Question on $\mbox{Ext}^1$

Question

I have 2 questions, one of them concerning the isomorphicity of quotient groups (rings), and the other is on $\mbox{Ext}^1$. It's pretty long, but somehow related to each other. So I just kinda put everything here, instead of dividing it into different posts.

Question 1

I know that given $A; B$ subgroups of an Abelian group $G$, then if $A \cong B$, then it's not necessarily true that $R/A \cong R/B$ (I can find a counter example of this). But is the reverse true? Like if $R/A \cong R/B$ then is true that $A \cong B$? I think that it's false, but I cannot find a counter example of it. :(

And moreover, how can I find counter examples of the above 2 statements in quotient rings? I've been searching for a counter example in common rings, like $\mathbb{Z}, \mathbb{Q}, \mathbb{Z}[x]$ but with no luck.

Question 2

So basically, I understand that, if you are trying to prove something of the line $X/X' \cong Y/Y'$, then showing that $X\cong X'$, and $Y \cong Y'$ is definitely not enough. Ok, here's my second question.

Given a ses of $R-$modules $0 \to M \xrightarrow{f} P \xrightarrow{g} X \to 0$, where $P$ is projective, and an $R-$module $Y$.

Here's a proof of $\mbox{Ext}^1 (X;Y) \cong \mbox{Hom}(M;Y)/\mbox{Im}(\mbox{Hom}(f;1_Y))$, that I don't really understand.

Firstly, construct a projective solution for $M$, i.e:

$$... \xrightarrow{\delta_2} P_1 \xrightarrow{\delta_1} P_0 \xrightarrow{\delta_0} M \to 0$$

Then, we can connect the above les with the ses as given to arrive at a projective resolution for $X$ as follow. $$... \xrightarrow{\delta_2} P_1 \xrightarrow{\delta_1} P_0 \xrightarrow{f\delta_0} P \xrightarrow{g} X \to 0$$

Now, drop the $X$ to obtain the following complex: $$... \xrightarrow{\delta_2} P_1 \xrightarrow{\delta_1} P_0 \xrightarrow{f\delta_0} P \to 0$$

Apply $\mbox{Hom}(-; Y)$ to the above complex to have yet another complex: $$... \xleftarrow{\mbox{Hom}(\delta_2; 1_Y)} \mbox{Hom}(P_1; Y) \xleftarrow{\mbox{Hom}(\delta_1; 1_Y)} \mbox{Hom}(P_0; Y) \xleftarrow{\mbox{Hom}(f\delta_0; 1_Y)} \mbox{Hom}(P; Y) \leftarrow 0$$

Now by definition, we have: $\mbox{Ext}^1(X;Y) = \mbox{ker} \ \mbox{Hom}(\delta_1; 1_Y) / \mbox{im} \ \mbox{Hom}(f \delta_0; 1_Y)$ (*)

---

Now, we'll try to express $\mbox{ker} \ \mbox{Hom}(\delta_1; 1_Y)$ in other terms.

Since $P_1 \xrightarrow{\delta_1} P_0 \xrightarrow{\delta_0} M \to 0$ is exact, apply $\mbox{Hom} (-; Y)$ to this sequence, to have an exact sequence: $\mbox{Hom}(P_1; Y)\xleftarrow{\mbox{Hom}(\delta_1; 1_Y)} \mbox{Hom}(P_0; Y) \xleftarrow{\mbox{Hom}(\delta_0; 1_Y)} \mbox{Hom}(M; Y) \leftarrow 0$, which means that: $\mbox{ker} \ \mbox{Hom}(\delta_1; 1_Y) = \mbox{im} \ \mbox{Hom}(\delta_0; 1_Y) \cong \mbox{Hom} (M; Y)$ (since $\mbox{Hom}(\delta_0; 1_Y)$ is injective).

---

And $\mbox{im} \ \mbox{Hom}(f \delta_0; 1_Y) = \mbox{Hom}(\delta_0; 1_Y)\left( \mbox{Hom}(f; 1_Y)(\mbox{Hom}(P; Y))\right) \cong \mbox{Im} \ \mbox{Hom}(f; 1_Y)$ (since $\mbox{Hom}(\delta_0; 1_Y)$ is injective).

---

And then, the author just kind of plug everything in (*), to have:

$$\mbox{Ext}^1(X;Y) = \mbox{ker} \ \mbox{Hom}(\delta_1; 1_Y) / \mbox{im} \ \mbox{Hom}(f \delta_0; 1_Y) \cong \mbox{Hom} (M; Y) / \mbox{Im} \ \mbox{Hom}(f; 1_Y)$$

This proof doesn't look good to me, since AFAIK we cannot prove isormorphicity of quotient modules like that. Or am I missing something here? If this proof is wrong, is there any way that I can fix it?

Thank you guys very much,

And have a good day, :*

Answer 1

Q1: Let $G = \mathbb{Z} = A$, $B = 2 \mathbb{Z} \cong \mathbb{Z}$. $G/A = 0$ but $G/B = \mathbb{Z}/2\mathbb{Z}$ (Edit: So this is an example that $A\cong B$ but $G/A \ncong G/B$, not what the OP was asking about (Thanks @Alex for pointing this out). I will leave this here for reference since I think it is a good example to keep in mind.)

Q2: Here is the way you want to prove this. Lok at the long exact sequence for $\text{Ext}^1$: Given a short exact sequence $0 \to M \to P \to X \to 0$ the Ext long exact sequence (the cohomology long exact sequence associated to the right derived functor $\text{Ext}^*(-,Y)$) is:

$$ 0 \to \text{Hom}(X,Y) \to \text{Hom}(P,Y) \stackrel{f^*}{\to} \text{Hom}(M,Y) \stackrel{\delta}{\to} \text{Ext}^1(X,Y) \to \text{Ext}^1(P,Y) = 0$$

Hence by exactness, $\text{Ext}^1(X,Y) \cong \text{Hom}(M,Y)/f^*(\text{Hom}(P,Y) )$ as required.

All of the stuff in the proof you provided kinda looks like material used to show that $\text{Ext}^*(-,Y)$ is a (cohomological) $\delta$-functor.

Answer 2

A counterexample for question 1: Let $R=\mathbb Z_4\times \mathbb Z_2, A=\mathbb Z_2\times \mathbb Z_2,B=\mathbb Z_4$. Then $R/A\cong \mathbb Z_2\cong R/B$, but clearly $A\not\cong B$. This holds with $R,A,B$ considered as rings or groups.