@anon I have a problem. help me out?

Poor Bal.

OK, @Pedro, you help me out.

I want to prove that a $k$-vector space $V$ equipped with the map $V \times V \to V$ as $(x, y) \mapsto x\cdot y$ satisfying $u \cdot (v + w) = u \cdot v + u \cdot w$, $(u + v) \cdot w = u \cdot w + v \cdot w$ and $a(u \cdot v) = (av) \cdot u = v \cdot (au)$ for any $u, v, w \in V$ and $a \in F$ is a $k$-algebra.

I have some ideas to begin with though :

What exactly do you mean by 'a vector space equipped with a map' ?

For the $(\Leftarrow)$ case, note that a $k$-algebra is a vector space $V$ over $F$ with a map $f : k \times V \to V$ satisfying the module axioms and the billinearity axioms. now this is equivalent to the existence of the ring homomorphism $g : k \to V$ with $g(k)$ in the ceneter of $V$. Maybe compose $f$ with $g \circ 1$ to get a map $k \times V \to V \times V \to V$? Is this the desired map?

@TheGame You won't get it. It's for higher mortals.

-___-

Hey @Pedro @Mike @anon. Help me out! Somebody!

@BalarkaSen at least name your map

@TheGame what map?

'equipped with the map V×V→V'

it's $V \times V \to V$, i have already mentioned that

@TheGame yes, that's it.

what do you mean by "name it"?

@Pedro @Mike @anon tapping fingers on the table

You don't get what I mean.

No, I don't.

@BalarkaSen I'm not going to read a bunch of garbage equations, my eyes slide right off the page. Tell me what axioms you assume and what axioms you need to prove.

$u:(x,y)\rightarrow u(x,y)$ is a map

$(x,y)\rightarrow \text{some expression}(x,y)$ is not named

lol

ignores @The

@MikeMiller Given : $V$ is a $k$-vector space. There is a map $V \times V \to V$ as $(x, y) \mapsto x \cdot y$ satisfying (1) $x \cdot (y + z) = x \cdot y + x \cdot z$ (2) $(y + z) \cdot x = y \cdot x + z \cdot x$ and (3) $a(x \cdot y) = (ax) \cdot y = x \cdot (ay)$

To Prove : $V$ is a $k$-algebra.

My Try : this garbage

ughh

@BalarkaSen Uh also, how do you even define the multiplication of vectors on a random vector space?

"a distributive map what preserves scalar products in both coordinates" or w/e

lol

@TheGame Just google vector space would you. they are all rings.

@MikeMiller wat

@BalarkaSen ' vector space over a field F is a set V together with two operations that satisfy the eight axioms listed below'

Where is it said it it said it inherits from some vector product ?

i'm not going to teach you basic theory of vector spaces, thank you.

a vector space is essentially a ring, that's all you need to know. thus multiplication is defined.

and vector spaces DO NOT consist of just vectors.

Uh ? Da hell ?

Then what else is included in a vector space ? (as a set)

depends on what you want it to include. as i said, i am not going to teach you the whole theory in here.

for example, $\Bbb C$ can also be a vector space. it's a $\Bbb R$-vector space in fact.

pretty much all field extensions are vector spaces.

But there are only vectors in the vector space $\mathbb{C}$ ...

@Mike you just LOLed twice and helped me nothing.

The definition of a vector is an element of a vector space

@TheGame Who the hell told you that

Everyone I know ...

what is your definition of vector spaces, then?

A set over a field with two laws +,* verifying 8 axioms

yes, so verify all those axioms for $\Bbb C$ over the field $\Bbb R$ and be done

Well, they're all verified.

-___-

so $\Bbb C$ is a vector space over $\Bbb R$.

$\mathbb{C}$ is indeed a $\mathbb{R}$-vector space.

So ?

that's what i said a few lines back

And ? I didn't say it was false

so what is your point?

i don't understand you

'and vector spaces DO NOT consist of just vectors.' - 'for example, ℂ can also be a vector space. it's a ℝ-vector space in fact.' - 'But there are only vectors in the vector space ℂ ...' - @TheGame Who the hell told you that'

^what do you even mean by multiplication? the multiplication is defined in those 8 axioms already.

@BalarkaSen No. The dot law is between scalars and vectors

Not between vectors and vectors

@TheGame i never heard explicitly referring to elements of vec spaces as "vectors" thus the comment

Oh ok

Anyway so if we say they are vectors

@TheGame wait you're not thinking of those geometric vectors when you say "vector spaces" are you?

No

I'm talking about the general definition

A member of $M_n[\mathbb{C}]$ is a vector

a vector space over a field is a ring.

and elements on the ring can be multiplied, IIRC

@BalarkaSen Why would a ring structure always be ?

@BalarkaSen That's not thde definition

that's the definition as i read it.

The basic definition is the one of the set with its two laws and 8 axioms

That's the official definition

then i am not familiar.

i was referring in general to modules over fields

You meant normed vector spaces maybe ?

who knows

i only know modules

-___-

You should know xD

nah

@TheGame I don't usually care about abelian groups anymore when taking rings to fields, which is what you seem to be not doing. F-modules are always rings to me =P

That is why i wanted you to name the map. Something was weird

To me a vector space isn't a ring, hence my opposition

@BalarkaSen That's what happens when you say you 'do not care about exceptions' :P

I remember you saying that a bunch if times

Or something like that

@BalarkaSen if a vector space over a field is a ring then how do you define a field?

