Mathematics - 2018-05-17 (page 2 of 3)

MatheinBoulomenos

15:02

@Silent that's wrong. $x^2+x+1$ has $\omega$ and $\omega^2$ as roots

Alessandro Codenotti

@MatheinBoulomenos derp of course $x^3-1$ isn't irreducible

Silent ignore my previous message, I wasn't paying attention :/

Also hi @Mathei

MatheinBoulomenos

Hi @AlessandroCodenotti

@Silent an example of such a matrix is $$\begin{pmatrix}0 & -1\\
1 & -1\end{pmatrix}$$

GFauxPas

lol here I am fiddiling with cut out pieces of paper trying to figure out how to make a $2$-sphere out of two $2$-simplices but it wasn't working. Google: hey bud you can't do that

o

oops

Ted would be happy that I tried to get a geometric intuition out of physical triangles, I hope he would be proud of me :3

Semiclassical

lol

I'm a little surprised you can't, actually. I'd have imagined you could get way with taking the two triangles and identifying the edges/vertices but not the faces

GFauxPas

the problem I'm running into is that you get a band around the equator that goes the wrong way

Semiclassical

15:13

hmm

Alessandro Codenotti

I'm a bit stuck on the calculation of a class number @Mathei, want to help me?

MatheinBoulomenos

okay

Alessandro Codenotti

So the field is $\Bbb Q(\sqrt{-6})$ and the class number should be $2$ (I'm asked to prove that it is $2$ so it better be!)

The Minkowski bound is 3.11

GFauxPas

semi, here's a fun thing

Alessandro Codenotti

So I'm looking at $(2)$ and $(3)$ in $\mathcal{O}_K$, which I think are both ramified, $(2)=\mathfrak p^2$ with $\mathfrak p=(2,\sqrt{-6})$ and $(3)=\mathfrak q^2$ with $\mathfrak q=(3,\sqrt{-6})$

GFauxPas

15:18

so you glue together two 2-simplices along their border, trying to make a sphere

except no matter how you traverse the equator, you're contradicting yourself in what direction you're going. you have to make an equivalence class for the space to work

so along the equator, going forwards and going backwards are identified, and you get $x \sim-x$

as the choice of equator is arbitrary you get $\mathbb RP^2$ @Semiclassical

:D

MatheinBoulomenos

@Alessandro so far, so good. The next thing to do is to compute $\mathfrak{p}\mathfrak{q}$

GFauxPas

fun stuff

Kaustabha Ray

A, B and C throw a die alternately and this keeps repeating.
What is the probability that A throws the first six, B the second and C the second?

MatheinBoulomenos

(and to convince yourself that $\mathfrak{p}$ and $\mathfrak{q}$ are not principal)

Alessandro Codenotti

@MatheinBoulomenos Wait why? I was thinking that now I want to decide whether there are $a,b\in\mathcal{O}_K$ with $(a)\mathfrak p=(b)\mathfrak q$

Kaustabha Ray

15:21

A, B and C throw a die alternately and this keeps repeating.
What is the probability that A throws the first six, B the second and C the second?

What is the probability that the first six is thrown by A, the second by B and the third by B?

Are these two statements equivalent?

Sorry I missed out the last part before.

I just why or why not the two statements are equivalent

MatheinBoulomenos

@AlessandroCodenotti if $\mathfrak{p}\mathfrak{q}$ is principal, then in the class group $\mathfrak{p}$ and $\mathfrak{q}$ are inverses of each other, but since they have order 2, they are actually equal

Alessandro Codenotti

Oh, right, that makes sense

MatheinBoulomenos

Note that $(\sqrt{-6})^2=(6)=(2)(3)=\mathfrak{p}^2\mathfrak{q}^2$. If you compare norms a bit, then this implies $\mathfrak{p}\mathfrak{q}=(\sqrt{-6})$ (you could also compute that directly)

