suppose I have a semisimple ring R. Then it is known that R is the direct sum of some R_i, where each R_i is the sum of all left ideals isomorphic to a representative simple left ideal $\mathfrak a_i$.

Also, this sum is finite, say $R=R_1\oplus \cdots\oplus R_s$, and each the representatives are $\mathfrak a_1,\ldots, \mathfrak a_s$.

If $1=e_1+\ldots +e_s$; then $e_i$ is a $1$ for $R_i$; which is a double sided ideal of $R$ and a ring itself, and $e_ie_j=0$.

I want to show each $R_i$ is a simple ring.

$R_i=\sum_{\mathfrak a\simeq\mathfrak a_i}\mathfrak a$.

@PedroTamaroff those are called isotypic components. :-)

Ah, thanks. I read the word is spanish, but never really used it.

Is a semisimple ring semihard? Hmm...

At any rate, say $\mathfrak b$ is a double sided nonzero ideal of $R_i$.

Since $R_i$ is a sum of ideals, $\mathfrak b$ has nonzero intersection with at least one.

Since $\mathfrak a$ is simple, $\mathfrak a\cap\mathfrak b=\mathfrak a$.

assuming you picked an $\frak a$ first

Yes, "...at least one."

"is a sum of simple ideals.

and possibly ${\frak a}\cap{\frak b}=0$

Yes.

Thus if $\{\mathfrak a\}$ is the class of ideals that cuts $\mathfrak b$, $\sum \mathfrak a\subseteq \mathfrak b$.

It suffices to show every ideal cuts $\mathfrak b$, then.

Every simple ideal, I mean.

But of course this implies it for all.

@blue so far so good?

yep

(I am not sure what the standard proof is, if you have any suggestions I'll ask at the end)

OK

you should probably justify that $\frak b$ is indeed an ideal of $R$ (not just $R_i$) hence $\frak a$s which nontrivially intersect it are contained in it

Right, this happens because $R_i$ interacts with itself only.

That is $R_iR_j=0$.

$i\neq j$.

right, other things annihilate it

my familiarity is with group algebras, in which the isotypic components are matrix algebras

OK.

So let's assume that $\mathfrak b$ doesn't cut some simple ideal $\mathfrak a$. Then $\mathfrak a\oplus \mathfrak b$ is a left ideal ideal, and it has a direct summand, since $R$ is simple.

Finally, someone uses OK like me instead of okay.

$k[G]\to \bigoplus_{V\in{\rm Irr}(G)} {\rm End}_k(V)$ is an iso. so you want a proof that at least absorbs the proof that matrix algebras are simple.

Right, but $M_n(k)$ has the lattice of ideals $k$.

fields, or division rings

the lattice of k?

Which makes it kind of trivial, right? I don't see it immediately here.

@blue Of ideals.

@PedroTamaroff and do you know how to prove that $M_n(k)$ has the lattice of ideals of $k$, in the special case that $k$ is a field? is the proof obvious?

@blue What we do is "pullback" ideals from $M_n(A)$ to $A$. We consider the ideal of elements in some coordinate.

mmhmm

Should I be clicking or something...? =/

iuhno

Oh. Do you have a proof?

My idea was just showing $\mathfrak b$ cuts all ideals.

well, go for that

any simple ideal is principal, and any ideal of a semisimple ring is a direct sum of simple ideals, so you have a basis-ish to work with for R_i. so you might be able to mimic the matrix algebra proof if you take this farther, is what I was thinking.

OK.

@blue A couple of very tricky theorems and Dedekind domain is now defined full-fledged.

@blue porpoises? that sounds like tortoises...

@BalarkaSen No shit Sherly.

@PedroTamaroff What d'you mean by Sherly?

Sherlock.

Ah.

@Chris'ssis What is the C constant?

Catalan.

@cc Catalan's constant.

ah ok

@blue Hmm.

Take the infinite field $\Bbb {\bar Q}$.

The Dedekind domain of this is precisely the algebraic integers, i.e., $\Bbb {\overline{Z}}$. But this is clearly not Noetherian.

