Conversation started Jun 5, 2012 at 14:51.
Matt N.
Matt N.
Jun 5, 2012 14:51
@tb Sorry, still confused. I don't think $d_1$ stays the same map after you remove $M$. In one case, $$\dots P_1 \xrightarrow{d_1} P_0 \to M \to 0$$ we don't necessarily have $im(d_1) = P_0$ but in the other case, $$\dots P_1 \xrightarrow{d_1} P_0 \to 0$$ we do. So how can I compute Ext using the wrong (exact!) sequence?
t.b.
t.b.
@MattN Hm... You just remove that $M$ from your complex. Nothing else changes, expecially not the cokernel of $d_1$. Only the first sequence is exact, the second isn't (at $P_0$).
Mark Dominus
Mark Dominus
"While being a teenager I for a limited amount of time was transformed into a superman with computer brain."
Eugene
Eugene
@Tim porton definitely satisfies a lot of criteria
and also this
Matt N.
Matt N.
@tb Noo. Third time I make the exact same mistake of forgetting that the other sequence isn't exact anymore. Thanks.
Mark Dominus
Mark Dominus
What I want is a list of early warnings signs that someone could use to see if they themselves might be turning into a crank.
Matt N.
Matt N.
Jun 5, 2012 14:54
Ah, since the maps don't change it's still exact in all other places.
Except $P_0$.
Ok. Getting there.
t.b.
t.b.
@MattN Yes. But not anymore after applying $\operatorname{Hom}({-},N)$ since that functor isn't exact (unless $N$ is injective).
Matt N.
Matt N.
@tb You mean unless $N$ is projective?
Tim
Tim
@MarkDominus Is that his quote too? Or what does that mean?
t.b.
t.b.
@MattN No, injective.
Matt N.
Matt N.
Ah, that's the other functor we didn't learn about.
Eugene
Eugene
Jun 5, 2012 14:57
@MarkDominus it's like an earthquake. there are no signs
Matt N.
Matt N.
Also, we don't know what an injective module is. Not so far, anyway.
t.b.
t.b.
@MattN Injective module, not functor. See fourth bullet point in the definition.
Matt N.
Matt N.
@tb Thanks for pointing this out.
@tb That's what I said.
J. M.
J. M.
@MarkDominus In lieu of that, may I suggest looking at Underwood Dudley's book for examples of what not to do? :)
t.b.
t.b.
@MattN I was referring to this.
Matt N.
Matt N.
Jun 5, 2012 14:58
@tb No you weren't : )
t.b.
t.b.
@MattN If $\operatorname{Hom}({-},N)$ were always exact, then the Ext functors would be pretty boring... :)
Matt N.
Matt N.
@tb Yes. : )
t.b.
t.b.
So, we've just proved that $\operatorname{Ext}^n(M,I) = 0$ for $n \geq 1$...
If $I$ is an injective module.
Gigili
Gigili
Eh, I was here but I wasn't here.
Mark Dominus
Mark Dominus
UD's book, like the links that Eugene posted, is directed at mathematicians who are trying to deal with cranks, not at people who want to know if they might be cranks.
Of course, a real crank is firmly convinced that he is not a crank.
But someone who is a semicrank might get the advice early enough to head off the problem.
Eugene
Eugene
Jun 5, 2012 15:02
@MarkDominus that is one of the signs.
Mark Dominus
Mark Dominus
What is?
Eugene
Eugene
a crank is convinced he is not a crank and all his peers are incorrect
Mark Dominus
Mark Dominus
Didn't I just say that?
Eugene
Eugene
@MarkDominus i know i was just highlighting it
of course einstein fulfilled the criteria of a crank initially
t.b.
t.b.
Can I prove that I'm not a crank by saying I'm firmly convinced that I am a crank?
Matt N.
Matt N.
Jun 5, 2012 15:04
@tb No. But we know that you're not so no need to prove anything.
t.b.
t.b.
@MattN Actually, I think this question might imply that I am a crank... Anyway. I should be going. See you guys later.
Tim
Tim
@tb Good proving skill
Matt N.
Matt N.
See you later!
Eugene
Eugene
@tb i don't think so
Mark Dominus
Mark Dominus
@tb It's not enough to say it; you have to actually believe it.
Eugene
Eugene
Jun 5, 2012 15:05
@tb bye!
Matt N.
Matt N.
Yes. Like the other Ext(P,N) thingies are equally boring if $P$ is projective in $Hom(P,N)$.
t.b.
t.b.
@MattN the other way around.
 
Conversation ended Jun 5, 2012 at 15:06.