Alessandro Codenotti

Well to go about it the naive way $\mathfrak p\mathfrak q=(6,2\sqrt{-6},-6,3\sqrt{-6})=(6,\sqrt{-6})=(\sqrt{-6})$

Skipping a few double inclusions here and there

MatheinBoulomenos

or you use that $\mathfrak{p}$ and $\mathfrak{q}$ are prime, so we get from $\mathfrak{p} \mid (\sqrt{-6})^2 \Rightarrow \mathfrak{p} \mid (\sqrt{-6})$ and the same for $\mathfrak{q}$

either way, if you know that $\mathfrak{p}\mathfrak{q}$ is principal, you're done

Alessandro Codenotti

15:30

Indeed, thanks a lot for your help!

Now I have $\Bbb Q(\sqrt{-31})$ to do, let's see if I can work it out by myself this time

GFauxPas

Actually now I'm doubting what I said semi, lemme think about it more

Alessandro Codenotti

This whole class number business is fantastic by the way, it's amazing how many smart ideas where developed to decide whether a ring of integers is an UFD

MatheinBoulomenos

yeah

if you try to prove that $\Bbb Z[\frac{1+\sqrt{-19}}{2}]$ is a PID with elementary methods, that's really tedious

with the class group and Minkowski bound stuff, that's really easy

Alessandro Codenotti

Do you get a bound lower than 2?

MatheinBoulomenos

you get a bound between 2 and 3, but $(2)$ is inert

Alessandro Codenotti

15:40

Oh, I see, that's still a very nice one

Ah, by the way, I have a curiosity, is there a simple way to construct a field with a class number bigger than some fixed $n$?

MatheinBoulomenos

that ring is the easiest example of a non-Euclidean PID

Alessandro Codenotti

I guess proving that it not Euclidean is far from straightforward?

MatheinBoulomenos

it's not too bad

the idea is this: suppose $R$ is an Euclidean domain with Euclidean function $\varphi$. Choose $c \in R$ non-zero non-unit such that $\varphi(c)$ is minimal among all non-zero non-units. Then by definition of a Euclidean domain, $\forall a \in A, \exists q,r \in R$ such that $a=qc+r$ with $r$ either $0$ or a unit. For $\Bbb Z[\frac{1+\sqrt{-19}}{2}]$, the only units are $\pm 1$, so the only possibilities for $r$ are $-1,0,1$.

Then there are some elementary calculations with different choices of $a$ to show that such a $c$ can't exist

Alessandro Codenotti

Well it does sound better than I expected

MatheinBoulomenos

@AlessandroCodenotti one can show that if for two natural numbers $n,m$ $d=n^m-1$ is square-free, then the class group of $\Bbb Q(\sqrt{-d})$ has an element of order $m$. I'm not sure how to show that you can always choose $n$ and $m$ with $m$ arbitrarily large such that $d$ will be square-free, though

Alessandro Codenotti

15:52

I see

another "vague" question, there is a very obvious and important difference between class number $1$ and class number $>1$, but what does the class number being $3$ rather than $2$ or $457$ tell me about the field?

MatheinBoulomenos

I think Zagier showed that for every number $n$, there are only finitely many imaginary quadratic fields with class number $n$, so if you just choose $\Bbb Q(\sqrt{-d})$ for $d$ large enough, you'll get the class number as large as you want

I'm sure about the result, but not sure if it is due to Zagier

@AlessandroCodenotti intuitively, the class group measures "how much" the ring is from a PID, so a small class number is somehow "closer" to being a PID than a large class number. If you do some calculations with ideals in the ring of integers, than knowing the class number can be helpful. For example, suppose you know that for some ideal $I$ $I^p$ is principal and $p$ doesn't divide the class number. Then you get essentially by elementary group theory that $I$ itself is principal

($p$ is prime)

this kind of argument is basically why Fermat's last theorem is easier to prove for primes $p$ such that $p$ doesn't divide the class number of $\Bbb Q(\zeta_p)$

Alessandro Codenotti

@MatheinBoulomenos Ah, I've heard about Kummer's proof for regular primes

GFauxPas

"We still have the cells $e_0 \cup e_0 \cup e_1 \cup e_1$ on the equator. We want 1-cells so looking down it looks like a pokeball."

lol topology

MatheinBoulomenos

yeah, it's really this reason why Kummer introduced regular primes. In the argument, you show that some $p$th power of an ideal in $\Bbb Z[\zeta_p]$ is principal.

