12:35 AM
Ugh, this problem appears daunting: Let $(X,Y)^\top$ have a two-dimensional normal distribution with means 0, variances 1, and correlation coefficient $\rho$, $|\rho|<1$. Compute the moment generating function of $(X^2-2\rho XY+Y^2)/(1-\rho^2)$.

1:13 AM
Honestly, I'm having difficulty figuring this one out

@Rithaniel express (X, Y) in terms of a two-dimensional independent normal distribution

How many functions can be represented as projections?

hyperbola, parabola, ellipse, what are the other 4?

So, in terms of $(X+Y,X-Y)^T$?
I feel as though another two-dimensional distribution might have been intended, actually

1:39 AM
hi

@Rithaniel that still isn't independent

Yeah they are, when Var(X)=Var(Y) (and here they both equal 1) where X and Y are normal random variables, X+Y and X-Y are independent.

that requires X and Y to be independent to start with

Nah, they can be dependent. Just calculate the covariance matrix
If $\sigma=Var(X)=Var(Y)$ and $\hat{\sigma}=Cov(X,Y)$ you have that $\left(\begin{array}{c c} 1 & 1 \cr 1 & -1 \cr \end{array}\right)\left(\begin{array}{c c} \sigma & \hat{\sigma} \cr \hat{\sigma} & \sigma \cr \end{array}\right)\left(\begin{array}{c c} 1 & 1 \cr 1 & -1 \cr \end{array}\right)=\left(\begin{array}{c c} 2(\sigma+\hat{\sigma}) & 0 \cr 0 & 2(\sigma-\hat{\sigma}) \cr \end{array}\right)$
Non-diagonal entries are zero and uncorrelated normal random variables are independent
Although, clearly this isn't what you had in mind, so perhaps I should keep looking for other potential independent representations

what do the numbers that appear below our icons in the chat window represent?\

1:59 AM
Alright, I've managed to get it into the form $y^2+\frac{(x-\rho y)^2}{1-\rho^2}$

2:10 AM
Now I've gotten it into the form $(\frac{y(\sqrt{\rho^2-1}+p)-x}{\sqrt{\rho^2-1}})(\frac{y(\sqrt{\rho^2-1}-p)+x}{\sqrt{\rho^2-1}})$

1

Let $K$ be the field with exactly $7$ elements. Let $\mathscr M$ be the set of all $2×2$ matrices with entries in $K$. How many elements of $\mathscr M$ are similar to the following matrix? $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ My attempt: Answer is given is $56$. We need to find the...

How did he use that formula?

Unfortunately I feel like I'm just chasing my tail

In a field, does the element $-1_F$ necessary have to exist?
or that is yes because that is the additive inverse?
actually yeah it has to nvm

Technically yes. Every element needs to have an additive inverse. However, it could equal $1_F$

yeah in F_2

2:22 AM
Or $F_4$ or $F_8$. Any field with characteristic 2

This is another strange question. If F is a finite field, it contains a copy F_p. But is F_p the only finite subfield?
can I have another subfield F_q where q is not p?

Not necessarily. I don't remember off the top of my head what the sizes of subfields can be, but a finite field can have multiple distinct, nontrivial subfields

so lets say it does, F_q, would that also imply F would be the finite extension of F_q?

I believe so

But does q have to be prime?
characteristics are unique right?

2:28 AM
So, a field of order $p^n$ has exactly one subfield of order $p^m$ for each divisor $m$ of $n$
Characteristics are unique, yes
A finite field must have prime power order

so q has to be prime still?
but that would imply q = p?

So, there exist fields of orders 2,4,8,16,32,64, etc, but no fields of order 6
q has to divide the order of F

so all my questions are because of the proof |F| = p^n. It just appears to me if we can find a subfield F_q where q isn't prime, we can show |F| isn't a prime power.

