12:01 AM
I'm encountering the above statement as part of a proof of this very theorem (roots vary continuously with the coefficients), which is leaving me quite confused

Hmm, that doesn't sound good.
So they're saying that mere boundedness of the set of roots should be more basic.
I see. You should be able to do a basic estimate to prove that.
The same way you can prove that $|f(x)|\to\infty$ as $|x|\to\infty$ when $f$ is any polynomial.

How does that guarantee a bound that works for all $t$ simultaneously?

All you need is bounds on the coefficients for this. So $t$ in a compact interval and the $a_i(t)$ continuous gives you that.

12:19 AM
Oh, I see. That works indeed. Thanks a lot.

You're welcome. Context helps :)

12:38 AM
Hello!

1 hour later…
1:58 AM
@AkivaWeinberger Yes. Sorry. $x^2-\alpha^2$ irreducible over $\Bbb Q$, hence it is field of dimension 2 over $\Bbb Q$. Is this correct answer?

1 hour later…
3:13 AM
Having trouble formulating an accurate linear program. I keep getting a result that the solution space is unbounded, which is absurd because the task is "find a position for a middle hub such that the distances to three nearby hubs in minimized"

1 hour later…
4:33 AM
(I was forgetting to make the basis columns elementary)

4 hours later…
8:29 AM
If $f$ is continuous function $f:\Bbb R\to\Bbb R$ then it has antiderivative. But, does continuous function $g:\Bbb C\to\Bbb C$ also has antiderivative?
OH!! Please verify this: a complex function that is continuous, but not differentiable, can't have antiderivative, because complex differentiable function infinitely differentiable.

8:56 AM
@Silent right

9:09 AM
In Brent's factorization method, what does $x_o$ represent? Is it an arbitrary integer that is usually taken as 2? maths-people.anu.edu.au/~brent/pd/rpb051i.pdf

9:34 AM
@Silent yes that's correct; an irreducible polynomial $f$ over $\Bbb Q$ generates a maximal ideal of $\Bbb Q[X]$ and the quotient by this ideal is a field containing a root of $f$

9:48 AM
To be checked: is there a way to invert a Möbius strip. What will inversion mean for a one sided object?

10:12 AM
1

Problem: Let $F$ be a field with $\text{char} F = p$ for some prime $p$. Show that if $X^p - X - a$ is reducible in $F[X]$, then it splits into distinct factors in $F[X]$ Solution in back of the book: This solution confuses me slightly. Why is $f(X)$ splitting as two monic polynomials? ...

1 hour later…
11:16 AM
@ÍgjøgnumMeg , LeakyNun, thank you very much.

11:49 AM
@Silent it's actually one of my favourite little proofs from field theory

12:00 PM
so... on facebook messenger there are well over a thousand groups that sell lsd, weed, you name it, where is SE on this you people have REALLY dropped the ball here
I mean just the traffic alone if you multiply that globally is a honeypot and it's all going to zucc
anyhoo numberphile have some great dvds on youtube, I gotta ask.. why.... why are his pupils so dilated? it's almost as if im peering into another parallel chronology of myself in another universe in the multiverse
@JennaSloan omg are you related to neal sLOan? him and I are like the best of pals he even replied to an email once albeit it ended in him threating to call the police but I feel as if our friendship can move past that

Because reality itself is unravelling thanks to Endgame

ur such a creep dude
don't start triggering end of days anxiety

oh. I officially delete previous comment

12:15 PM
and I nuked mine lol

i don't mind either so long as get a ticket on on a space station to watch all the idiots die on live stream but it seems I'm not going to be getting much from Elon seeings i troll the #@$# out of him i retract that comment also The death of humanity is not enough for me, I want the permanent death of totality itself. Only by the erasure of God himself will this boring cycle stops repeating (even death is too light a word, no words can describe that) but real life hunger games.... even if i get left on the surface I'm totally going win guerrilla warfare with dark ages regulations i mean... yeah i think ive got a handle on that being on welfare for past few years I'm not actually kidding here people will literally try to stab you for a$20 note