Alessandro Codenotti

I see, I want to read the proof, but maybe after the exams

well I have to go for a while now, thanks a lot! @Mathei

MatheinBoulomenos

16:02

you're welcome @Alessandro

Secret

16:18

vocabulary.com/dictionary/laurel

#yannylaurel

Yanny or Laurel? Hear both! (Pitch shifting)

►

Secret

16:36

Bleeeeeeeeep

Semiclassical

lol

Secret

AKS Dies not exist

Abhas Kumar Sinha

Hello, what you do when you are unable to solve a mathematical problem and feel frustrated?

Semiclassical

putting a voltage across pencil graphite was an old high school amusement

Secret

Take a bath and sleep

Silent

16:41

@MatheinBoulomenos Thank you so much! Is it true that if $\alpha\in K$ is algebraic over field $F$, where $K$ is field extension of $F$, then any polynomial $p$ with coefficients in $F$ and $\deg p=n$, we can find an $n\times n$ matrix with entries from $F$?

Abhas Kumar Sinha

I'm too much frustrated and I don't know what to do in order to be good in that

Alessandro Codenotti

@Silent That question looks incomplete

Secret

You don't need to be good to take a bath and sleep

Abhas Kumar Sinha

@Secret hehehe

Secret

lol

Tuki

16:45

helo

Abhas Kumar Sinha

I took a coaching for JEE and brought a of material of 32067 pages which is to be completed in a month, but I'm not able to solve a single problem properly

Life sucks

Silent

@AlessandroCodenotti no i mean that, as above , $w$ and $w^2$ are roots of $x^2+x+1$, and we get a matrix of real matrix does this happen always?

Alessandro Codenotti

@Silent So you want the minimal polynomial of $F$ to have $\alpha$ as a root? Who's $n$?

watchme

Why is there no common language in math like latin?

Isn't it "not good" to have German people with their own vocabulary for math-things and English people with theirs? Why is there no common language?

in medicine this is no problem...

Silent

@AlessandroCodenotti yes, i think that's what i want to ask

So, does that always happen?

we always get matrix?

Alessandro Codenotti

17:00

Math notation is largely international @watchme

Silent

@AlessandroCodenotti, for example, $3\times 3$ real matrix exists with char polynomial $x^3-1$. Does this always happen?

Alessandro Codenotti

Yes, look up the so called companion matrix of a polynomial

Silent

Wow! This was amazing jackpot :) Thank you so much

Secret

17:19

I wonder if the cube of companion matrix has any significance :P

Mary Star

We have the weights of coins: 50ct ~7,8g, 1€~7,5g, 2€~8,5g.

If we have that the total weight is 143,5 g, what is the value of coins?

Let x,y,z the number of each coins (50cents, 1 Euro, 2 euro), then we have that 7.8x + 7.5y + 8.5z = 143,5, right?

What we want to calculate is x/2 + y + 2z.
How can we caluclate x, y, z?

Secret

Impossible Active Audio Noise Cancelling by Muzo

►

Mary Star

Do we have to solve 7.8x + 7.5y + 8.5z = 143,5 for one variable and then use inequalities, so that x,y,z are positive?

Does someone of you have an idea?

Secret

uh, that's one equation in 3 unknowns, so it is not going to have a unique solution?

