@JonasTeuwen I would like him not being here, in short.

@Ilya Am I blind or what in saying that $(x^2 + 1)$ is not maximal in $\Bbb{Z}_4[x]$?

@KannappanSampath 8-).

@BenjaminLim not all Russians are good in field theory :)

@Ilya horosho

tvai

nice!

@BenjaminLim It's maximal if and only if it's prime.

@BenjaminLim And prime it is if and only if it's irreducible.

@MattN No maximal implies prime but not the other way

@Ilya oh, that's fun; I put on the ChatJax goggles and now the world looks different!

@Fx does it? :)

@BenjaminLim No I meant the polynomial not the ideal.

how can a polynomial be maximal?

@MattN You're using "it" to mean two different things in the same sentence...

@BenjaminLim The ideal generated by p(x) is maximal if and only if p(x) is prime = irreducible

What Zhen said : ) sorry for the confusion

@MattN $\Bbb{Z}_4$ is not a field

so, now with correct chatjax markup: in $f(\vec{a}) = S_{ijkk} a_i a_j$, should the $k$ index be summed over?

And to finish what I was saying: p(x) is irreducible if and only if it has a zero in Z_4 since it's of degree 2.

render

What Ilya said.

Now $x^2 +1$ has no zero in $\Bbb{Z}_4[x]$ hence is irreducible

@BenjaminLim Oh. True dat.

@BenjaminLim No. Now I need to think about whether what I said also applies in rings that aren't integral domains.

@BenjaminLim: The underlying ring need not be a field, only the quotient ring modulo the ideal $(p(x))$ needs to be one.

@KannappanSampath what? ah, that's how you rendered :D

@MattN, I believe it does apply in non (integral domains).

@FortuonPaendrag I am confused, everything is telling me that the ring $\Bbb{Z}_4[x]/(x^2 + 1)$ is a field, but what is its characteristic??

@MattN The following result is true in any commutative ring:

@Fx unfortunately, I can't help - doesn't it fit physics more?

$p$ is irreducible iff $(p)$ is maximal among all principal ideals in your ring.

@Ilya I would have tried on physics’ chat, but it's silent there :)

@Fx hah, nerds :)

@BenjaminLim Well then. : ) That answers your question, no?

@Jonas: do you know about Einstein summation?

@MattN What is the characteristic of the field I named above?

It can't be 4

Why not, @BenjaminLim ?

@BenjaminLim Why not?

@MattN Because 4 is not prime

By definition it's the number of times you have to add the neutral element of multiplication to get 0, the neutral element of addition.

I am guessing now that the thing there is not a field

@Fx: do you know at least the dimension of $f$?

@MattN How is the characteristic of that ring 2?

Ah screw this. Sorry.

@MattN Any field has characteristic 0 or characteristic $p$. You're confusing that with a field having $p^n$ elements

@Ilya $f$ is a scalar, $a$ a vector, $S$ a tensor of rank 4

@BenjaminLim Indeed I do : )

@Ilya if $f$ were a vector, I'd say it's a typo and it means $f_k$, but I'm sure $f$ is a scalar

@MattN Help with this: math.stackexchange.com/questions/126242/…?

@Fx so if it is a scalar, it should be independent of indices - that's the only trick I can think of

:4029920 Well to me as well. But I have a feeling there is something funny going on.

@MattN: that didn't ping me :(

@BenjaminLim Reading.

@Ilya same here, which is why I think it must be summed over, like the other indices; but I don't think it's standard Einstein notation to have implicit sum on terms in the same variable

anyway, thanks for your help; I've asked a question on Physics as you said it'd be more appropriate there

at least a brief googling for examples does not find such cases

pesky mathematicians wanting their hands clean :)

@Fx what do you mean? :)

(you don't have to reply to me rhyming as well)

okay, my physics question is getting close votes as “too localized” :(

@BenjaminLim Is $a_p$ a typo?

@BenjaminLim: Characteristic can be non-prime, but only in a non-integral-domain.

anyone here good with identifying patterns?

@ZhenLin, but we "established" that the ring we were dealing with was a field

@Fx Thanks for stopping by and sharing your concerns and opinions with us.

@Fx well, that's weird

Then the characteristic must be 2.

Aren't we in $\mathbb Z_4$?

Yes, the coefficients are from $\mathbb{Z}_4$.

@BenjaminLim Sorry, I can't help you off the top of my head.

And I have to go do some stuff. See you later folks.

@MattN Haha, since you've said that the lower part of my right wrist started hurting like hideous monkey's. I wonder if it is psychological.