So, for $F_{64}$ (The field of 64 elements), there is a subfield of order $2$, order $4$ and order $8$
Well, there $F_8$ in $F_{64}$ and $8$ isn't prime.
Heya Semi

well its a prime order

2:34 AM
Well yes, the order of a subfield has to divide the order of the parent field

that's only because we know ahead of time the finite filed must have prime power

Well technically it falls from the fact that fields are, first and foremost, groups, and a theorem by Lagrange that the order of a subgroup must divide the order of the parent group

well i understood that
the issue is the proofs that show |F| = p^n assmes F contains only the subfield F_p.

Well, are you comfortable with the thought that the characteristic must be prime?

yeah
n*1_F = 0 has to be prime because otherwise n would not be the smallest positive integer

2:48 AM
Alright, let $p$ be the characteristic of $F$. Since $1$ has order $p$, we know that $p$ divides $\vert F\vert$. Now assume that $q$ divides the order of $F$ where $q$ is a prime distinct from $p$. Then, by Cauchy, there exists an element $x\in F$ of order $q$. So $qx=0$ and $px=0$. Also, by bezout, there exist integers $a$ and $b$ such that $aq+bp=1$. So $x=(aq+bp)x=a(qx)+b(px)=0$, but this is a contradiction, because $x$ is assumed to have positive order.
So, the only fields that exist require that they only have one prime divisor

@TedE waves frantically.

@Semiclassical If you have a moment, I have this probability problem which is giving me a real headache: Let $(X,Y)^\top$ have a two-dimensional normal distribution with means 0, variances 1, and correlation coefficient $\rho$, $|\rho|<1$. Compute the moment generating function of $(X^2-2\rho XY+Y^2)/(1-\rho^2)$.

Hi folks. The answer to this question uses that $(A \cup B) \cap C=\emptyset$. But how do they know that?

Let A, B, and C be events in a sample space S such that S = A ∪ B ∪ C. Suppose that
P(A) = 0.3, P(B) = 0.6, and P(A ∩ B) = 0.2.

@Rithaniel in the last step, x = 0_F, you say contradiction is because it is assume positive order. Is the order not 1?

@Hawk $0_F \ne 1_F$ is one of the field axioms

2:56 AM
Yeah, I misspoke. The order is 1.
Then the contradiction would be that 1 is not prime

i think th contradiction is that the characteristic is not p
lol

That was something we assumed off the bat

@Jeff what is the question?

the contradiction is that 1*x = 0_F, instead of the characteristic p. So q does no exist. So what if we don't assume q is prime? Then q is composite, and we just let another prime, say b, divide q?

Yeah, that's the stuff. Now you just repeat the steps from there
By Cauchy there is a y of order b in F

3:02 AM
if jus leave q as composite, can bezout's equation force out the prime b? or do i have to make that argument ahead of time? i am trying to see if i can reach a contradiction by allowing q to be composite

You don't need to. If you have a composite q, you just pick out a prime divisor of it: b, and work exclusively with it.
You can then find an element of order b, and show that it has to equal 0
Just pick b such that $b\neq p$. If you can't, that means $q=p^n$ for some $n$

Let A, B, and C be events in a sample space S such that S = A ∪ B ∪ C. Suppose that
P(A) = 0.3, P(B) = 0.6, and P(A ∩ B) = 0.2.

@Jeff that is a bunch of information. It doesn't have a question.

hang on if b does not exist. this means q is not composite since we assumed it has a prime divisor, so q is prime. Therefore q also does not exist by repeated argument. So the remainder factor m in |F| = m*p^n is also prime but it cannot be distinct from p, so it must be p.
@Rithaniel i think i forced it out!

@anakhro The question has multiple parts. The part I'm answering is "Find $P(C)$". But my question is "Why is $(A \cup B) \cap C$ the emptyset?

3:09 AM
because if it is distinc from p, it has only m and 1 as divisor and if proceed with Cauchy and Bezout, we would arrive at the same contradiction that 1*x = 0_F

Alright, so you agree that all fields have order p^n?

if what I wrote is correct

@Jeff can you show the full solution as written?