Semiclassical

equivalently, you've got 78x+75y+85z=1535 where x,y,z are integers

Mary Star

17:27

Yes! But how can we caluclate from that x,y,z? @Semiclassical

Semiclassical

@Secret the fact that there's an integer number of coins is probably the saving grace

Secret

Ah, so this is a linear diophataine equation

Semiclassical

yeah, that's the question

i'm not entirely convinced there's going to be a unique solution

Mary Star

Accoring to wolfram there is a unique integer solution

But how can we find it?

Semiclassical

hmm

Secret

17:29

Diophantine equation

In mathematics, a Diophantine equation is a polynomial equation, usually in two or more unknowns, such that only the integer solutions are sought or studied (an integer solution is a solution such that all the unknowns take integer values). A linear Diophantine equation equates the sum of two or more monomials, each of degree 1 in one of the variables, to a constant. An exponential Diophantine equation is one in which exponents on terms can be unknowns. Diophantine problems have fewer equations than unknown variables and involve finding integers that work correctly for all equations. In mor...

Semiclassical

well, you can rearrange that to 78x=1535-75y-85z

Secret

You need to find one solution, and then the other solutions are given by integer multiples of their quotient or something

Semiclassical

and the RHS is divisible by 5, so x has to be a multiple of five

so let x=5x' and then divide through by 5 to get 78x'+15y+17z=307

Secret

hmm... the remaining coefficients are pairwise coprime, I am not sure how that helps

Semiclassical

no, they're not. 78 and 15 have 3 as a common factor

Secret

17:32

ah right...

Semiclassical

i guess you can take that equation mod 3 to get 2z=1 mod 3

and multiplying both sides by 2 gives 4z=z=2 mod 3

so z=3z'+2

in which case you can simplify that to 78x'+15y+51z'=273 -> 26x'+5y+17z'=91

and then you really do have pairwise coprime coefficients

Alessandro Codenotti

Hi @Ted

Secret

I need to revise whether bezout's identity works for 3 terms

gcd(26,5,17) =/= 91 though... hmm...

gcd(x',y,z')=91?

Samuel Yusim

boop

Semiclassical

i guess at this point you can note that 0 <= x' <= 3 since 26(4)=104

so there's not so many choices possible for x', and similarly 0<= y <=17, 0<= z' <= 5

Mary Star

17:40

I understand!

Is this only the way? I mean is there also an other way, to see this as a diophantine equation?

Semiclassical

so in principle one could do the brute force task of plugging in all possible x',z' into 91-26x'-17z' and see which choices are a multiple of 5

Daminark

Hey everyone!

Secret

brilliant.org/wiki/linear-diophantine-equations-one-equation

hmm....

7.8x + 7.5y + 8.5z = 143,5 should be 78x + 75y + 85z = 14350?

Mary Star

Wy not 1435 ? @Secret

Secret

is 143,5 143.5?

Mary Star

17:46

143.5

Secret

ah ok, so the equation after multiply by 10 is 78x + 75y + 85z = 1435

Mary Star

Yes

Secret

So according to diophataine analysis in that link, I need to first check gcd(78,75,85)

and check if it divides 1435

Mary Star

Is the only way to solve that seeing it as a diophantine equation?

Can we maybe get an additional information that 2 Euro = 2*1 Euro = 4*50 cents ?

Secret

are x,y,z all euros or some euros or cents?

Mary Star

17:49

x is the number of 50 cents coins, y the number 1 euro coins and z the number of 2 euro coins

Can we get an infomation from that? Do you have an idea?

Secret

so that means, converting all of them into cents, they will become x,y,z=x',2y',4z'

and this give us the equation:
78x'+150y'+340z'=1435

Mary Star

Ok! And what do we have then?

Secret

wait a minute, is it supposed to be that complicated. I think I need to look at the original question again (why do we have $7.8 of 50 cents to start with?)

because logically if only one of these is cents, then the decimal part can only be contributed by cents

so I found it strange why the number of each type of coin is multiplied to some decimal number

Mary Star

We have the weights of coins: 50ct is 7,8gram, 1€ is 7,5gram, 2€ is 8,5gram.