Later Matt.

@MattN See you!

Hi

@JonasTeuwen everything which deals with nerves can be affected by psychology

anyway, thanks for your help, and regarding chat moderation issues, if the problem is recurrent, the best is to discuss it on meta or contact your moderators

see you

@Fx: thanks a lot, see you later and sorry for not being able to help

@Ilya I hope that is the case! 8-).

I'm not convinced that $\mathbb{Z}_4 [x] / (x^2 + 1)$ is a field.

@ZhenLin I'm not convinced either. So now we need to show that $(x^2 + 1)$ is not maximal

Well, $(2, x + 1)$ is certainly maximal and contains $(x^2 + 1)$.

@ZhenLin, there we go

Then that's it

@ZhenLin However it is true that $(x^2 +1)$ is maximal among all principal ideals in the ring

because $x^2 +1$ is irreducible

That's not obvious to me either.

Theorem $(p)$ is maximal among all principal ideals iff $p$ is irreducible

True over any commutative ring

Yes, but how do you show $x^2 + 1$ is irreducible?

casebash

0^2 + 1 = 1

Exhaustion

1^2 + 1 = 1

2^2 + 1 = 1

3^2 + 1 = 2

That method doesn't work since $\mathbb{Z}_4$ is not an integral domain.

no roots hence irreducible

The exhaustion I meant was, assume there was a breakup into linear factors, and check all cases.

How do you know it doesn't have factors of higher order?

For example, $(2x^2 + 1)(2x^2 - 1) = -1$.

Oh, right..

@ZhenLin So the result that a polynomial of degree 2 having no roots with irreducibility is only true over an integral domain?

It's only obvious in an integral domain. It might still be true without that hypothesis, but then the proof is not obvious to me.

hmmmmm

all the things I've dealt with before are nice

@ZhenLin can you help with this? math.stackexchange.com/questions/126242/…

:/

You have a typo. "$f$ is separable", not "$F$ is separable".

edited it

Thanks for all the new ideas, Zhen and Benjamin! I will think about them. I must leave now, however

@FortuonPaendrag Good bye and good night. Till we meet again.

@ZhenLin any ideas?

The only thing I can think of is that $p$-th roots are unique in characteristic $p$.

where are all the pros on math.se when you need them.....

@ZhenLin what do you mean by $p$ - th roots?

Meaning solutions to the equation $x^p = b$

Anyway, I've figured out a proof. Just do it by hand.

oh ok

I think that is because if we have one solution, we have them all right?

Let $g$ be the minimal polynomial of $a^p$, and let $f$ be the minimal polynomial of $a$; suppose $f$ is separable, then show $g(x^p) = f(x)^e$ for some $e$.

you mean positive power?

Yes.

The point is, show that $a^p$ and $a$ have the same minimal polynomial, if $f$ is separable.

right.

i'm doing stuff now

we know $e > 1$

@ZhenLin I am really tired, going to bed now

I will think about this tomorrow

probably a good idea.

thanks.

you the boss

got a field theory exam next week

man this characteristic zero or characteristic $p$ thing is messing up ma mind

nite'

is anyone here decent with combinatorics

@FortuonPaendrag Could you please explain to me the reasoning behind your flagging of my comment directed at Kannappan, that is if he found it offensive then shouldn't he be the one to flag it, rather than because it seemed to me to be offensive to Kannappan.?

@Rob nope, anybody can flag you - that how the chat works

@Ilya nope, anybody can pretend be over sensitive -that is how the world works

you may think whatever you want, that is how your mind works

thank you

@Rob I am with Fortuon this time. He did the right thing. While I should have flagged, I did not because, you would be going away for eternity, then.

@KannappanSampath what do you mean?

I think Rob was suspended enough number of times that one another suspension, he is going to be banned by the system @Ilya.

(Makes sense right?)

sure

I am off for now.

There is such a feature?

@anon: talking about suspensions - maybe

Yes, I think I read it once at meta. SO

@Kan: but " I would like him not being here"

so what has stopped you this time? don't want to be an executioner?

It would look like I was responsible for his staying away and whatnot. Let someone else execute.

@Rob: we don't ask much from you, don't ask much from Kannappan

It's obviously different now, but I thought that was weird.

time paradox has you

why is there so much drama on a math forum

does not make sense to me

Life is a stage.

doesn't mean what goes on has to be silly

@Ilya Thank you so much for telling Rob that.

@WhatsInAName drama-rama. Don't take it seriously

@Rob Sorry, I didn't catch that.

