groups-rings-fields

[crystal vale]

Then use another result if H is proper subgroup of finite group G, then the set union x^(-1)Hx( x runs over G) is not whole of G

By using this result I will have an element xN which is not contain in any conjugation of H/N

My doubt is here H/N is proper subgroup of G/N ..... How?

And what is Normal subset of group G?

[mighty kiln]

Is this the statement you're trying to prove?

[mighty kiln]

This result is just a rephrasing of that

[crystal vale]

Yes

[crystal vale]

Yes but I used this to prove given statement

[tender wharf]

a normal subgroup? idk what else it could mean. do you have the original question at hand

[rocky cloak]

The coefficients of a polynomial are exactly the elementary symmetric polynomials evaluated in the roots. And the elementary symmetric polynomials are an algebraic basis for the symmetric polynomials.

[mighty kiln]

Is that be tautological tho

[hidden wind]

does anyone know of a neat way to order the permutations?

[crystal vale]

If H1 and H2 are cyclic Subgroup of a finite group G then if gcd( |H1|, |H2|)=1 , then is it correct that only element that belong to intersection of H1 and H2 is e?

[rocky cloak]

Think about what the order of a common element could be

[crystal vale]

Okay got it..... if I am correct if there is any element which belongs to both set H1 and H2 then it's order also divide the order of H1 and H2 thus order of that element is 1 which is e

[formal ermine]

yes

[crystal vale]

If G is infinite and H is non-empty finite subset which is under closure operation.....is H subgroup?

[formal ermine]

take any finite field and its algebraic closure

wait

are you asking if ONLY the closure is guaranteed, then it is a subgroup?

[crystal vale]

Yes

[formal ermine]

or are you asking if an infinite group can have a finite subgroup

[crystal vale]

Any counter example

[formal ermine]

or wait

i revoke the no obviously

[formal ermine]

its true

if its closed under closure then every element has finite order

so we have inverses and identity

had a brainfart there

[crystal vale]

Okay thank you

[opal osprey]

i mean ye, but I'm trying to show that fact i mentioned of S(x_1,...,x_n) being some polynomial expression on the coefficients without using what you said

namely, i want to understand that statement highlighted on the wikipedia page

that S(x_1,...,x_n) being a polynomial expression on the coefficients is a consequence of the fact that the fixed field of a galois group is the galois group itself

[crystal vale]

Torsion subgroup of Z×(Z/nZ) is (0,Z/nZ) ?

[coral spindle]

Yes

[cloud solar]

Hello. Where can I find a full proof for this formula. The number of distinct necklaces with n beads and m colours for each bead.

I know it is an application to Burnside's orbit counting lemma and I did the n=8, m=2 case without this formula

[barren sierra]

F2 is the only finite algebraically closed field right?

[rustic crown]

nope, all alg closed fields are infinite

[barren sierra]

???

F2 tho?

What polynomials aren’t satisfied lol

[rustic crown]

in general if k is finite, then the poly f = prod_{a in k}(x - a) + 1 has no roots

[barren sierra]

Hm

Ah yea ok

I was being dumb

[rustic crown]

for F2, x^2+x+1 is a standard irred poly to construct F_4

[barren sierra]

Yea

[crystal turtle]

デットかわいい uwu

[rustic crown]

https://tenor.com/view/cute-eevee-gif-25670637

[dim widget]

have you tried proving it yourself? What if n is prime?

[dim widget]

There is an action of S_n on Q(X_1, …, X_n) by permitting the variables, and the fixed field is Q(Y_1, …, Y_n) where the Y_i map to the elementary symmetric polynomials in the X_i

So this is a galois extension of degree n!, it’s the splitting field of the polynomial \prod (T - X_i)

these are theorems of Newton

they are asking about c_1 and c_2 arbitrary

so it’s good to break it up between c_1 = c_2 and c_1 \neq c_2

two cases

[stark helm]

if it can be like c1 contains interval between [0,0.4], c2 contains interval between [0.6,0.8]

[dim widget]

c_1 and c_2 are points

They don’t contain anything

[stark helm]

I am considering it maps from interval [0,1] to 0 for Ic, so how will Ic1 and I c2 possibly look like, if we can say Ic1 maps from [0,0.4] to 0 as an example?

[stark helm]

Do you mean we can give examples like like c1=0.3 and f maps from c1 to 0 in I_c1? and c2=0.5 and f map from c2 to 0 in I_c2?

[formal ermine]

[barren sierra]

I had a dumb moment lol

[dim widget]

can you share why you thought F_2 would work but not F_3?

[barren sierra]

I literally went "oh yea x^2 + 1 is satisfied"

[dim widget]

It’s also satisfied in F_5!

[barren sierra]

oh true

[wraith cargo]

I'm curious where you got this
Did you read that meme proof?

[barren sierra]

No I was just bored in complex analysis

What meme proof

[wraith cargo]

there was a blog post like for april fools a few years ago with this proof
It equates the 0 polynomial and a polynomial that equates to 0 on all elements

[slim kayak]

Breaking news: local mathematician proves that F_p[X] is finite

[stark helm]

for part b, can someone tell me how will I_c1 look like as an example?

[coral spindle]

Idk what you mean 'look like' here, or what you're looking for. There's not any better way to describe I_c other than the definition given in the question. Maybe just draw a few functions in this ideal to make yourself happier with its elements.