In a box we have a weight of 143.5 g.

We want to know the value of coins in the box.

Secret

ah grams... ok, there's no shortcut then

Mary Star

17:56

So do we have to convert everything into cents?

Or what do we have to here, exept using diophantine equations?

Secret

I thought the stuff on the RHS is monetary value.
No, the weight of each coin has no easy/integer correlation with its weight (as you can see in the given question) thus conversion is useless here

Mary Star

Ah ok

Secret

So yeah, the only efficient way to solve it is to solve 7.8x+7.5y+8.5z=143.5 as a diophataine equation

So that means, the first step is to find gcd(78,75,85) and see if it divides 1435

78=2*3*13
75=3*5*5
85=5*17

thus gcd=1 uh... that's not very useful...

yeah... I think Semiclassical's way is the most efficient we have there

1435=5*7*41

hmm, 78 and 85 are coprime with each other so 78x+85y=1. Meanwhile we have 78x+85y=1435-85z thus perhaps 1=1435-85z and z = ....
ugh, that does not work

GFauxPas

18:14

Yay done with my last assignment

MatheinBoulomenos

@Silent you can always construct a $n \times n$ matrix with a given monic polynomial $f(x)=x^n+a_{n-1}x^{n-1}+\dots + a_0$ as a characteristic polynomial:

$$\begin{pmatrix}0 & & &\dots & 0 & -a_0\\
1 & 0 & &\dots & 0 & -a_1\\
0 & 1 & 0 & \dots &0& -a_2\\
\ddots &\ddots&\ddots &\ddots &\ddots &\vdots\\
0 & &\ldots & 0 & 1 &-a_{n-1}
\end{pmatrix}$$

Silent

@MatheinBoulomenos Thank you so much for this!

MatheinBoulomenos

you will need Laplace expansion and induction if you want to prove this

Secret

78x + 75y + 85z = 1435
3x + 0y + 0z = 0 mod 5
0x + 0y + 1z = 1 mod 3
hmm...

so 5 | x and 3 | z

Thus we have:
78x' + 25y + 17z = 287

Daminark

Hey @Mathein!

MatheinBoulomenos

18:22

hey @Daminark

Secret

where x'=5x

Mary Star

Ah ok!

Secret

there should be some other ways to reduce it. Still thinking

correction z = 1 mod 3 means z is in the equivalence class 3n + 1

so z = 4,7,10,13,16,... and x = 5,10,15,20,25,...

0x + 10y + 7z = 5 mod 13

10x + 7y + 0z = 7 mod 17

now does bezout identity work in modular arithmetic?

user193319

If $f \in L^p(\Bbb{R}$ (i.e., $|f|^p$ is integrable), does this mean that $f$ vanishes outside some bounded set almost everywhere?

Secret

So we start with the linear diophataine equation:

78x + 75y + 85z = 1435

and this can be resolved into the following system of congruence relations:

3x + 0y + 0z = 0 mod 5
0x + 0y + 1z = 1 mod 3
0x + 10y + 7z = 5 mod 13
10x + 7y + 0z = 7 mod 17

Thus we have:

x = 5m
z = 3n + 1
(solving)
(solving)

MatheinBoulomenos

18:35

@user193319 consider $f(x) = \frac{1}{x^2}$ if $|x|>1$ and $f(x)=1$ if $|x| \leq 1$ this doesn't vanish anywhere and it's in $L^p(\Bbb R)$ for all $p \in [1,\infty]$

Secret

Using x we can go further and get:

78x' + 25y + 17z = 287

which we should be able to repeat the same game to get:

MatheinBoulomenos

@user193319 or you can take a Gaussian $e^{-x^2}$ that doesn't vanish anywhere and is in $L^p$ for all $p\in [1,\infty]$

user193319

@MatheinBoulomenos Ah, I see. Thanks for the examples!