"the dedekind domain of this"? you mean the integral closure of Z in it.

Yes.

That's how D.D.s are defined for algebraic number fields.

@blue

LEL

mmmm. one proves number rings are dedekind domains. it's not "how DDs are defined for" them.

yes pedro?

@blue potato pohtato. the guy who is teaching us is defining them as DDs first and then define in general DDs.

Lang defines a module to be simple iff it is semisimple and has one isomorphism class of simple left ideals. Then $R_i$ is simple using this definition. =P

A ring is simple if it is not hard.

@JasperLoy I've never heard of a hard group.

@JasperLoy is it the same for boxes?

boxes?

Yay. I got another upvote in a not-so-high hanging fruit.

I am officially retiring after 2 days. I will come to chat though.

@blue (my message is up there, blue)

about simplicity being trivial with his definition?

Yes.

I didn't say trivial, though. hehehe.

@FernandoMartin Mirá quien llego.

I think my enemy might attain 200k this year...

I don't care much about reps obtained through LHFs.

So I literally almost always comment rather than to post an answer.

Can someone point me to a copy of Folland's Advanced Calculus? I still haven't found it on the Russian servers...

@robjohn Are GRE scores very important? I think I might not do well on them.

@Chris'ssis That looks similar to the integral I was looking at earlier.

@robjohn Which integral?

@JasperLoy GRE were necessary when I was applying for college. I don't know what is required today.

@robjohn They are still necessary now. I was just asking about their weight.

@Chris'ssis I posted something a while ago. Let me look....

@robjohn OK

@Chris'ssis $$\int_0^1\frac{\log(1-x^2)}{1+x^2}\,\mathrm{d}x=\frac\pi4\log(2)-C$$ where $C$ is Catalan's constant

I know that one.

But I am too lazy to solve it.

Hello

@robjohn Yeah.

@Chris'ssis it was the Catalan's constant that caught my attention

can someone help me please

Is your life in danger?

no no

It is now his favourite phrase, lol.

Can someone help me please?

@JasperLoy thank you

@robjohn that integral arises in the series below (that can be computed in many ways)

@Vrouvrou You should thank the one who helps you, not me, lol.

It depends on how you tackle the series in terms of the integrals.

if $u=u^+-u^-$ where $u^+$ and $u^-$ are the positive and negative part of $u$

if $||u||\rightarrow+\infty$

is $||u^+||\rightarrow+\infty$ ?

@Vrouvrou You should give more context. Without context, nobody knows what the situation may be. I suspect the context is your recently deleted question, in that case the answer is "Maybe, maybe not, it depends on $u$".

@DanielFischer He probably means it as a general question.

@Vrouvrou what does $\|\cdot\|$ mean?

$||u||^2=\int_{\Omega} u'^2(t) dt$

@Vrouvrou The $L^2$ norm of the derivative of $u$? is this over $\mathbb{R}$ or $\mathbb{R}^n$?

what is the difference ?

@Vrouvrou well, the question arises about how you define the derivative on $\mathbb{R}^n$.

it is defined on $R$

$\int_\Omega|\nabla u|^2\,\mathrm{d}x$ ?

@Vrouvrou okay

no on $R$ so u' not \nabla u

@Kaj Hey.

Hey there

@Sawarnik, no idea

@Kaj I have a problem for you.

Maybe I'll be clever enough to solve it...

@Vrouvrou In any case, consider $u(x)=-e^x$ for $x\ge0$.

Here comes Matt Damon, lol.

:P

@KajHansen Compute $\text{Aut}(Q_8)$. Do it in a clever way.

i.e., less computations.

Oh God.

You see, the quaternions and I have a love/hate relationship.

But mostly a 'hate' one.

haha, why?

yes $u\rigtarrow +\infty$ and $u^+\rightarrow 0 $right ? @robjohn

I think it started when we had to determine the automorphism group of a degree-8 extension whose base field was not $\mathbb{Q}$. Ended up being $Q_8$, and a real pain in the ass.

@KajHansen What was le problem?

@BalarkaSen You like French too much.

@JasperLoy I don't know any French.

