From a short exact sequence $\mathbb{Z}_k\cong \mathbb{Z}/ \mathbb{Z}$, $k>1$?

Question

I just can't get it. In a short exact sequence $$0\rightarrow A \mathrel{\mathop{\rightarrow}^{\mathrm{\varphi}}} B \mathrel{\mathop{\rightarrow}^{\mathrm{\psi}}} C\rightarrow 0$$ am I right at considering both

$$C\cong B/A$$ $$C\cong B/Im(\varphi)$$

for $A,B,$ and $C$ abelian groups and $A,B,$ and $C$ modules over an arbitrary ring?

I know the 2nd one is true based on the first isomorphism theorem, but then I only see the second one being true when $\varphi$ is an isomorphism...
And further in the case of: $$0\rightarrow \mathbb{Z} \mathrel{\mathop{\rightarrow}^{\mathrm{\cdot k}}} \mathbb{Z} \mathrel{\mathop{\rightarrow}^{\mathrm{mod\,k}}} \mathbb{Z}_k\rightarrow 0$$ the above will imply:

$$\mathbb{Z}_k \cong \mathbb{Z}/ \mathbb{Z}$$ $$ \mathbb{Z}_k \cong \mathbb{Z}/ k\mathbb{Z}$$

is that right?
Thank you in advance!

Answer 1

Since $\phi$ is injective, certainly $A\simeq \operatorname{Im}(\phi)$. Note, though, that when you write $B/A$ you are identifying $A$ with a submodule of $B$ (otherwise it wouldn't even make sense). In the case of $$ 0 \to \Bbb Z \xrightarrow{\cdot k} \Bbb Z \xrightarrow{\operatorname{mod} k} \Bbb Z/k\Bbb Z \to 0 $$ you get $C \simeq \Bbb Z/k\Bbb Z$, with $k\Bbb Z\simeq \Bbb Z$. Indeed, $k \Bbb Z$ is a principal ideal for every $k\in \Bbb Z$ and taking the integers $\operatorname{mod} k$ is equivalent to take the quotient of $\Bbb Z$ by $k\Bbb Z$ (think about it, as a good exercise).

As a side note, consider avoiding the notation $\Bbb Z_k$ for the integers $\operatorname{mod} k$ whenever possible, since it is commonly used for $k$-adic integers when $k$ is prime.

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Edit: A short exact sequence $$ 0 \to A \xrightarrow{\varphi} B \xrightarrow{\psi} C \to 0 $$ gives you that $C\simeq B/\varphi(A)$. You can simplify the notation and write $B/A$, but only if you silently assume that you are identifying $A$ and $\varphi(A)$ through $\varphi$.

By saying that $k\Bbb Z=(k)\simeq(1)=\Bbb Z$ we are just saying that they have the same structures as modules over $\Bbb Z$. The only thing that we are measuring is an internal property of those two modules. They both happen to be ideal in $\Bbb Z$, though, so we can take quotients. But in doing so we are using an external property, namely the density of their elements in $\Bbb Z$, i.e. their "position" with respect to the other elements.

If you think of taking a quotient of a ring as "nullifying" the elements of one of its ideals, then you can see that, as sets, $\Bbb Z/\Bbb Z=\{0\}$, while $\Bbb Z/k\Bbb Z=\{0,1,\dotsc,k-1\}$. This is a nice example, as you noticed yourself, that $A\simeq A'$ doesn't necessarily imply $B/A\simeq B/A'$.

Answer 2

In a exact sequence, we have that $\ker(\psi) = \operatorname{Im}(\varphi)$. And we have by exactness that $\varphi$ is injective, so $\operatorname{Im}(\varphi) \cong A$, so that gets us both of them.

Something good to think about also is that taking all sorts of these quotients makes sense (i.e., we stay in the category of abelian groups or the category of modules) is because we are dealing with so called Abelian categories.

Edit: With regards to your second question about the specific example, but also in general (I don't have a way with words)... You are jumping ahead too much. We just have that $C/\operatorname{Im}(\varphi)$ for that specific map $\varphi$.