Secret

0x' + 1y + 2z = 2 mod 3

3x' + 0y + 2z = 2 mod 5

10x' + 8y + 0z = 15 mod 17

MatheinBoulomenos

@Secret @MaryStar the systematic approach to solving a linear diophantine equation $a_1x_1+ \dots +a_nx_n=c$ if $\gcd(a_1, \dots, a_n)$ divides $c$ is this:
- use the extended Euclidean algorithm to find a solution for $a_1x_1+ \dots +a_nx_n=\gcd(a_1, \dots, a_n)$
- multiply that solution by $\frac{c}{\gcd(a_1, \dots, a_n)}$

Secret

18:43

the problem here is that for 78x + 75y + 85z = 1435, the gcd is 1

hmm...

uh... 78x + 75y + 85z = 1??

and x,y,z > 0?

MatheinBoulomenos

yeah, you won't always have positive solutions with that approach

Secret

hmm...

MatheinBoulomenos

but once you have one solution for 78x + 75y + 85z = 1435, you can get the other solutions if you add solutions of 78x+75y+85z=0

Secret

right

mercio

19:16

o. .o

geocalc33

can you have a function that has a real variable in it and a complex variable in it?

for example: (w)^(1/log(x)), where w is a complex number and x is a real number

Leaky Nun

sure, why not

geocalc33

and are the roots of a complex function where the real part and complex part both equal zero at the same time?

Leaky Nun

@geocalc33 real part and imaginary part

and yes

GFauxPas

19:43

@geocalc33 just remember that that expression is really an infinite set of functions unless you choose a convention as to which branch you choose

jcora

hey all I'd like to know whether this would be a suitable question here.

I have an equation for a signal x, given as dx/dt = (B - x) * max(0, input) + (D - x) * max(0, -input). the purpose is to control the gain of the signal x so that it's in the interval [D, B]. what the above expression does is basically, for positive inputs, increase x until it reaches B, and afterwards decrease it. vice versa for a negative input that drives x below D

my question is whether this equation can be written without using max, i.e. only via basic algebraic operations, and why (not). for example, I thought …

also I should add that the exact behavior of the equation is not that important, all that matters is that the signal is driven in the correct direction and driven back when it passes the bound (possibly proportionally to the distance from the bound)

basically is this an ok thing to ask here or is it too engineer-ish? I am actually interested in the math behind it otherwise I wouldn't really bother:)

GFauxPas

20:17

Sounds mathy enough to me, to ask on the site @jcora

They can always migrate it to physics.se if ppl don't think so, but it seems fine to me

jcora

20:31

awesome I'll post it later!

jcora

20:54

here math.stackexchange.com/questions/2785551/…

if someone feels like it, help me format:)

I don't know latex

Ted Shifrin

21:13

@jcora: Perhaps you didn't know this — $\max(x,y) = \frac12\big(x+y+|y-x|\big)$.

Balarka Sen

Yeah that's a weird identity.

Alessandro Codenotti

Hi @Ted and @Balarka

Leaky Nun

@TedShifrin interesting

Ted Shifrin

Hi @Alessandro, mr @Balarka, and @Leaky

Balarka Sen

Hi @Alessandro, @Ted.

Alessandro Codenotti

21:14

Hi @Leaky

Leaky Nun

Hi

mercio

h

Ted Shifrin

Well, at least Kasmir is still alive ...

MatheinBoulomenos

Weird fun fact: there exists a normal subgroup $N$ in $\operatorname{SL}_{2}(\Bbb Z)$ such that $\operatorname{SL}_{2}(\Bbb Z)/N$ is isomorphic to the monster group

Ted Shifrin

Agh. That's too monstrous.