@Alizter fields are defined by the usual axioms.

a commutative rings with nonzero elts having multiplicative inverse

@BalarkaSen Which is a special type of ring right?

yes, so?

so..nicated

flees

Does anybody here know anything about the shadethm package and how to get rid of the numbering in the title?

ok, so nobody actually helped me and @TheGame just nitpicked to death

anything else?

@BalarkaSen At least you learned something :P so rare for me to teach stoof

I know what modules are @TheGame. I just don't care for lesser structures than ring.

Then don't say vector space casually :)

No wonder why no one helps if they don't understand

@Khallil tex.stackexchange.com/questions/198820/…

@TheGame It's most usual and common to assume that vector spaces are rings over fields.

@BalarkaSen Why don't you ask on MSE ?

Thanks, @TheGame. ^_^

@BalarkaSen Well obviously, not common for everyone :P

you don't count eeh mean >:c

not for people in France, yes.

hehe

I have two exercises sheets full of exercises where no vector spaces are normed

actually now that i see it i have mentioned that $V$ is a ring.

yeah, but i mentioned it rather cryptically, yeah =P

-_______________________________-

shadow table bomb

@BalarkaSen Be happy, I gtg to sleep :P

byes

but nobody helped me. cries

@BalarkaSen ask on MSE

nah.

i was looking for hints

Ask for hints on MSE

I do that often

Here a little pic before leaving :D

Ask for hints on MSE

And they'll give you hints which will turn it into a totally trivial problem

old, @The

e_____e

scattering in the wind

What do you need hints with @Balarka? Not that I think I will be able to help, but worth the shot..

@BalarkaSen Diz shiz easy bro.

WAT @Karl

Hippa ("The Game" now) babbled me to death for not mention that my vector space over fields were rings.

Halp me, @Karl !!!11!

I have to prove that $V$ is a $k$-vector space and all $k$-vector spaces $V$ satisfy the above axioms.

I "think" I worked out the second part :

Oops I mean $k$-algebra, not vector spaces.

I have to prove that $V$ is a $k$-algebra and all $k$-algebras $V$ satisfy the above axioms.

hehe..

@BalarkaSen Dude.

Get you s**t together.

OK, I won't ask anymore.

Bleh.

@PedroTamaroff You still didn't give me a problem to think about.

First you need to get your s**t together.

Pardon, sir? =P

OK.

Let $R$ be a ring, let $S$ be another ring.

mmhmm, then?

Then $S$ is an $R$-algebra if there is a ring homomorphism $\eta: R\to S$ such that $\eta(R)\subseteq Z(S)$.

sigh I know.

I was actually asking questions about those above.

OK. Classify all nonabelian groups of order 63.

For the honor of 'Nam.

C'mon,

You just.

Do eet.

sigh

OK.

$63 = 3^2 \cdot 7$

Oh hey that's one $7$-Sylow we've got there.

Well, if there is only 1 3-sylow and only 1 7-sylow then we get a split exact sequence 1 --> N_3 --> G --> N_7 --> 1.

$N_3$ and $N_7$ are both abelian, so there we get two abelian groups. $\Bbb Z_9 \times \Bbb Z_7$ and $\Bbb Z_3 \times \Bbb Z_3 \times \Bbb Z_7$

if there are 7 3-sylow and 1 7-sylow then things get complicated. having a single sylow subgroups implies that it's normal thus the sequence would be 1 --> N_7 --> G --> H --> 1

searches for the copy of Artin evidently I forgot how to handle these

Hello @ccorn

Hi @Balarka

Actually I had a problem for you @ccorn. Pleasant coincidence.

Do you know of a closed form for sections (partial sums) of jacobi thetanulls?

I have been busy for quite a while, and will drop by only infrequently for the time being.

or even closed form for the coeffs would do.

Hi Professor @TedShifrin

@ccorn hi, long time no see :-)

Hi skull and mr @Pedro and @Balarka

What I mean is this : We can explicitly write down the coefficients of $\left ( \sum_{n \in \Bbb Z} x^{n^2} \right)^2$ (which was done by Jacobi). Can we also do it for $\left ( \sum_{n \leq \ell} x^{n^2} \right)^2$? @ccorn

Hello @TedShifrin

@BalarkaSen Well, up to $\ell$ surely :-)

@ccorn That's a relief. How?

Erm, @TedShifrin, did you see the link?

@Balarka: Take a generator of the subgroup of order 3 and see how it acts by conjugation on the subgroup of order 7. There are fancier ways to say this ...

No, what link?

@TedShifrin Yes, that's what I was looking for. I don't want semidirect voodoos, just explicit representations. =)

@TedShifrin The ping is a link.

@BalarkaSen The difficulty arises when you want to get coeffs with index higher than $\ell$

Yes, that's what I want actually grins

What ping is a link?

Ah, never mind. this

LOL smack Have you nothing better to do with your time?

That's the reply I was aiming for.

Or I could reignore ...

No, no.

I'll just... go. Don't ignore, @TedShifrin.

It's bed time in any case.

bye

Had an interesting numerical experience: Solving a banded triangular Toeplitz system. Easy, right? Well, the data flow is then the same as in an infinite impulse response filter, and such filters can become unstable. And that's what happened.

Hi @Karl!

Hi @Ted

How was your weekend?

You were in DC, right?

@TedShifrin Hello!

Yup, @Karl. Tiring but fun. Heya mr @Pedro

@ccorn I know some of those words!

@TedShifrin You weren't the guy with the knife on the white house property, I hope.

That'd be tiring but fun, I admit.

Nope. That may have happened before my arrival?

I saw it in the news either Friday or Saturday. The secret service is being accused of not reacting well, since he made it virtually to the front door.