He asked you to elaborate on your comment to his answer on meta.

Could you be more specific? I didn't leave anything out, so to speak.

@DylanMoreland That's ok it wasn't important.

@anon which meta tread?

It's linked on the sidebar, Chat Etiquette Rules

@Rob wasn't important? that's not surprising

How to use MathJax, I want to post some Mathematical questions which has Squareroot etc .. I dont know, how to type Squareroot on question..

@Ilya Nice bit of sarcasm.

@Medex: Do you know LaTeX?

And you mean post on chat, right?

@anon, Nope i am relatively new for all these concepts

hmm

@Medex: Okay. Are you planning to post here in chat, or on the main site?

I have bunch of Mathematical questions which has Square roots and other Mathematical derivations ..

@Medex $\sqrt{x}$ - and always use $ $ to write a formula inside, even if it the name of the variable - otherwise difference fonts are confusing

@anon, I wish to post it in the Main Site

thanks @Ilya

For laplace, sqrt to the power of 2 or 3 etc

@Medex \sqrt[n]{x} where n is a power

You have to put inline equations inside single dollar signs, like $\sqrt{}$. Display equations (that go in the middle and have their own line) go inside double dollar signs, like $$a+b$$. For help with LaTeX markup, see Wikipedia's reference guide. You can right click equations in answers/questions, hover over "Show Math As" and then click "TeX Commands," you can see the markup code other people have used. @Mendex

Okay great @anon and @Ilya - Thank you for your help !!

@anon and @Ilya - inside single dollar signs, is a new form of Syntax !! I love to explore a lot. Thank you for guiding me

I'm in the train :)

@Ilya: I rewrote 3 and 7 so that the links appear on one line for me. Are they better for you, too?

@JonasTeuwen nice :) where are you going?

@Robjohn: yes, thank you :) I deleted my annoying comment

Utrecht. There is a slow coffee workshop for which I am invited.

cool that

@JonasTeuwen what is slow coffee?

@Robjohn: it's when are smoking marijuana while cooking coffee

@Ilya ahh.. thanks :-)

@robjohn: Basically any coffee not made with an espresso machine.

@JonasTeuwen the only coffee should be such coffee, I would say

There are hundreds of methods.

I don't like coffee from machine. I never tried a good one

@Ilya I don't like coffee, machine or not.

@robjohn sometimes I come to CA and cook you a coffee if you provide a stove :)

8-]

@Ilya You're welcome to do so, though I think you'd be wasting your time :-)

I will talk to you and take a look at your pets - is it wasting a time?

@Ilya that is not a waste of time. You might skip making the coffee :-) (more time for the talk and pets)

yeah, you know - you're right

I only should find a nice group to visit - and to do this I should determine what would I like to do there: probability, formal verification, biology or financial math - sometimes it's hard to work on everything this together

@Ilya I can make you coffee; my wife says I make good coffee. Or I could join you in a pot of tea (a bit crowded, but hey...)

mmmm, the tea

it's just so far away as for now

some guys want me to come to Boston

@Ilya We have a large tea selection in our cupboard.

I think I won't resist

can anyone help me with a counting problem?

i have a program that delivers output based on two parameters and i am trying to find an effective way to count valid solutions

pastebin.com/Nm3rH0z8

i want to find a way to be able to deliver the count number without needing to iterate all possible winnable positions

so i am looking for patterns within the valid solutions themselves

to see if there is some way to get the count knowing the pattern

pastebin.com/VftDBxS1 n=12 c=5 output

@Ilya: did you see the Example that I added to my binomial answer. It confirms the value of $\alpha$ I got.

@WhatsInAName does that have to do with Mastermind?

no it is a variant of a Nim game

where you move coins along a strip of paper

@WhatsInAName are the rules posted?

tinyurl.com/4x5qace

it is a problem to solve for fun

but it is very challenging indeed

my program will output valid counts using brute force but obviously it will be infeasible for high n and c

the trick is usually to take advantage of mathematical patterns

in this case i am reasonably confident it will involve a few generating functions

The usual Nim falls to the binary representations of the numbers in each pile

right

the spaces between coins are typically seen as the "piles" in Nim

but in this case it doesn't quite hold

simply ensuring xor=0 does not guarantee winnable position

no, because you are essentially moving coins from one pile to the next

right, but we're also potentially removing piles altogether

(i also wrote a xor function for the winning positions and they're all over the place)

yeah, if you consider the last pile an abyss :-) the coins are just separators between the piles

yes

pastebin.com/fmTwbaks <-- n=12 c=4 with XOR listed

Out of curiosity, what does your brute-forcing look like? I'm not sure I could even think of an easy way to do that.

dynamic programming

actually my xor may be off

nvm it's fine

does anything stand out to the mathematically-minded that I am totally missing here?