I wish 2020 will arrive sooner. I need that Global Financial Crisis to happen

man that amazon game is killing me i need to turn my music off i don't get it

12:25 PM
When the current dysfunctional economy that fuel all these inequality crumble, humantity will finally get to their sense
Only that GFC is powerful enough to crumble the global market, the lifeblood of this economy

ah it wouldn't crumble if the control freaks accepted the that although fraught with potential problems, a free market is the optimal solution for humanity
you guys do a GFC scare every decade its not a fake news swindle that works ffs it's almost on the same level of lame as Africanised killer bees
anyhoo you're disturbing my deep state of focus on amazon

lol true, for the thousand of years, the economy always recover after it has crumbled

Does anyone have an example of a field of characteristic $p$ that isn't isomorphic to $\Bbb{F}_p$ or $\Bbb{Z}_p$ (these are basically the same fields)?

1:15 PM
@user193319 $\Bbb F_{p^2}$, $\Bbb F_p(t)$

1:29 PM
Suppose I need to design an experiment where I can test a set of people all at once, but the cost of each test increases as the number of people in the test increases. I also need to make sure that, for the total population of N people, every subset of this population of size k is in at least one of the test sets.

Is this a type of problem that's already been studied? Ideally, I'd like to find a rough bound for the maximum number of tests I need to run as a function of N and the size of each test set.

@user193319 Try GF(p)?

google-fu tells me this might be called "covering design"

2:15 PM
@user193319 $\Bbb Z/(p)$ is a much better notation than $\Bbb Z_p$ in this context (which is just $\Bbb F_p$)

I feel like the universe occasionally gets into its head that I should understand Teichmüller theory
But I really don’t.

It's interuniversal
you have to commune with your counterpart in another universe

Amusingly enough, I am running into the phrase “universal Teichmüller space” lately

lol

Gotta make sure to use quotes when googling that lest I get IUT results as well

2:29 PM
@ÍgjøgnumMeg That's too much latex code to type.
@Secret What is GF(p)?

Galois field of p elements
tbh I don't really understood them, other than have read the cayley table of GF(4) before

Isn't a field of $p$ elements just $\Bbb{Z}_p$?

Residue field of the ring of integers of a local field has char p
or rather
non arch local field

3:24 PM
Hello
5

Hi.
5

Given $$a + bx + cx^2 +dx^3 + x^4 = [(x-p)^2 + A(x-q)^2][B(x-p)^2 + (x-q)^2]$$, where $a,b,c$ and $d$ are integers. Find $p,q,A$ and $B$ in terms of $a,b,c$ and $d$. I have expanded both sides and compared which got me a messy system of equations. I wanted to know if there is any easy method to solve this or any key observation. Thanks.

3:55 PM
Hey all. If I have A and B, and performed A + B, and I can prove I have the correct A and B, how can I prove that I actually did the operation A+B?
This is a simple case for a more complex idea related to homomorphic operations

what
"performed A+B"??
what is a "correct" A or B?

Let A = 5 and B = 4, I can prove that they are exactly 5 and 4, and I did the summation of the two, but how can I prove that I did the summation and not some other operation

What do you mean that you "can prove that they are exactly 5 and 4"?

I can prove that the integrity of these numbers before summation is maintained (i.e. they have not been changed to another integer)

Why would they be changed?

4:06 PM
I feel like this may be the wrong chat room for homomorphic operations

4:21 PM
Well if you wanted help really all that badly, you could explain the context.

4:41 PM
Let $F:\mathbb{R}^2 \rightarrow R$ be continuous and suppose there exists $C>0$ such that $|F(t,s_1)-F(t,s_2)|<C|s_1-s_2|$ for all $(t,s_1),(t,s_2) \in \mathbb{R}^2$.

