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Alizter
21:26
@BalarkaSen What exactly does the center of $\Bbb H$ look like?
anon
quaternions?
Alizter
From a ring point of view
anon
Z(H)=R
Alizter
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
anon
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)
Agawa001
21:32
hey @Chris'ssistheartist wats up today
Alizter
But I can show that C is not in Z(R)
Chris's sis the artist
@Agawa001 Hey! Struggling with math problems and allergy. :-) You?
anon
it'd be easier to just show anything not in R is not in Z(H) directly
Alizter
hmm true
Agawa001
@Chris'ssistheartist same struggle we are cursed with (apart allergy)
Alizter
21:34
Thank you @anon
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's sis the artist
@Agawa001 :D
Agawa001
@Chris'ssistheartist i have allergy to dust but its not its time to show off yet
Balarka Sen
Z(H) is just R
anon
the formula $\bf ab=-a\cdot b+a\times b$ is very useful in $\Bbb H$
Karim Mansour
21:35
Z(H)
what is H
wouldn't it be C not R ?
Balarka Sen
hamiltonian quaternions
Chris's sis the artist
@Agawa001 In my case it might be a problem related to the dog hair also.
Karim Mansour
Z(H) is just the center of the hamiltonian right ?
anon
yes (quaternions)
Agawa001
@Chris'ssistheartist do you own a doggy which kind ?
Chris's sis the artist
21:36
@Agawa001 cross-breed.
Agawa001
@Chris'ssistheartist a poodle ?
Chris's sis the artist
@Agawa001 lol, no. Some mixed breed dogs.
Agawa001
i rather like kittens, but dont mind own a dog
Alizter
Also how do you prove intersections of non-empty subrings are also subrings?
anon
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
Alizter
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
anon
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.
Agawa001
enough wrestle, gonna sleep now for waking up earlier morning. lot of stuff to do
take care
Alizter
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
Karim Mansour
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
anon
22:44
probably "Are T_2 topologies closed under intersection? Are T_0 topologies closed under intersection?"
Tien-Cheng Huang
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".
morphic
Yea I think my friend uses "Topics in Algebra" for his undergrad course
robjohn
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.
Antonio Vargas
23:47
I heard your call @robjohn.
How's it going?
Josh
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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