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21:26
@BalarkaSen What exactly does the center of $\Bbb H$ look like?
quaternions?
From a ring point of view
Z(H)=R
Oh I messed up
That was quite simple
I just proved that the center of a division ring is a field
So the center of H must be a field
There are two fields in H
C and R
C is not the field you are looking for
so it is R
there are infinitely many copies of C inside H btw
also, Q is a field in H. so is every other subfield of any copy of C (or of the copy of R)
21:32
hey @Chris'ssistheartist wats up today
But I can show that C is not in Z(R)
@Agawa001 Hey! Struggling with math problems and allergy. :-) You?
it'd be easier to just show anything not in R is not in Z(H) directly
hmm true
@Chris'ssistheartist same struggle we are cursed with (apart allergy)
21:34
Thank you @anon
any quaternion looks like a+bu where u is a pure imaginary quaternion. pick any pure imaginary quaternion v perpendicular to u, so that uv=-vu, which means a+bu does not commute with v
@Chris'ssistheartist i have allergy to dust but its not its time to show off yet
Z(H) is just R
the formula $\bf ab=-a\cdot b+a\times b$ is very useful in $\Bbb H$
21:35
Z(H)
what is H
wouldn't it be C not R ?
hamiltonian quaternions
@Agawa001 In my case it might be a problem related to the dog hair also.
Z(H) is just the center of the hamiltonian right ?
yes (quaternions)
@Chris'ssistheartist do you own a doggy which kind ?
21:36
@Agawa001 cross-breed.
@Chris'ssistheartist a poodle ?
@Agawa001 lol, no. Some mixed breed dogs.
i rather like kittens, but dont mind own a dog
Also how do you prove intersections of non-empty subrings are also subrings?
step (1): recall the checklist of things that make up the definition
step (2): check things off the checklist
in fact, you can use that procedure for a large number of verification exercises
I've patented it, so every time you use it you owe me 1 bitcoin
21:42
So we have that a subring S is an abelian addative group and closed under multiplication
Elements a, b in S1 $\cap$ S2 are necesserily closed under multiplication?
I do not see why
Given: a is in both S1 and S2, and b is in both S1 and S2.
Desired conclusion: ab is in both S1 and S2.
enough wrestle, gonna sleep now for waking up earlier morning. lot of stuff to do
take care
Eugh
I am too stupid
I have spent hours thinking about it
I should probably just focus on school so I can go to university and learn how not to be stupid.
"How about you read the definitions of the things you are doing before you use them"
Oh yes good point
sorry
22:04
Hey @anon I don't understand the following wording of the question Are $T_2$ and $T_0$ topologies closed under intersection ?
what does it mean does he mean if we get two topologies which are $T_0$ show that their intersection is either $T_0$ or not $T_0$
or does he mean it is closed under arbitrarily intersection
22:44
probably "Are T_2 topologies closed under intersection? Are T_0 topologies closed under intersection?"
23:00
Hello! Herstein's "Topics in Algebra" and "Abstract Algebra" which one is suitable for a learner with some mathematical maturity?
OK I guess "Topics in Algebra".
Yea I think my friend uses "Topics in Algebra" for his undergrad course
23:30
There are so few people here this afternoon... of course, in other places it is probably later at night.
In any case, usually at this time, there are more people here.
23:47
I heard your call @robjohn.
How's it going?
Hi
I ran across the user "Julian Rachman"
I saw his blog
I don't understand his blog post here: julianrachman.wordpress.com/2015/10/17/…
Can you tell me what he is trying to do?
Thanks!

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