@BalarkaSen, came from Ted's algebra. Determine $Aut(K/F)$ given that $K = \mathbb{Q}[\sqrt[8]{2}, i]$ and $F = \mathbb{Q}[i\sqrt{2}]$.

That's an eighth root of $2$ if you can't see it.

But no worries, @Kaj, it's not at all a problem about quaternions. It's a problem about discrete group of isometries of a cube.

@KajHansen Yes, I see it.

And what is supposed to be the intuition behind this problem at all?

Oh, the problem actually said to confirm that $Aut(K/F) \cong Q_8$, but that's it.

@BalarkaSen it's an exercise, to practice use of one's tools

No previous parts or anything like that.

@blue Compute me $Aut(\Bbb Q(\sqrt{2+\sqrt{3+\sqrt{5}}}- \sqrt[3]{79}, \zeta_3)/\Bbb Q)$

That too is an exercise.

That's probably a lot easier tbh

Because I doubt that's a splitting field

@KajHansen Gah, right.

I forgot to adjoin cube roots of unity.

=P

Which means that $|Aut(K/F)|$ will be strictly less than, but also need to divide $[K:F]$.

Which will probably make it something easy like $\mathbb{Z}_2$.

Contrarian. (noun) /kənˈtɹɛɹiən/ A contrarian is a person who takes up a contrary position, a person who seems to be "contrary for the sake of being contrary."

@blue sighs

I think theory-related problems are much better, with a good observation behind you and a great lot of interesting conclusion.

Motivations are always needed to do some particular problem. Otherwise it just becomes an olympiad problem.

But, well, it's true that motivations are relative. Varies person-to-person.

As far as $Aut(Q_8)$, it might become a counting argument depending on the possible places $i$ can map to.

@KajHansen Just visualize the elts as vertices of a cube.

Given that $i$ has order $4$, it would need to map to something that also has order $4$.

Symmetry group $S_4$ acts on $Aut(Q_8)$, so that's sitting inside it.

Cubes are great. You can do a similar argument to show that $Cube \cong S_4$.

@Vrouvrou no, $u^+=0$ yet $\|u\|=\infty$

@KajHansen My favorite is the dodecahedron (a.k.a. the icosahedron)

That $Dodeca \cong Icosa$, or that $Icosa \cong A_5$?

$A_5$-symmetry. There is a neat proof of simplicity of this group using a fair bit of geometry.

@KajHansen The former is trivially true. The latter.

@BalarkaSen umm... dodecahedron has 12 faces, but the icosahedron has 20 faces

@BalarkaSen using basic tools just learned in a section does not constitute an olympiad problem, that's utterly ridiculous. just because an exercise is not immediate or requires multiple uses of multiple tools does not mean it is elaborate for the sake of being elaborate (as your "exercise" suggested). I find your pedagogy something of an unrealistic pipe dream. students need to be able to use theory, and if they never practice then they can't.

@robjohn Yes.

IIRC, it's easier to show that $A_5$ is simple by taking the geometric approach with an icosahedron than a more algebraic approach working directly with $A_5$. It's been quite some time though.

But look at the vertices.

@BalarkaSen they are duals, but not aka

One is the dual of another.

@robjohn I was speaking in terms of symmetry groups.

An easier argument is to inscribe an Icosahedron in a dodecahedron.

@BalarkaSen okay... move along..

From there, it is easy to show that $Icosa \subseteq Dodeca$.

@KajHansen Righto. It's in Artin.

From there, you finish up showing that $|Icosa| = |Dodeca|$.

@KajHansen Dodeca \cong Icosa is trivial to prove but it's not a trivial fact. It has a fun consequence which I think is very important whenever you work with quintics.

What's the easiest way you know how to prove $A_5$ is simple?

I think A5 is very hard.

@KajHansen The geometric proof is better.

I remember doing it using a geometric argument and fiddling with an icosahedron months ago.

Don't remember what I did lol

@blue Oh well. I (I, not anyone else) like to have some intuition for doing a math problem.

It's far better to work your way through constructing some $F/\Bbb Q$ with galois group $Q_8$ than verify that it is for an arbitrary field mod an arbitrary field.