I haven't solved this, but I'm going to speculate a bit. Please stop me if it gets annoying.

I'm asking myself what is so special about the silver coin. It looks like you're generating solutions by fixing the silver coin in one location and then placing the other coins, but the other coins act exactly the same, so why place the silver coin first?

?

it's what wins the game

the list of layouts i posted above are winning positions

which means no matter what your opponent does

you can guarantee you can get the silver

it's not like standard Nim where you just ensure you take the last stone by making moves that set the xor of all stacks to 0

Well, I know that for the game's purposes, we want the silver coin, but what if we decided to play our own game, where we needed the coin to the left of the silver coin? What if we were presented with a configuration of coins first, and then one was flipped over to reveal that it was the gold coin?

i don't understand

if all you care about is a coin to the left of an arbitrary coin then why not just make that the silver coin

It seems like it would cut down on computation a bit, to only have to examine a coin configuration once.

well when you have high n and high c there are too many configurations to iterate through directly

it would take years

the ideal goal is to identify patterns that allow you to more efficiently count valid solutions

so many configurations you won't have to examine at all

Sorry, I'll cut the pretend socratic stuff.

Trying to figure out a way to phrase this; sorry for the long pauses.

no problem -- i have to step out for a couple minutes anyhow, brb

back

@anon You do the \rm thing too?

When I feel like it.

@WhatsInAName I was looking at your solution list for n=12, c=4, and I noticed that #2 and #331 are considering the same coin configuration, but with a different silver coin.

@DylanMoreland not as much as Bill Dubuque does

But that's Bill's thing.

@DylanMoreland indeed.

imitation, flattery, etc.

I notice that the completely separate entity Math Gems has not been active lately.

Ah well, people do stupid stuff all the time. And usually they aren't this helpful.

:|

@Srivatsan: is it 22:55 (10:55 PM) there?

@WhatsInAName Actually, I've realized that I oversimplified my model too much. Nevermind, disregard.

hm, Math Gems also (?) uses /rm. didn't that user state he was a student of Bill or something?

@anon I have never gone to Math Gems...

yeah it is a hard problem

This site?

I even showed it to one of my old professors and he couldn't solve it

@robjohn: This user.

Ah. a user, not a web site :-)

Wow, it looks as if all of the comments, and perhaps the downvotes, on this answer are because of the word "obvious".

Obviously, a bad word. :-)

Don't make me flag you for offensive language.

@anon oops :-p

@robjohn Yes, robjohn. (Sorry for the delayed response; I am using IE right now.)

@anon Now, that is really suspicious. :)

@Srivatsan My condolences (on IE). I just realized that your time is off by a 30 minute phase

what do you guys do for livings?

@WhatsInAName I write software.

I have to eat and sleep to live.

@robjohn It's 11.11 now. It is GMT+5.30. I can't remember the conversion from the US times.

@Srivatsan that comports with what I saw. you are 12:30 ahead of us

Yeah, we are UTC-7 because of Daylight Savings

@robjohn Ok. [I don't try to remember that either. :) Googling is quicker...]

@Srivatsan I use this site.

Ha, that works too. But google can answer queries like "California time right now"; very convenient

What about him?

It seems he has caused quite a stir again.

?

@Srivatsan You mean Rob (he has the peace sign as a gravatar)

Hi, just lurking... I've just learned that's OK :)

@robjohn: not at the moment, see the upper right

causing a stir how

Let me refresh

@robjohn Well, Rob is unfortunately too common a name. Imagine if I had asked you: "Do you know what's up with Rob?" :=)

@Srivatsan yeah, but Skullpatrol doesn't have a main site character, nor does he use that chat character any more that I have seen. I didn't know if you knew that Rob was the same person.

I used Skullpatrol to disambiguate. I know/knew that his name has changed to Rob.

@robjohn Sure. I just realised that (See my previous comment).

I see he has the Gauss-Einstein Band as his gravatar now :-)

@Srivatsan I know that now, but not before :-)

They rock!

@Srivatsan I need to see if they have a CD on Amazon.

This is a funny site

A cover for their album: i.sstatic.net/MkjZl.png

@WillHunting Hi Billy!

@robjohn I liked the Tesla-Edison one :)

Can someone help me understand something about this answer mathoverflow.net/questions/40309/…