For $f \in C([a.b])$, define the function $g$ on $[a,b]$ by $g(x)=\int_a^x F(t,f(t)) dt$. How do I show that $g \in C([a.b])$?

4:54 PM
Let $(t_n)_n$ be a real sequence such that $t_n\rightarrow t$. Consider $|F(t_n,f(t_n))-F(t,f(t))|\le|F(t_n,f(t_n))-F(t_n,f(t))|+|F(t_n,f(t))-F(t,f(t))|\le C|f(t_n)-f(t)|+|F(t_n,f(t))-F(t,f(t))|$.

@Thorgott How does this help?

You can use this to conclude that $F(t,f(t))$ is continuous in $t$ and the statement then follows from the fundamental theorem of calculus

@Thorgott Ok, but how do I control $|F(t_n,f(t))-F(t,f(t))|$ so that $|F(t_n,f(t_n))-F(t,f(t))|\leq \epsilon$?

$F$ is continuous

5:10 PM
@Thorgott Yes. Ok, thank you I got it.

5:20 PM
Actually, we have $(t_n,f(t_n))\rightarrow (t,f(t))$ regardless due to the continuity of $f$, so this should already imply $F(t_n,(f_n))\rightarrow F(t,f(t))$, hence the continuity, meaning that the assumption on the existence of $C$ could be dropped entirely. Not sure whether I'm overlooking some subtle detail or the assumption is actually unnecessary.

@Thorgott I think this is required in the second part.
(b) Show that $T:C([a,b]) \to C([a,b])$ defined by $Tf(x)=\int_x^a F(t,f(t))$ is a continuous map.

5:56 PM
I see. That part certainly uses the constant.

@Thorgott The first part also show that $Tf$ is well defined.

Indeed. That's a neat exercise to be honest.

@Thorgott I am now fairly convinced that part $(a)$ does not the use the condition on $C>0$.

So am I. The thing that made me skeptical is that not using a hypothesis is usually a very good indicator of doing something wrong, but since it gets used in part $(b)$ and I believe the argument above is sound, I'm convinced now.

6:13 PM
How to visualize a 3 x 2 matrix as transformation between dimensions ?

6:24 PM
If $f(x)=\frac{\sin[x]}{[x]}$ if $[x]\neq 0$ and $f(x)=0$ if $[x]=0$, where $[x]$ denotes the greatest integer less than or equal to $[x]$, then $\lim_{x\to 0} f(x)$ is: (A) 0 (B) 1 (C) $\infty$ (D) does not exist
I got the left hand limit to be equal to $\frac{\sin(-1)}{-1}$ and the right hand limit to be equal to be to $\frac{\sin(0)}{0}$.
What do I conclude from this?

What is $[x]$ for $x$ close to $0$?

Depends from which side with respect to 0 you are approaching 0.
If you are approaching from the left hand side of 0 then $[x]=-1$ and if you are approaching from the right hand side of 0 then $[x]=0$.
@Thorgott

Right. Then what is $f(x)$ when $x$ approaches $0$ from the right? You already got the left-hand limit correctly.

@MrAP I think the answer should be (D) i.e. does not exist as $RHL = 1$

The right-hand limit is not $1$

6:33 PM
What do I make of a situation where the limit come out to be $\frac{0}{0}$.
Is it undefined, indeterminate or does not exist or what?

Well, $0/0$ is not defined and strongly indicates you made a mistake somewhere. Look at the definition of $f(x)$ again. It makes a special exception so that this case actually does not occur.

But the right hand limit does come out to be $\frac{0}{0}$

@Thorgott There is a third part as well. It says that show that $T(B)$ is an equicontinuous family of functions in $C([a,b])$ provided $B \subset C([a,b])$ is a bounded set. I am having trouble estimating $\int_x^y |F(t,f(t))| dt$ provided $|f(t)|<M$.

For $x\in[0,1), [x]=0$ and so $\frac{sin[x]}{[x]}=\frac{0}{0}$
Well how did you differentiate the greatest integer function?

