ(1).

suppose φ(1) = n. then n^2 = φ(1)φ(1) = φ((1)(1)) = φ(1) = n, so n^2 - n = n(n-1) = 0.

hence n = 0 or 1 (we ruled out the 0 case above implicitly because we were only considering isomorphisms).

this gives us precisely two homomorphisms Z-->Z, the identity, and the 0-map.

5. Got it. I was uncertain about finding all homomorphisms because I was questioning what one is 'allowed' to use as a homomorphism. I understand the above argument, but how does it rule out that there are no homomorphisms that do not involve a ring operation and ring element? I don't doubt that there are no others for the integers because of the simple structure, but how do we know they don't exist?

6. re-read my post. φ(1) HAS to be 0 or 1. if φ(1) = 0, every element gets mapped to 0: φ(2) = φ(1+1) = φ(1) + φ(1) = 0 + 0 = 0, etc.

if φ(1) = 1, every element has to be mapped to itself: φ(2) = φ(1) + φ(1) = 1 + 1 = 2, etc.

if we require that rings have unity, this leaves just the identity map. in fact, if R is any ring with unity, there is exactly one ring homomorphism Z-->R,

namely, the homomorphism that sends 1 to 1_R.

one is "allowed" to use any map that satisfies φ(r+s) = φ(r) + φ(s); φ(rs) = φ(r)φ(s). the fact that the identity map and the 0-map are the only ones that

qualify for the ring Z, is a combination of the properties of the specific ring Z, and the properties a ring homomorphism must have.

usually, translations: additive maps of the form f(x) = x + a, are not morphisms in additive structures, precisely because they don't preserve 0.

sometimes dilations (also sometimes called translations): maps of the form f(x) = ax are morphisms, often this depends on what "a" is, and what

kind of structure you're considering. Z has a unique type of structure, which is based on "adding 1 over and over". one way of expressing this is by

saying Z is a minimal ring of characteristic 0. the behavior of 1 completely determines the behavior of Z (this is what lets us do proofs by induction, for example).

this isn't true of rings in general, in a polynomial ring, for example, the behavior of p(x) is determined by the behavior of 1 and x.

7. ok, I understand now, thank you.