Balarka Sen

21:22

Jesus fuck

Alessandro Codenotti

How's that fun

Leaky Nun

@MatheinBoulomenos :o

really

mercio

I think it's fun

Leaky Nun

same here

@MatheinBoulomenos citation-needed

mercio

isn't $SL_2(\Bbb Z)$ freely generated by a $2$-cycle and a $3$-cycle

Balarka Sen

21:23

Yep

Leaky Nun

so then the monster group is di-generated?

mercio

so we only need to show that the monster group can be generated by a $2$-cyc

welp

MatheinBoulomenos

yes

wait, I think that's $PSL_2(\Bbb Z)$

Leaky Nun

@MatheinBoulomenos still citation-needed

GFauxPas

Wilson asserts that the best description of the Monster is to say, "It is the automorphism group of the monster vertex algebra". This is not much help however, because nobody has found a "really simple and natural construction of the monster vertex algebra".[1]

MatheinBoulomenos

21:25

24

A: Generating finite simple groups with $2$ elements

Since I happen to know the OP is number-theoretically inclined, let me add the following remark: For "most" finite simple groups $G$ it is indeed the case that $G = \langle x, y \rangle$ where $x$ has order $2$ and $y$ has order $3$. Equivalently, $G$ is a quotient of the free product $\mathbb{...

Leaky Nun

never mind

MatheinBoulomenos

Pete Clark doesn't give a reference on the fact that the monster group is generated by an element of order 2 and an element of order 3, but apparently that's "well-known"

Alessandro Codenotti

For some broad definition of well-known

mercio

that's a pretty large definition of "well-known"

Alessandro Codenotti

sniped!

mercio

21:27

>..>

MatheinBoulomenos

also note the geometric consequence: if let that normal subgroup act on $\overline{\mathcal H}$ by Möbius transforms, the quotient is a modular curve that is a branched covering over $\Bbb P^1(\Bbb C)$ with the monster group as the group of deck transformations

Ted Shifrin

That's enough to make me retire completely.

MatheinBoulomenos

don't you like Riemann surfaces? :P

Balarka Sen

Lmao

mercio

but

would the normal subgroup contain a congruence subgroup

MatheinBoulomenos

21:37

no idea. That subgroup probably isn't very easy to describe

I kind of want to study modular forms wrt this subgroup just because it's so weird

Leaky Nun

the subgroup is just the kernel of the projection :)

Alessandro Codenotti

@Mathei do you have five minutes to think about $\Bbb Q(\sqrt{-31})$?

MatheinBoulomenos

okay

Alessandro Codenotti

Ok, the minkowski bound is between 3 and 4

$(2)$ is split, $(2)=\mathfrak p_1\mathfrak p_2$ with $\mathfrak p_1=\left(2,\frac{1+\sqrt{-31}}{2}\right)$ and $\mathfrak p_2=\left(2,\frac{1-\sqrt{-31}}{2}\right)$. $(3)$ is inert.

anakhronizein

@TedShifrin is there a (simple) way to deduce from the tensor algebra quotient definition of the exterior algebra that $(\alpha\wedge\beta)_p(v,w) = \alpha(v)\beta(w) - \alpha(w)\beta(v)$?

For 1-forms $\alpha,\beta$

MatheinBoulomenos

21:44

$\mathfrak{p}_1$ and $\mathfrak{p}_2$ are inverses of each other, so the class group is generated by the class of $\mathfrak{p}_1$. You only need to figure out the order

Alessandro Codenotti

I know that $\mathfrak p_1$ and $\mathfrak p_2$ are each other inverse in the class group, so I only need to decide whether they are the same, that is whether $\mathfrak p_1^2$ is principal or not

well the order of $\mathfrak p_1$ actually

right

anakhronizein

I thought it would be inherited from the tensor product of maps

$(\alpha\otimes \beta)(v,w) = \alpha(v)\beta(w)$

Akiva Weinberger

38

Q: General request for a book on mathematical history, for a VERY advanced reader.

I am aware that there are answered similar questions on here, however I am specifically after a text that would be engaging for a professor of mathematics, also FRS. He is unwell and in the hospital, and I would like to get him something to pass the time. However anything aimed at undergraduate ...