20. a. By the formula for the probability of a general union,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.3 + 0.6 − 0.2 = 0.7
b. Because (A ∪ B) ∩ C = ∅, then
P(C) = P(S) − P(A ∪ B) = 1 − .07.
c. By DeMorgan’s law for sets, Ac ∪ Bc = (A ∩ B)c = S − (A ∩ B). Hence, by the formula for
the probability of the complement of an event,
P(Ac ∪ Bc) = P(S − (A ∩ B)) = P(S) − P(A ∩ B) = 1 − 0.2 = 0.8.

I don't see any reason to conclude that that's an empty intersection.

3:21 AM
I believe it is, @Hawk, just needs to be more thoroughly stated if it's for an assignment or anything like that

No not an assignment lol
I am literally self studying galois theory
i think this would be a sick assignment question though

Actually, I'm taking Galois theory next semester.
I'm familiar enough with a few results

@anakhro Oh? That's the textbook's answer.

Weird.

hahah

3:32 AM
lol i think u know more then me

I found some fun points
$(a,b)$ and $(1/b,1/a)$

3:55 AM
@Rithaniel neat. So you've got $E[XY]=\rho$
(the joy of having zero mean and unit variances)
Note that this does uniquely specify the normal distribution, i.e. you can write down the pdf

Say for all $m<1,$ you take all points $(a,b)$ and $(1/b,1/a)$ such that line $y=mx$ contains those points.
Does that make sense so far?

4:10 AM
I think I figured it out.

Nice.
mathematica gives the answer as just 1/(1-2t)

$\frac{X^2-2pXY+Y^2}{1-\rho^2}=\left(\begin{array}{c c} X & Y \cr \end{array}\right)\left(\begin{array}{c c} 1 & -\rho \cr -\rho & 1 \cr \end{array}\right)^{-1}\left(\begin{array}{c} X \cr Y \cr \end{array}\right)$

Right.
Wait...
there you go
well, probably shouldn't have the minuses when taking the inverse

@Ultradark still up to study tonight?
Sorry pinging you back late, today got some work from my dad

By Lotus, we just get that this is the integral of the pdf of $(X,Y)^T$ with an extra factor of $(1-2t)$ which pops out when you integrate

4:15 AM
ya
A cute observation to be had in this vein: Let $A=\begin{pmatrix} 1 & \rho \\ 0 & \sqrt{1-\rho^2}\end{pmatrix}$

@shi probably too late for tonight but can you help me with a question

Then $A^\top A=\begin{pmatrix} 1 & \rho \\ \rho & 1\end{pmatrix}$
So you can factor the above as $(X,Y)A^\top A(X,Y)^\top=|A(X,Y)^\top|^2$

@Ultradark I can only help you if you communicate fast enough
It's frustrating sometimes waiting for responses

Oh snap

4:17 AM
okay
@shi

Let's start a room

4:28 AM
@LeakyNun Hello
@LeakyNun chess games?

sorry not now

okay no problem
i was trying to solve this
gf = hf then h=g
i could not find a couter example

Context?

f: A--> B
g , h : B--> C
the compositon is same
If I argue by contradiction
i need to find b in B

well, that means that g=h on the image of f

4:31 AM
such that g (b) is not equal to h (b)

so you want g,h which do have the same image on some subset of B but not on all of it

I did not find a way to prove or disprove it

The simplest counterexample I see is this: f(x)=x^2, g(x)=x, h(x)=|x|
g and h agree on positive x, so they'll agree on the entire image of f

@LeakyNun I looked at Stewart. That's a cool, elementary proof. I should have noticed that.

@BalarkaSen nice

4:35 AM
@Semiclassical I see thank you !
but how does one prove this

@BalarkaSen it's "essentially the same idea"

without giving counter example
I keep thinking thiws
which is wrong
for gf : A-->C

Alright, I have this expression $ne^{-x}(1-e^{-x})^{n-1}-\text{log}(n)$ and I'm trying to find a value as $n$ goes to $\infty$. How should I try and manipulate this?

in my mind i think of , all of B must have an image
but that is clearly not the case for the composition , we need only to work with image of f
@Semiclassical You see what confuses me?

@LeakyNun Yeah I see the point.

4:41 AM
I like your proof better.