@MrAp As you correctly noted, when $x$ approaches $0$ from the right, $[x]$ is eventually $0$, not only approaches it. How is $f(x)$ defined when $[x]=0$?

6:37 PM
@Apptica

@MrAP, I made a mistake here.

The greatest integer function is not differentiable at the integers, so applying L'Hospital is nonsensical.

@Thorgott Exactly, don't know what I was thinking

I made a mistake there above. It would be for $x\in (0,1)$
For $x=0$ the function is defined.
But still the right hand limit is $0/0$.

The definition you posted says $f(x)=0$ when $[x]=0$.

6:42 PM
Oh shit!
I did not realize that.
Then the right hand limit is 0.
@Thorgott
?

Yup, that's right

So the limit does not exist.

@MrAP How ?

Because L.H.L is not equal to R.H.L.

How right hand limit is 0 ?

6:47 PM
Because whenever $[x]=0$, $\frac{\sin [x]}{[x]}=0$ as per the definition and $[x]=0$ for $x\in [0,1)$

@MrAP yeah the definition!

@Thorgott, I wonder what would be the case if this was not a piecewise function. I mean $f(x)=\frac{sin[x]}{[x]}$ for all $x \in R$. What do you think will be the R.H.L in that case.

The definition does not say $\frac{\sin[x]}{[x]}=0$ whenever $[x]=0$; it says $f(x)=0$ whenever $[x]=0$. This is an important distinction as that fraction is definitely undefined.
That $f(x)$ would not be defined in the interval $[0,1)$ so it makes no sense to ask what its limit is (at least the RHS limit; the LSH limit remains $\sin(1)$).

Oh yes Thorgott. It should be $f(x)$
Would it be undefined or indeterminate?
I always get confused between the two.
As per my knowledge any real number divided by 0 is undefined and there are seven indeterminate forms.
Like 0/0, $0.\infty$,etc.

@Thorgott when we are calculating the right hand limit of $f(x)$ then it means that $x$ is a positive number close to $0$ , which means that $[x] = 0$, and if $[x] == 0$ then $f(x)$ is also $0$. So don't you think the right hand limit will be $0$ ?

6:54 PM
@Thorgott Does this result follow from part $(a)$ as we showed that $F(t,f(t))$ is continuous over $[a,b]$? Then, $\int_x^y |F(t,f(t))| dt \leq int_x^y |F(t_0,f(t_0))| dt=N|x-y|$. But then, where did we use the fact that $|f(t)|<M$.

Undefined, because both the nominator and denominator would be constant $0$ in the interval $[0,1)$. Indeterminate is a term only used to describe limits and an indeterminate form like $0/0$ means that we are looking at an expression $f(x)/g(x)$ as $x$ approaches something and we have that both $f(x)$ and $g(x)$ individually go to $0$. The expression $f(x)/g(x)$ in those cases is usually not defined at the point $x$ approaches, but only in a neighborhood thereof.
@Apptica Yes, in the original question, the right-hand limit is $0$.
@user330477 The issue is that we need an upper bounded that works for all $f\in B$ simultaneously. Give me a minute to think about it.

@Thorgott Sure. Thank you.

If $G \simeq S_3$ and $g_1,g_2 \in G$ are of order $2$2 and $3$, respectively, can I say that $g_1 \mapsto (1,2)$ and $g_2 \mapsto (1,2,3)$ defines an isomorphism?

@user330477 I believe the following should work. Since $B$ is uniformly bounded, we have an $M$ s.t. $|f(t)|<M$ for all $t\in[a,b]$ and $f\in B$. So $f(t)$ lies in the compact interval $[-M,M]$ for all $t\in[a,b]$ and $f\in B$. Then $(t,f(t))$ lies in the compact set $[a,b]\times[-M,M]$ for all $t\in[a,b]$ and $f\in B$. Thus, we can bound $|F(t,f(t))|$ above for all $t\in[a,b]$ and all $f\in B$ simultaneously. This should suffice after choosing a sufficiently small $\delta$.