@TedShifrin

Book recommendation for OP to give to a friend, a professional mathematician in the hospital

Michael.P

Guys. Silly question. How do I prove something as simple as this?
$$\sum_{k=0}^n a_k=\sum_{k=m}^{n+m} a_{k-m}$$

Akiva Weinberger

Define $k'=k+m$.

Michael.P

21:49

I don't follow

Akiva Weinberger

We have $k=k'-m$. Substituting it in, we get:

$\displaystyle \sum_{k=0}^na_k=\sum_{k'-m=0}^{k'-m=n}a_{k'-m}$

or $\displaystyle\sum_{k'=m}^{n+m}a_{k'-m}$

and then we drop the primes (i.e. substitute in $k$ for $k'$)

@Michael.P Alternate idea: Use induction

Prove it for $n=0$, and prove that if it's true for $n$ it's true for $n+1$

That seems potentially simpler

$\displaystyle\sum_{k=0}^0a_k=a_0$ and $\displaystyle\sum_{k=m}^ma_{k-m}=a_{m-m}=a_0$, so it's true for $n=0$

Michael.P

@AkivaWeinberger, I belive I explored similar proof, and found it to be erroneous (it relies on self-reference; e.g. like when you substitute in equation a variable from that same eq.) I'm talking about substitution

Ted Shifrin

DogAteMy: I'm not sure I know of any "advanced" math history texts.

Balarka Sen

History of Inter Universal Teichmuller Theory maybe?

Michael.P

@AkivaWeinberger I tried that. I can't properly 'unfold' the right hand side so that part of it is clealy the original eq.

Akiva Weinberger

21:55

If $\displaystyle\sum_{k=0}^na_k= \sum_{k=m}^{m+n}a_{k-m}$, then \begin{align}\sum_{k=0}^{n+1}a_k&= \sum_{k=0}^na_k+a_{n+1}\\&=\sum_{k=m}^{m+n}a_{k-m}+a_{(m+n+1)-m} \\&= \sum_{k=m}^{m+n+1}a_{k-m},\end{align} finishing the induction proof.

anakhronizein

@BalarkaSen perhaps you have an idea as to my question above? Let me recopy it.

Ted Shifrin

It's not about induction. It's about algebraic substitution.

Akiva Weinberger

You can't replace $k$ with $k^2$ and expect it to work, though

anakhronizein

is there a (simple) way to deduce from the tensor algebra quotient definition of the exterior algebra that $(\alpha\wedge\beta)_p(v,w)=\alpha_p(v)\beta_p(w)−\alpha_p(w)\beta_p(v)$, $\alpha,\beta$ being 1-forms.

Akiva Weinberger

whereas the substitution proof would seem to suggest that you could

so I think this is better

Balarka Sen

21:57

@anakhronizein Hmm. I'd just prove that basis-wise. Hmmm.

Ted Shifrin

@anakhronizein: Well, with the quotient definition, you need to see that $\alpha\otimes\beta-\beta\otimes\alpha$ represents the equivalence class.

Akiva Weinberger

(That is, $\displaystyle\sum_{k=0}^na_k\ne\sum_{k=0}^{\sqrt n}a_{k^2}$ in general.)

Balarka Sen

lol @Akiva

anakhronizein

Which equivalence class?

Ted Shifrin

Well, DogAteMy, we need a bijection from one interval of integers to the other, thanks.

Akiva Weinberger

21:58

Besides, $\sum$ is defined inductively, so.

Ted Shifrin

Blah.

I've never been that pedantic.

Akiva Weinberger

In any case, it is proven.

Balarka Sen

Proven it is

Ted Shifrin

@anakhronizein: Elements of a quotient are equivalence classes.

anakhronizein

Yes.

Ted Shifrin

21:59

So spaketh DogAteMy.

Balarka Sen

Thus Spake Akivathustra

anakhronizein

In this case we are quotient-ing by the double sided ideal generated by the relations $v\otimes w + w\otimes v$

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$Mathematics$ Mathematics