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10:00 PM
Yeah, the problem I posed is not exactly the problem I want to solve
 
you might look at how jordan canonical form works in the case of 2x2 matrices over Z. that is mostly about existence of C, but if you look at the proofs they may hint at where lack of uniqueness can arise.
 
I know there exists C s.t. the above is true, and I want to find $C'\in GL_2(Z)$ such that $B = C'AC'^{-1}$
Also you can assume $A\in Sl_2(Z)$ with distinct eigenvalues
 
Are there sufficient conditions for similarity to hold over a restricted ring?
I think I thought about this many decades ago.
 
Oh that's a good idea
 
What idea?! 🤷‍♂️🤷‍♂️
 
10:14 PM
HI
 
10:33 PM
Hi :)
I feel dumb. I have a 9 month review coming up for my PhD and, while I get the concepts behind them, I keep forgetting things.
I've had a lot of time off for my health.
It's scary stuff.
During my 1st attempt at a PhD, I had loads of time off for my health, then, too. I'm concerned that the same thing'll happen again here.
I think the amount of insights and eureka-moments I've had so far are thin on the ground compared to last time too.
I just haven't done as many exercises.
It's a beautiful area to study, much more structured than combinatorial group theory. But there's less experiments I can do on a computer when I'm stuck.
Linear algebraic groups, I mean.
I could rant for a while . . .
I still have COVID.
Also, I feel like I'm past it. I'm 32. I haven't got my PhD yet. I have only one published article. That puts me behind in academia, if I'm to have a career.
If I fail this attempt or get an MPhil, I'll try elsewhere. My psychiatrist said he knows people on their third PhD attempt. He encouraged me.
Sometimes it just doesn't work out, y'know?
@geocalc33 Hi :)
Mathematics can be cruel.
I'm not sure I even get what an algebraic variety is, as sad as it sounds, and it's fundamental!
Then again, algebraic geometry is notoriously difficult.
 
11:12 PM
Is anyone here?
Never mind. Good night, all :)
 
I'm not here
 
Is anyone really here?
Oops, be quiet, Shaun is back
 
Oh no, do I chase people away!? :D
 
It's just a quiet Saturday here.
 
You're all AFRAID OF ME! Mauhahaha!
 
11:23 PM
I will be taking my dog to the park in a few minutes.
That's my story, and I'm sticking to it.
 
What kind of dog have you got?
 
She is 3/8 AmStaff, 1/8 Shepherd, 1/8 Boston Terrier, 1/8 Rottie, and 1/4 mixed breeds.
She is black with brown highlights
and white spots on her paws and chest.
and the tip of her tail
Anyway. I'm off for a while.
 
Cool. See you later :)
 
Hello, Shaun.
 
Hi @noballpointpen :)
How are you?
 
11:28 PM
Doing logic exercises. Completed section on completeness theorems.
Good in terms of life-health-etc.
 
That sounds fun :)
@noballpointpen Excellent :)
 
There was a weird exercise and its statement seems to come directly from one lemma present. Do not know why they put it.
 
Sometimes you see things; sometimes you don't. Maybe you just understood the lemma better than anticipated.
 
Do you agree on your end?
Lemma: if a theory has a model of cardinality $m$ then it has one with cardinality $\geq$ than $m$.
Ex: there is no a theory $K$ whose models have finite domain.
Sorry to hear your struggles with PhD.
 
I can't comment. I haven't studied logic properly on a while. And thank you.
I signed up for a challenge. If it was easy, I'd be disappointed.
But still: I could be doing better and I'm concerned.
Maybe if you explain why it's trivial, I could help, @noballpointpen :)
 
11:40 PM
Mental health problems put huge weight on shoulders of any person having it, so it is understandably. I would be proud for myself if I worked 9+ hours on math in my past (like you in your undergrad years).

Ok. If such a theory existed, then we could have some model whose domain is infinite. If the statement is correct (the book had at least one typo, I even posted a quesiton here about one), and I understood it -- that it has _only_ models with finite domain. But I am suspicious that it is not what is meant.
 
Thank you, @noballpointpen :)
That does sound like the "only" is intended.
 
"Prove that there is no theory $K$ whose models are exactly the interpretations with finite domains."
This is the exact statement.
 
Yes, the "exactly" does the same work as "only" in your restatement.
 
Good, then :)
 
I should get some sleep. G'night :)
 
11:48 PM
Good night, Shaun.
 
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