@KajHansen So where are you going for grad school?

we aren't always in the (of course nice) position of having intuition guiding us, which is why we need to develop the ability to use tools.

@JasperLoy, no clue. I'm about to start third year undergrad.

@KajHansen Me neither. I don't usually bother remembering geometric proofs. I think you do it by embedding the symmetry group in special orthogonal groups and finishing though some conjugation argument.

I have to reconstruct it, sorry.

@JasperLoy, a lot depends on how much I can raise my GPA in my final two years. MY GPA is pretty low right now.

Not terrible, but I feel like it won't be competitive for grad school applications as it stands.

@KajHansen Mine is 4.4 out of 5. Is that OK to get into a decent school?

No clue. We're out of $4$ in the states. I have a $3.3$ out of $4$ right now, and it seems like most people have $\geq 3.7$ when applying. I might go for a masters program first or something.

Bear in mind I have no clue what I'm talking about.

@BalarkaSen, I presume you've seen the result that $S_4 \cong Cube$.

@KajHansen Yes.

I am very set on grad school though. I want to do math for a living, hopefully as a researcher and not applied stuff for some factory, etc.

@KajHansen Me neither.

@KajHansen I would be happy to get a teaching position somewhere in the US after grad school.

The funny thing is Ted's responsible for my lowest major-level math grade so far.

@KajHansen What is your worst math grade? Mine was B+

Which is?

I actually think teaching might be fun. Even as a high school teacher. But it would have to be at a private school; I don't think I could stand all the bureaucratic red tape in public school.

Haha, we won't go there :)

@PedroTamaroff Eh, I got B+ for set theory, real analysis, and geometry, and approximation theory, lol.

@JasperLoy, are those four separate classes?

I was already mad when I went to uni.

What's your lowest grade, @Pedro? An A+?

@KajHansen Yes.

Of course, if I did not go mad, things would be different.

I've gotten a B+ in combinatorics, Ted's geometry and his first semester multivariable, and first semester abstract algebra.

It's OK. Stephen Smale got poor grades as well.

I was actually pretty upset over the abstract algebra. I worked my ass off for that course. In the long run, I feel like I benefited a lot from it though.

I suspect he went mad as well.

@BalarkaSen We don't have letters.

We have 1-10.

What's that?

That is called a number @BalarkaSen

@JasperLoy No it's not. It's called an integer.

"number" is a highly vague word.

From Wikipedia:

"Earlier in his career, Smale was involved in controversy over remarks he made regarding his work habits while proving the higher-dimensional Poincaré conjecture. He said that his best work had been done "on the beaches of Rio"."

LOL

haha

@KajHansen Have you watched Good Will Hunting?

The time bandit dudesth called me and thed letsth thmoke sum purpth, bro, and maybe thum white widow

thamoking purpths bra

j/k

I'm just trying to sound like I have a lithp

Your joking would be better without j/k

^_^

Why did you remove your face?

That's heavy

@EnjoysMath (If (you are (trying (to (sound like you (have (a lisp)))))), you need more parentheses.

I never bothered to look up lisp.

This chat is dead.

Yes.

I would like to ask this question again.

What if one day you want to do something to improve society but it is illegal? That is, what you consider good others consider bad and even made it illegal.

Like genocide?

Well, I can't say what it is here. But it is not genocide I am trying to do.

Drug dealing?

Human trafficking?

No, I can't say it, and you will never guess.

Singapore, wasn't it? So probably spitting chewing gum on the pavement.

lol

@JasperLoy If you can't say what it is, then it is probably fucked up and you shouldn't do it.

I could write to the police and ask them if I can do it.

I was also thinking of writing to the minister to see if laws could be changed.

I think I will ask my best friend what he thinks when he returns from Oxford.

Jasper what the fuck. Email me.

@PedroTamaroff Let me think about that first... Anyway, how is school?

@JasperLoy Awesome.

Good good. I hope you find a gf soon.

I am going to sleep. Good night.

@JasperLoy good night

@JasperLoy, Yep