@Thorgott How did you get that $B$ is uniformly bounded?

7:06 PM
Wasn't $B$ being a bounded set a hypothesis?

@Thorgott Yes. So, this part also does not use the condition on $C$.

It doesn't seem like it. I also don't see any point in this part where $C$ would be applicable, let alone helpful.

@Thorgott I am not very convinced with the proof of the third part.

@user193319 What about the other elements of $G$?

Take the homomorphic extension...but that presupposes that $g_1$ and $g_2$ generate $G$...
Hmm...

7:13 PM
Well, which part does not convince you?

@Thorgott I meant to say this is probably not the expected answer. I know it is rigorous and convincing. But it requires knowledge of topology, which we have not talked to much about in my course.

Yeah, you haven't specified whether $\langle \{g_1,g_2\}\rangle = G$.

No, I don't know that, actually...But I feel like it should be true because I already know that $G \cong S_3$.

You mean whether any two elements of order 2 and 3 in $S_3$ must generate the group?

I think that's what it comes down to.

7:18 PM
That the continuous image of a compact set is compact (in particular, bounded) is a standard result. I'm not sure what other approach there even is to bound $F(t,f(t))$ as all we have to work with are continuity assumptions.

6

As James noted in his comment, generating sets are not unique, since if $A$ is a set that generates the group, then any set containing $A$ will also be a generating set. However, I assume you are trying to find a smallest set of generators. If you allow an element $g$ to be a generator, then eve...

@Thorgott And bounding $|F(t,f(t))|$ is the only way of showing that $T(B)$ is an equicontinuous family of functions in $C([a,b])$ provided $B \subset C([a,b])$ is a bounded set.

Does anyone know where I can find a proof that the symmetry group of a complete graph $K_n$ is isomorphic to $S_n$?

Do sin, cos, log work the same as they do in base 10 for some different base?

@user193319 Isn’t it kind of trivial?
Since the graph is complete, for any pair of vertices u and v, they are connected, and their permutes s(u) and s(v) must also be connected, so the biconditional is true (since both sides are always true).

7:30 PM
However, the fact I'm using can be proven rather succinctly (if you've seen a proof of the extreme value theorem, this is the same idea). Let $F:[a,b]\times[c,d]\rightarrow\mathbb{R}$ be a continuous function. Assume $F([a,b]\times[c,d])$ is unbounded. Then pick a sequence $(x_n,y_n)$ such that $|F(x_n,y_n)|\rightarrow\infty$. By Bolzano-Weierstrass, there is a subsequence s.t. $(x_{n_k},y_{n_k})\rightarrow(x,y)$. But by continuity, $F(x_{n_k},y_{n_k})\rightarrow F(x,y)$. Contradiction.

@user76284 I don't think so. Here's what I have. Let $\sigma \in S_n$. Define $\tilde{\sigma} : K_n \to K_n$ to be $\tilde{\sigma}(v_i) = v_{\sigma (i)}$ and $\tilde{\sigma}(e_i) = e_{\sigma (i)}$, where $v_i$ is a vertex, $e_i$ an edge. it is easy to show that $\tilde{\sigma}$ is a bijection, but showing that it preserves adjacency is a little tougher. I need to show that $Ends(e) = \{v_i,v_j\}$ implies $Ends(\tilde{\sigma}(e)) = \{\tilde{\sigma}(v_i),\tilde{\sigma}(v_j)\}$.
@user76284 Your proof isn't rigorous enough.

I’m going off the definition here: en.wikipedia.org/wiki/Graph_automorphism

does anyone have an answer to my question?

“an automorphism of a graph G = (V,E) is a permutation σ of the vertex set V, such that the pair of vertices (u,v) form an edge if and only if the pair (σ(u),σ(v)) also form an edge”

Well, I am going off the definition in the book I am using (Groups, Graphs, and Trees by John Meier).

