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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!

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Well, when $\varphi$ is an isomorphism you can easily prove that $C=0$. Anyway, notice that the category of Abelian groups IS a category of modules. Furthermore, the three isomorphism theorems have their generalizations for modules over arbitrary rings (and also for abelian categories)-- – Simone Apr 14 '13 at 7:21

2 Answers 2

up vote 3 down vote accepted

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.

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 $B/A\simeq B/A'$.

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Thank you. so in conclusion why do $C \simeq \Bbb Z/k\Bbb Z$ and $k\Bbb Z\simeq \Bbb Z$ do not imply $C \simeq \Bbb Z/\Bbb Z$? – dktr.k1 Apr 15 '13 at 2:32
I tried to clarify my point. – A.P. Apr 15 '13 at 7:03
Thanx a lot! :) – dktr.k1 Apr 15 '13 at 9:09

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 $\varphi$.

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hmm.. as a basic example consider $k>1 \in \mathbb{N}$ and the sequence $0\rightarrow \mathbb{Z} \mathrel{\mathop{\rightarrow}^{\mathrm{\cdot k}}} \mathbb{Z} \mathrel{\mathop{\rightarrow}^{\mathrm{modk}}} \mathbb{Z}_k\rightarrow 0$. This sequence is a s.e.s., the function: $\cdot k$ is injective, $modk$ is surjective, $Im(\cdot k)=ker(mod k)$and, and thus $\mathbb{Z}_k \cong \mathbb{Z} / 2\mathbb{Z} \cong \mathbb{Z} / \mathbb{Z}$? – dktr.k1 Apr 14 '13 at 7:23
Ok, so for the even numbers ($k = 2$) each $m$ can be written as $m = 2l$, and $l$ can be any old integer. So we can identify $\mathbb{Z}$ with its image under the multiplication map in this way. Is that what you were getting at? – AlexM Apr 14 '13 at 7:29
kind of...if we have that sequence then both $\mathbb{Z}_k \cong \mathbb{Z}/ k\mathbb{Z}$ and $\mathbb{Z}_k \cong \mathbb{Z}/ \mathbb{Z}$ are true? – dktr.k1 Apr 14 '13 at 7:32
No quite, but this is good to hash out! The image of the multiplication map is $k\mathbb{Z}, not $\mathbb{Z}$. If a module is large enough, you can embed it into itself as a submodule, as you found. Infinity is weird like that. – AlexM Apr 14 '13 at 7:39
A good exercise to think about now is that any finite integral domain is a field. – AlexM Apr 14 '13 at 7:39

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