@Thorgott Ah, you wrote down Stewart's proof (essentially). Right, the key point is any embedding of $K$ in a bigger field restricts to an automorphism of $K$ again, since $K$ is a splitting field.
Thanks @Leaky :)

@BalarkaSen is K/F normal equivalent to "any two F-embeddings K -> L have the same image"?

They both have image K, right?
Oh, equivalence.
Let's see.
(Don't tell me if you know the answer, I should be able to figure this one out)

I know the answer.
Another (irrelevant) question: are all purely inseparable extensions normal? @BalarkaSen

So in particular any F-embedding K -> bar{K} has image K. Suppose f was an irreducible polynomial in F[x] which had one root, say a, in K. Then for any other root b, there is an isomorphism K = F(a) -> F(b) \subset \bar{K}. This is an F-embedding in the closure, so must have image in K, so F(b) = K = F(a), so b belongs to F(a).
So all roots of f are in K, and f splits completely in K
I know algebraic closure can be avoided here but it's slightly easier for me to think in that term

4:56 AM
nice
I don't object to the use of algebraic closure
it is an important philosophy

@LeakyNun Hmm. That's an interesting question.

btw K is not F(a)

Just ignore that "= F(a)" everywhere, it doesn't affect the solution.
$F(a) \to F(b) \subset \bar{K}$ extends to $K \to \bar{K}$, which has image $K$, so in particular $F(b) \subset K$.

how do you extend the map?

First extend to an automorphism $\bar{K} \to \bar{K}$ by what I wrote earlier, then restrict to $K$.

5:00 AM
nice
what extension can you replace $\overline{K}$ with?
I call this philosophy "reflection"
the proposition containing "for all L" is equivalent to the proposition where you substituted $L$ with $\overline{K}$
a universal proposition is reflected by a particular instance
in a ring, $\forall x, x = 0$ is reflected by $1$

That makes sense
@LeakyNun I'm thinking the splitting field of $f$.

$\forall a, \forall b, (a+b)^2 = a^2 + 2ab + b^2$ is reflected by $\Bbb Z[a,b]$
@BalarkaSen I don't think you can map K to the splitting field of $f$
K can be very big

Splitting field of $f$ over $K$, I mean. If that's $K$ I am done, otherwise it's some bigger extension $L/K$. $F(a) \to F(b)$ extends to $L \to L$ because automorphism group of splitting field acts transitively on roots. Restrict to $K$ to get $K \to L$.

cool
what if I want the field not to depend on $f$
(so it would have to be bigger)

Oof. Let me think.

5:12 AM
4 mins ago, by Balarka Sen
Splitting field of $f$ over $K$, I mean. If that's $K$ I am done, otherwise it's some bigger extension $L/K$. $F(a) \to F(b)$ extends to $L \to L$ because automorphism group of splitting field acts transitively on roots. Restrict to $K$ to get $K \to L$.
wait I don't think this works

Why so

take $F = \Bbb Q$, $K = \Bbb Q(2^{1/8})$, $f = x^4-2$, $L = \Bbb Q(2^{1/8}, i)$, $a = 2^{1/4}$, $b = i2^{1/4}$
it's because $L/F$ isn't the splitting field of $f$
"automorphism group of splitting field acts transitively on roots" requires the polynomial to be irreducible
$f$ may not be irreducible when transferred to $K$
wait does it work for my example

Splitting field of $x^4 - 2$ over $\Bbb Q$ is $\Bbb Q(2^{1/4}, i)$, a strict subfield of $L$. I see the point.

maybe replace $2^{1/8}$ with $2^{1/1024}$ just to be sure

Ok but this shouldn't be a problem I think.