7:33 PM
@Mathphile $\sin$ and $\cos$ are defined without any reference to a number base whatsoever, so I'm afraid the question does not make sense. Same applies to the logarithm unless you mean the base of the logarithm, in which case they are different functions.

Okay, but how does what you said prove that $Sym(K_n) \cong S_n$? And how do you know that the edge gets sent to the right place?

@Thorgott what i mean by base is the numeral system

According to my book, an automorphism (symmetry) is a map on the vertices AND edges.

for example binary numbers have base two number system

Are the edges labeled in any way? I find it easier to think purely in terms of vertices and then ask whether a particular pair is in the “edge set” or not.

7:36 PM
I don't know. But the person who answer my question here math.stackexchange.com/questions/3197468/…
said I have to worry about how the map sends vertices and edges.
I can't just look at how the automorphism maps vertices.

Ok, in that case I do not understand the question. The way we define the functions $\sin,\cos$ and $\log$ is completely independent of any numeral system.

I do know that in $K_n$ that there is a unique edge between any two vertices, so perhaps uniqueness will guarantee that $Ends(\tilde{\sigma}(e)) = \{\tilde{\sigma}(v_i),\tilde{\sigma}(v_j)\}$

The fact that sin(0)=0 doesn’t depend on us agreeing to use base-10

In simple graphs edges don’t really have any ontological status of their own. An edge is just a pair of vertices that is in the “edge set”.

The fact that we write sin(1)=0.017... may, at first glance, seem to depend on base-10

7:43 PM
I think edges matter according to my book. The book I am using is a book on Geometry Group Theory and their conventions are different from those of graph theorists. E.g., to a geometric group theorist, a graph is a 1 dimensional CW complex.
So it's actually a topological space.

But the only role that base-10 is giving in there is to give a convenient label for the two quantities involved
If you replace those labels by their base-2 equivalents, the statement remains true
As a mapping, tho, sine doesn’t care a whit about the representation you choose

Does this have symmetry?

Looks like it’s odd with respect to the middle of the interval?
Equivalently: if you take that figure and rotate it by 180 degrees, the purple part looks to be unchanged
Easiest way to check that, if you’re plotting a list of y-values, is to reverse that list and add it to itself. If it really is symmetric in that manner, then the resulting sum of lists should be identically zero

Yeah it is odd wrt to the middle

Neat.

8:00 PM
so that would mean $f(-x)=-f(x)$

hi @Semiclassical
Given $$a + bx + cx^2 +dx^3 + x^4 = [(x-p)^2 + A(x-q)^2][B(x-p)^2 + (x-q)^2]$$, where $a,b,c$ and $d$ are integers. Find $p,q,A$ and $B$ in terms of $a,b,c$ and $d$. I have expanded both sides and compared which got me a messy system of equations. I wanted to know if there is any easy method to solve this or any key observation. Thanks.

That looks wretched
@Shobhit only thing I see immediately is that (1+A)(1+B)=1, as can be seen by matching the highest powers on both sides

Yeah, this is one of the equations.
do you want me to post the rest?

plz no

you might spot something
$$1+A+B+AB=1$$
$$(1+AB)(p+q) + 2qA + 2pB = -d/2$$
$$(1+AB)(p^2+q^2+4pq) + 6q^2A + 6p^2B = c$$

8:17 PM
I think “plz no” was the correct reaction

$$(1+AB)(pq^2 + qp^2) + 2q^3A + 2p^3B = -b/2$$
$$(1+AB)(p^2q^2) + Bp^4 + Aq^4 = a$$
Just have a look

Where are you getting this problem from?

Its an assignment

Yuck

Any other ideas?

