Ext

Conversation started Jun 5, 2012 at 13:34.

t.b.

Jun 5, 2012 13:34

waiting for a question

Matt N.

So if I get asked what Ext is in the exam and I say it's the functor mapping an $R$-module $M$ to a homology group $Ext^n_R (M, N)$ as follows:
Take a projective resolution of $M$:
$$\dots P_1 \to P_0 \to M \to 0$$
chop off $M$ to get a chain complex
$$\dots P_1 \xrightarrow{d_1} P_0 \xrightarrow{d_0} 0$$ then apply $Hom(-,N)$ to it to get
$$ 0 \xrightarrow{\overline{d_0}} Hom(P_0, N) \xrightarrow{\overline{d_1}} Hom(P_1, N) \dots $$
and define $$ Ext^n_R(M,N) := ker(\overline{d_n}) / im(\overline{d_{n-1}})$$

Fortuon Paendrag

Hello All!

Matt N.

Also: How do I ever manage to remember what the indices are in the RHS of $Ext^n$?

t.b.

@MattN oral or written?

Matt N.

Oral :,(

Unfortunately.

t.b.

Jun 5, 2012 13:37

First thing to say: the right derived functor of Hom.

(in either variable)

Matt N.

He never used the word functor or derived anywhere in the notes.

And he only defined it for $Hom(-,N)$, not for $Hom(M,-)$.

He didn't give it a name either.

t.b.

Its name is Ext

Matt N.

Or extension functor?

Or perhaps, Ext group?

Anyway, is what I wrote above more or less correct?

The definition in the notes I copied doesn't contain much information.

t.b.

Yes, that's correct. I wouldn't worry about the indices and stuff. You take a projective resolution of $M$, apply Hom and take homology.

Matt N.

I'm terrified about messing up the indices.

But ok.

Ah but wait.

The reason why I'm so worried about the indices is that I wanted to "see" how $Ext^0 (M,N) = Hom(M,N)$.

And then I failed.

Let me post my failed attempt:

t.b.

Jun 5, 2012 13:43

The thing is that $H^0(\operatorname{Hom}(P_\bullet,N)) = \operatorname{Hom}(M,N)$ because $\operatorname{Hom}({-},N)$ is left exact. That is, it sends cokernels to kernels.

Matt N.

But...:

t.b.

...what?

Matt N.

$$ Ext^0 (M,N) = ker(\overline{d_0}) / im (\overline{d_{-1}}) = ker(\overline{d_0}) = ?$$

I don't see what this kernel is.

t.b.

You forgot to apply Hom

Matt N.

No, hence the overline.

Gigili

Jun 5, 2012 13:45

@leo: I said?

t.b.

Oh, that's implicit in $\overline{d_0}$.

Matt N.

Yes.

Doh.

I messed up the sequence, forgot to chop off $M$:

t.b.

You have a right exact sequence $P_1 \to P_0 \to M$.

Matt N.

$d_0 : P_0 \to 0$ is the zero map.

So the kernel of $\overline{d_0}$ is all of $Hom(M,N)$.

Thank you!

t.b.

@MattN Wait.

Gigili

Jun 5, 2012 13:47

Hi, Matt.

Matt N.

Don't know why I had to ask a question in order to notice that I'd made a mistake.

Gigili

Still snake lemma?

Or is it worm?

Matt N.

Hi Gigili. No, Ext functor today.

t.b.

@MattN the $d_0$ being the zero map is not the point.

(unless I misread what you said)

Matt N.

No, you read right. It doesn't seem to make sense what I wrote since $M$ is not in the chain complex.

t.b.

Jun 5, 2012 13:49

exactly.

Look at the right exact sequence $P_1 \xrightarrow{f} P_0 \xrightarrow{g} M \to 0$.

Matt N.

Then chop off $M$?

t.b.

No. Apply $\operatorname{Hom}({-},N)$ to it.

Matt N.

But that's not how Ext is defined.

t.b.

Wait!

Matt N.

You get:

t.b.

Jun 5, 2012 13:52

Get an exact sequence $0 \to \operatorname{Hom}(M,N) \xrightarrow{f^\ast} \operatorname{Hom}(P_0,N) \xrightarrow{g^\ast} \operatorname{Hom}(P_1,N)$.

Matt N.

Yes.

Eugene

hi all

t.b.

So $\operatorname{Hom}(M,N)$ is the kernel of $g^\ast = \overline{d_1}$

Matt N.

But Ext of an exact sequence would all be zero.

J. M.

@Eugene That was quick...

Eugene

Jun 5, 2012 13:53

@JM = )

t.b.

@MattN We're not taking Ext of that. I just identified the kernel of $\overline{d_1}$.

Matt N.

@tb Ok.

t.b.

Now you can remember that you chopped off $M$ and that you wanted to compute $\operatorname{Ker}\overline{d_1}$.

Matt N.

No. It was $Ker \overline{d_0}$ that I wanted. Maybe I missed up indices again.

t.b.

Giving you the desired identification $\operatorname{Ext}^0(M,N) = \operatorname{Hom}(M,N)$.

@MattN you're interested in the entire homology of the complex $\operatorname{Hom}(P_\bullet,N)$. The $0$th homology is the kernel of what you called $\overline{d_1}$ and that's what we computed before.

Matt N.

Jun 5, 2012 13:58

But $$ Ext^k_R (M,N) := Ker \overline{d_k} / Im \overline{d_{k-1}}$$
No?

t.b.

Again: forget the indices of the maps. The interesting part of the complex starts at $\operatorname{Hom}(P_0,N) \to \operatorname{Hom}(P_1,N) \to \cdots$ and it's the kernel of that first map you're interested in because that gives the first nonzero homology.

I think in your notation that should be $\overline{d_1}$.

Matt N.

@tb Yes.

t.b.

If that's right then you should shift the indices the other way: $$\operatorname{Ext}^k (M,N) = \operatorname{Ker}\overline{d_{k+1}}/\operatorname{Im}\overline{d_{k}}$$