5:22 AM
I just need $\zeta_{16} \notin \Bbb Q(2^{1/16}, i)$
(this fails for $8$ because $\zeta_8 = (1+i)/\sqrt2$)

We have some irreducible $f \in F[x]$ with a root $a \in K$ and we want to show all roots of $f$ are in $K$. Suppose not. Then $f$ has a nontrivial splitting field $L$ over $K$. The splitting field of $f$ over $F$ is contained in $L$, let's call that $E$.
For any other root $b$ of $f$, there is an element $\sigma \in \text{Aut}(E/F)$ such that $\sigma(a) = b$. But how to get from this $\sigma : E \to E$ to an embedding of $K$ in a larger field, I guess.

why not take $F = \Bbb Q$, $K = \Bbb Q(2^{1/4})$, $f = x^2-2$, $L = K$, $E = \Bbb Q(\sqrt2)$, $a = 2^{1/2}$, $b = -2^{1/2}$.
then $E \to E$ cannot be extended to $K \to L$

Because $E \to E$ need not leave $K$ fixed pointwise
Or rather $K \cap E$.

right

That's annoying.

5:30 AM
c'est la vie

Oh. But every $F$-automorphism $L \to L$ (where $L$ is the splitting field of $f$ over $K$) must preserve $K$, because you can restrict that to $K$, and the image of the restriction will be $K$.
Hm, does that really accomplish anything.
I want to somehow say $\text{Aut}(L/F)$ still acts transitively on the roots of $f$ over $F$.
$\text{Aut}(E/F)$ is a subgroup, which acts transitively on the roots of $f$ over $F$. Does that not imply so?
It's not a subgroup, it's a quotient
Very annoying.

it's a subquotient

yeah lol
I am getting confused by the picture. Need to smoke and think
It's like you can go to a higher extension but with lesser symmetries, I guess
@LeakyNun Which was the point of this example

I guess

Thanks for your patience!

5:42 AM
what does this mean geometrically
np

Non-normal covers can cover normal covers, I suppose.

Galois theory <-> covering space?

I got a definition question. If a polynomial is separable, meaning it has no multiple roots. Does that immediately imply it is irreducible since we know it is separable it must mean it is completely broken down to its linear factors right?

I have never actually deeply thought about normal coverings.

can we construct an explicit example?
@Hawk more context needed

5:44 AM
@LeakyNun what is the missing info i need to supply?

under my interpretation, separable means "no multiple roots in an extension containing all the oots"
well I don't know how your source defines separable

oh

and whether your polynomial is over any arbitrary field

let me just grab dummit and foote

I suspect it is, since you were asking about fields previously
but in any case (x-1)(x-2) is separable but reducible

5:45 AM
okay let p(x) be in F[x] (F is a field)
wait (x-1)(x-2) is reducible?

yeah

what does it reduce to?

x-1 times x-2
@BalarkaSen do you have an example of a normal cover and a non-normal cover?
(abnormal cover?)

@LeakyNun Yeah this shouldn't be hard to cook up. Consider the subgroup $\Bbb Z * 0$ of $\Bbb Z * \Bbb Z$ and the corresponding covering space of the wedge of two circles.

cool
so ...-o-o-o-o-o-...
or rather, ..._O_O_O_O_O_...

5:48 AM
I don't think it corresponds to that
That would be the subgroup generated by a^n b a^{-n}

oh you're right

Z * 0 corresponds to just "opening up" one of the petals of the figure 8
It's like a loop attached to a tree
That has fundamental group Z indeed so it checks out

or rather consider the Cayley graph of F2 and then modulo translation one unit to the "right"

Yeah
It's two copies of Cayley(F_2) attached to a loop
That loop projecting to the a-circle of 8

what does normality mean?

5:54 AM
@LeakyNun The way to immediately tell if a cover of 8 is irregular is if it's not even a regular graph, yeah?
because the action of the subgroup on Cayley(F_2) has to be transitive

hmm

Oh I mean if it's regular every vertex has to have equal valence
I think
The converse need not be true

4

A covering is normal or regular or Galois if the group of deck transformations act transitively on the fibers. It's called "normal" because this is the case if and only if the fundamental group of the covering space is a normal subgroup of that if the base. See wikipedia for more details.

interesting
and _O_O_O_ is not normal because...

yeah

hi @AkivaWeinberger

5:58 AM
Hey

@BalarkaSen why isn't _O_O_O_ normal
the action looks transitive to me