8:19 PM
Getting a,b,c,d in terms of p,q,A,B is simple enough
But at first glance I see no reason to expect the other direction to be tractable

Can you say anything about p,q,A or B, like they are integers or something'

No idea, but I wouldn’t count on it

Ok. i'll continue thinking
thanks

9:17 PM
@user193319 saying $\tilde{\sigma}(e_i)=e_{\sigma(i)}$ doesn't make sense. $S_n$ acts on $\{1, \dots, n\}$ and there are $\binom{n}{2}$ edges in $K_n$, so you can't index them by elements from the set on which $S_n$ acts
for a graph with no multiple edges, the notions for graph homomorphisms coincide, any map between vertices that preserves adjacency induces a unique compatible map on the edges

Hi.
Is true that $\displaystyle \int_C \frac{z}{1-\exp(z)} dz$ where $|z| = 4$ is 0?

9:41 PM
@MatheinBoulomenos can you help me with my question?

@Shobhit depends on the question

Given $$a + bx + cx^2 +dx^3 + x^4 = [(x-p)^2 + A(x-q)^2][B(x-p)^2 + (x-q)^2]$$, where $a,b,c$ and $d$ are integers. Find $p,q,A$ and $B$ in terms of $a,b,c$ and $d$. I have expanded both sides and compared which got me a messy system of equations. I wanted to know if there is any easy method to solve this or any key observation. Thanks.
If you look just 10 messages above, i have wrote what i got
after comparison

oh god, that looks horrifying
I have no idea

ok :(
This question is of 4 marks, and i have been doing it for 2.5 hours now

A function can be analytic even is not defined at a point, right?
In the complex sense.
E.g: $\dfrac{z}{1-exp(z)}$ is analytic even is not defined at $z=0$ since the limit at 0 exists, right?

9:51 PM
@Odestheory12 I wouldn't say that. I'd rather say that it can be analytically continued to $z=0$ or that it has a removable singularity there

10:04 PM
Yeah I see, ty.

10:18 PM
Anyone know of a ring with unit group isomorphic to $\mathbb{Z}/5\mathbb{Z}$?

@Rithaniel there is none

Really? How does one go about showing that?

Suppose $R$ is a ring with $R^\times$ cyclic of order $5$. Then $\mathrm{char}(R)=2$, because else $-1$ would have order $2$ in $R^\times$ contadicting Lagrange's theorem
so $R$ contains $\Bbb F_2$. Let $u \in R^\times$ be a generator and consider the subring generated by $u$, this a quotient of $\Bbb F_2[x]$ (via sending $x$ to $u$) and the kernel contains $x^5-1$, since $u$ has order $5$, so it is in fact a quotient of $\Bbb F_2[x]/(x^5-1)$
the factorization of $x^5-1$ over $\Bbb F_2$ into irreducibles is $(x-1)(x^4+x^3+x^2+x+1)$, so there are three possibilities for this subring:
- it is isomorphic to $\Bbb F_2[x]/(x-1) \cong \Bbb F_2$
- it is isomorphic to $\Bbb F_2[x]/(x^4+x^3+x^2+x+1) \cong \Bbb F_{16}$
- it is isomorphic to $\Bbb F_2[x]/(x^5-1) \cong \Bbb F_2 \times \Bbb F_{16}$
in the first case, the unit group of $\Bbb F_2$ is trivial, so it doesn't contain an element of order 5, in the other two cases, the unit group has order 15

11:00 PM
Alright, a chunk of that went over my head, I won't deny. Though, what I'm most curious about is why it is that $\text{char}(R)\neq 2\implies (-1)^2=1$

well, $(-1)^2 =1$ always holds, but we need $\mathrm{char}(R) \neq 2$ to show that $-1 \neq 1$

(Also, $\mathbb{Z}[\sqrt{-5}]$ has a unit group isomorphic to $\mathbb{Z}/10\mathbb{Z}$, correct?)
(Or is it $\mathbb{Z}/5\mathbb{Z}\oplus\mathbb{Z}/5\mathbb{Z}$?)

