Mathematics

12:56 AM

Currently trying to prove that all irreducibles of $\mathbb{Z}[\sqrt{-3}]$ are primes in $\mathbb{Z}[\frac{-1+\sqrt{-3}}{2}]$. It is showing itself to be difficult.

I know that I should use the units of the two rings, somehow

Lukas Heger

1:33 AM

@Rithaniel three steps:
- show that for all $a \in \Bbb Z[\frac{1+\sqrt{-3}}{2}]$, there is some unit $u \in \Bbb Z[\frac{1+\sqrt{-3}}{2}]^\times$ such that $ua \in \Bbb Z[\sqrt{-3}]$
- conclude from step 1 that all irreducible elements in $\Bbb Z[\sqrt{-3}]$ are irreducible in $\Bbb Z[\frac{1+\sqrt{-3}}{2}]$
- show that $\Bbb Z[\frac{1+\sqrt{-3}}{2}]$ is a PID and hence every irreducible is prime

which step do you have trouble with?

Rithaniel

Ah, that'd be the first step, actually

With that outline, it becomes fairly straight forward, actually

Lukas Heger

oh for the second step, you also want to use that every unit $u \in \Bbb Z[\frac{1+\sqrt{-3}}{2}]$ that lies in $\Bbb Z[\sqrt{-3}]$ is actually a unit in $\Bbb Z[\sqrt{-3}]$

Rithaniel

Yeah, I got that part, the first one is what I might have some trouble with. But I think I can whip up some algebra to figure it out

Lukas Heger

it's not really very conceptual, take an arbitrary element, make some case distinction and multiply with the right unit in each case

Rithaniel

It seems like I should do cases based on the oddness/evenness of the components?

Lukas Heger

1:47 AM

right

Rithaniel

Well, that sounds like I need to figure out the full list of cases. Could I ask for a how many I should expect to see?

Lukas Heger

2:04 AM

@Rithaniel two cases

Rithaniel

2:26 AM

So, if b is even or if b is odd. If b is even, it's already in the overring. If b is odd, multiply by $1+(\frac{-1+\sqrt{-3}}{2})$. Correct?

That is, where $a+b\frac{-1+\sqrt{-3}}{2}$ is an element of $\mathbb{Z}[\frac{-1+\sqrt{-3}}{2}]$

Wait, no, what if both $a$ and $b$ are odd

Lukas Heger

yeah I didn't count the case b even, cause that's obvious

you just need to multiply with one unit if a is odd and with another one if a is even

Rithaniel

Alright, yeah. I always get lost when distributing these things. Like $(\frac{-1+\sqrt{-3}}{2})^2=\frac{-1-\sqrt{-3}}{2}$ took me longer than I should probably admit

Jack Ohara

6:39 AM

@LeakyNun Hello

Arvin Ding

8:42 AM

hello?

northerner

10:19 AM

What is the term used to describe when someone computes the single value of an arithmetic expression? e.g. someone finds out that 2*2+3-1 is 6? Is it called evaluate, simplify, solve or something else?

Leaky Nun

10:35 AM

evaluate

northerner

thanks

adesh mishra

11:39 AM

@StupidQuestionsInc Please help me over here

Mathphile

12:07 PM

Can anyone help me prove $\int_0^{\pi/2} (\tan x)^{2/3}dx=\pi$

I am able to find the indefinite integral but still am not able to prove this

Even WA only gives an approximation wolframalpha.com/input/…

Rithaniel

12:36 PM

I have a couple of problems I could use some pointers on, if anyone could help:
First one is "Let $F\subseteq K$ be fields. Show that the domain $F+xK[x]$ is integrally closed if and only if $F$ is algebraically closed in $K$."
Second one is "Let $F\subseteq K$ be fields. Show that the domain $F+xK[x]$ is completely integrally closed if and only if $F=K$"

Thorgott

"$F+xK[x]$" being the polynomials over $K$ with constant coefficient in $F$?