no, the unit group of $\Bbb Z[\sqrt{-5}]$ is isomorhic to $\Bbb Z/2\Bbb Z$
there are only $1$ and $-1$

Wait, got the wrong notation.
I meant to say $\mathbb{Z}[\sqrt[5]{-1}]$
As you can probably tell, I'm new to algebraic rings/fields.

the unit group of $\Bbb Z[\sqrt[5]{-1}]$ is actually $\Bbb Z/10 \Bbb Z \times \Bbb Z^3$
this follows from Dirichlet's unit theorem, but that's probably over your head right now

11:14 PM
Hmm, now that's wild.

no wait I made a mistake
it's $\Bbb Z/10 \Bbb Z \times \Bbb Z$

Still, that extra copy of $\mathbb{Z}$ catches me off guard.
(Also, what is $\mathbb{Z}^3$? Would that be $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$?)

(yeah, that's what I meant, but it doesn't apply here)

(Gotcha. I just haven't seen that notation before so had to make sure)

@MatheinBoulomenos
here ?

11:23 PM
yes

0

The complete graph on $n$ vertices has exactly one edge joining each pair of distinct vertices, and is denoted by $K_n$ A symmetry of a graph $\Gamma$ is a bijection $\alpha$ taking vertices to vertices and edges to edges such that if $Ends(e) = \{v,w\}$, then $Ends(\alpha(e)) = \{\alp... I want to ask a question from my final in measure theory today Let G be a locally compact group and u Haar measure on it Perhaps someone could clarify what exactly a symmetry of a graph. The definition my book gives isn't terribly clear (e.g., what exactly is a edge? Is a symmetry a function on the vertices, edges, or both?) consider$\phi : G \rightarrow [0,\infty)$given by$x \mapsto u(E \cap xE)$consider the subset$H \subset G$given by$g \in H$such that$g \mapsto u(gE) = u(E)$then H is a subgroup because$g_1,g_2 \in H$we have$u(E \cap g_1g_2 E) = u( E \cap g_1(g_2E)) = u(E \cap g_1E) = u( E \cap E) = u(E)$right? @user193319 I'd say it is a function on both vertices and edges. But if there are no multiple edges (as graph theorists like to assume), then the information what the symmetry does on the edges is redundant, as it is determined by the action on the vertices, so in this case authors usually omit the action on the edges 11:31 PM @MatheinBoulomenos agree? @Newbie do you know the modular function?$H$is the kernel of the modular function$\Delta:G \to \Bbb R_{>0}$which is a group homomorphism ohhh shit yeah but my argument work no ? Okay, so in the case of multiple edges (such as in this math.stackexchange.com/questions/3197468/… question), how does one deal with edges? How does one label them? Without multiple edges, to denote an edge formed by vertices$v$and$w$I would just write$\{v,w\}$(as does the author of the book I'm using); but with multiple edges this is ambiguous... why are you looking at$u(E \cap g_1g_2 E)$? you just need to show that$u(g_1g_2E)=u(E)$for the definition of$H$yeah it is just exam stress making me write this 11:35 PM @user193319 I think there is no universal way to label them but yeah that is good but$u(g^{-1}E) = u(E)$because u is left invariant haar measure @Newbie okay I was thinking that$u$is right invariant. If$u$is left-invariant, then why don't we have$H=G$? The reason H is closed is because$H = \bigcup_{x \in H} xV$where V is closed nbhd of G. sorry u is right invariant or alternatively, H is closed because it is the kernel of$\Delta$and$\Delta\$ is continuous

yeahhh

11:43 PM
Howdy @Mathein and Karim.

Hi @Ted!

Ugh, just finished typing up a review of a book I hated spending time on. Why do I agree to these things?

@TedShifrin I am officially done with all my graduate courses
good riddance
now I can just focus on research

Meh. I took courses all the way through, and I encouraged students to do it, too. Even doing research, you should continue to learn some stuff out of your area ... or more deeply in your area.

oh