Rithaniel

Yeah, exactly

Rithaniel

1:37 PM

Well, one thing of note is that any UFD is completely integrally closed, so I know that $F+xK[x]$ isn't a PID as I would have suspected. That's just a corollary to these results, though, and doesn't actually help me prove them

topologicalmagician

Hi

is it a convention

for $0$ (in the integers) and any element of a ring $r$, to satisfy $0r=0$?

Leaky Nun

2:14 PM

@topologicalmagician it is a theorem

@Rithaniel which implication are you stuck in?

adesh mishra

Leaky would like to help me in propositional Logic?

Stupid Questions Inc

@TedShifrin Thanks Ted, it's from Hoffman - Kunze's Linear Algebra (page 195).

adesh mishra

@StupidQuestionsInc Hi

Did you see my ping?

Stupid Questions Inc

@adeshmishra Yes i'm writing a question here and then i'll check it

adesh mishra

Okay no problem

Stupid Questions Inc

2:29 PM

Another question from that book,the $T$-conductor of $\alpha$ into $W$ is defined as $S_T(\alpha,W)=\{\text{polynomial} g\in\mathbb{F}[x]: g(T)\alpha\in W\}$ where $W$ is an invariant subspace for $T$ and $\alpha\in V$. While I do understand why the minimum polynomial of $T$ will always lie in that set (since it maps to $0$), I don't see why a polynomial $f\in S_T(\alpha,W)\iff$ the minimal polynomial divides $f$?

@adeshmishra that's more abstract logic even for me ^^ But you can see what it's talking about, for instance for 3, if the set of propositions contains a proposition $\varphi$ then it must also contain its negation

@adeshmishra btw are u done with the logic parts of the book that I sent you?

adesh mishra

@StupidQuestionsInc yes

Stupid Questions Inc

@adeshmishra ok, will you then move to basic naive set theory?

adesh mishra

@StupidQuestionsInc Yeah, we can prove the distribution of “or” by the truth tables, ha?

Stupid Questions Inc

@adeshmishra yup

adesh mishra

2:47 PM

@StupidQuestionsInc I saw your profile many times but then also I couldn’t understand the reason behind your user name

😁

Stupid Questions Inc

@adeshmishra Read my description :)

adesh mishra

@StupidQuestionsInc How can I fight with your description? :-)

Stupid Questions Inc

> Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis? (Paul Halmos)

adesh mishra

Yes, but how is it related to username in any way?

@StupidQuestionsInc

Stupid Questions Inc

@adeshmishra So the idea is that you have to ask a lot of "stupid" questions, tada you get my username

adesh mishra

2:57 PM

@StupidQuestionsInc hahahaha

But why Inc, you opened a whole factory of questions?

Stupid Questions Inc

@adeshmishra you can say it that way :P

adesh mishra

@StupidQuestionsInc Why your profile picture shows only front part of your face and rest of it is a cartoon image?

geocalc33

3:44 PM

Can you map $\Bbb R^{1,1}$ to $T^2$

anakhro

3:55 PM

What is $\mathbb R^{1,1}$ again? I always think it looks like the supermanifold notation. @geocalc33

geocalc33

@anakhro oh, it's a two dimensional pseudo riemannian manifold

In differential geometry, a pseudo-Riemannian manifold,[1][2] also called a semi-Riemannian manifold, is a differentiable manifold with a metric tensor that is everywhere nondegenerate. This is a generalization of a Riemannian manifold in which the requirement of positive-definiteness is relaxed.

Every tangent space of a pseudo-Riemannian manifold is a pseudo-Euclidean vector space.

I don't know if this question really makes a ton of sense

because a pseudo-riemannian manifold is inherently different from a riemannian manifold, specifically the requirement of positive definitenss is relaxed for the pseud-riemannian case.

Maybe, $T^2$ has to be a pseudo-riemannian manifold but I don't know if a torus can be a pseudo-riemannian manifold

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