[stark helm]

I mean lfor arbitrary I_c1 and I_c2, if both of them are different, then can I have the case when Ic1 describe a map f between an interval [0,0.4] and 0. And I_c2 describe a map g between an interval [0.6, 0.8] to 0? Since Ic describe a map from interval [0,1] and 0 such that f(c)=0

[coral spindle]

I have no idea where you're getting this from

I_{c_1} := {f in R | f(c_1) = 0}, nothing has changed about the definition.

I have nothing to add here.

[stark helm]

it seems to me that I_c1 and I_c2 are the same set? then where is the difference between I_c1 and I_c2?

[coral spindle]

c_1 and c_2 are real numbers in [0, 1]

They may be different real numbers.

Is that clear?

[stark helm]

OK, now it is clear, really appreciate your response

[reef trench]

Is the last part a typo?

I don't see how nZ[x] could be an ideal of Z?

Or are we talking about (Z/nZ)[x]?

[wooden ember]

Yes they mean (Z/nZ)[x]

[south patrol]

I wouldn't really call it a typo, but it is suboptimal typesetting for (Z/nZ)[x] yeah

[reef trench]

Yeah, it wouldn't kill to add parentheses lol

[crystal turtle]

Sure, parenthesis might help, but is should also be clear from the context of what you said that it can't be that nZ[x] is an ideal in Z, so it must be (Z/nZ)[x]

[keen loom]

Crazy, I was literally just using this in my work

It is a consequence of polya enumeration...which is a very powerful and interesting generalization that works in tons of different scenarios.

A proof is on this page https://en.m.wikipedia.org/wiki/Pólya_enumeration_theorem

You could consider this a big Burnside generalization. The "weights" idea is very flexible

I have never seriously studied counting before...and find this theorem very exciting

[crystal vale]

Is there any H subgroup of order>2 which is not in Centraliser of G but Normalizer of H in G is exactly G?

[coral spindle]

Yes

[dim widget]

If you just started naming all groups ordered by size you would immediately find an example

[crystal vale]

S3

[coral spindle]

Indeed, the smallest non-Abelian group provides a counterexample.

[south patrol]

S3 always good to check

[deft pecan]

is there anything wrong with this argument?

good to know -- most other approaches i saw felt a little more involved to me so i wondered if this was off. thanks!

[slim kayak]

yeah, altho a minor thing

okay maybe major - G is not commutative

(abc)^p = a^p b^p c^p is generally false. But (yxy^-1)^p happens to be yx^p y^-1 (verify this)

[deft pecan]

oh yeah, that does seem pretty significant! i'll play with that -- thank you

[glossy crag]

I figured you must've known this

[celest furnace]

So you said “nope” sarcastically?

To a person asking if there was anything wrong with their argument?

[slim kayak]

things heating up on the groups and rings channel 🍿

[glossy crag]

Oh shoot, I just now noticed y^q and y^-q! I parsed this as y and y^-1.

[celest furnace]

Yeah i did that the first read too

Honest mistake then that makes sense

[deft pecan]

yeah, all good

[celest furnace]

But if you want to prove an exercise related to this you can show that if n is the order of x, and m is the order of y and gcd(m,n) = 1 AND most importantly xy = yx, then the order of xy is m*n

This has the following consequence: If g is an element of maximal order n in a finite abelian group G, then for any other g' in G, |g'| divides n

[deft pecan]

ooh that's the next one on my list

[celest furnace]

Oh awesome

Almost accidently spoiled that one glad i didnt lol

That exercise shows up more than you would think

[slim kayak]

its pretty essential to know tbh

[celest furnace]

Agreed

[slim kayak]

poor mans structure theorem

[deft pecan]

i'm speedrunning trying to pick up enough to learn some baby projective geometry and invariant theory stuff for a research project, so trying to pick problems that seem important/useful but not spend too much more time than i need to

[tepid rune]

Hello, I'm new here. How are you? Is it possible to ask any questions here or just in the help forum?

[celest furnace]

Feel free to ask here too

[slim kayak]

No, this channel requires a vow of silence

[slim kayak]

do you have a non-math background?

bruh

[tepid rune]

alright, thanks

[deft pecan]

nah i'm doing pure math but i'm a little weird. like i've taken mostly analysis, like the full undergrad and graduate analysis sequence at my school, and complex analysis. i'm an undergrad technically but mostly i just take grad courses or stuff that seems interesting/relevant to my research

algebra is the last class i have to take

not aggressively trying to graduate cause i'm doing all this part time, got a full time job and busy life

[celest furnace]

U gonna do Ph.D?

[tepid rune]

In my undergraduate I lived a same situation

[slim kayak]

more of an algebro so I have no idea how you dodged it this long lol

Especially since you usually should take the courses in sequence which means you gotta fill the rest of your semester with something

okay the full time job hypothesis enters here lol

[deft pecan]

yeah 🙂 and i'm not getting financial aid, just self-paying at a cheap school. fewer constraints as an undergrad, like don't have to take qualifying exams within a set amount of time and don't have to meet credit hour minimums. might do phd, not sure

[celest cairn]

Would Q(zeta_4) be considered a trivial subfield of Q(zeta_8)?

[next obsidian]

Wdym trivial?

[coral spindle]

Where are you seeing the term "trivial subfield." Show us the context.

[celest cairn]

I just kinda made it up but is there a good word for when a field extension and it’s subfield are equal?

[coral spindle]

You can just say it's equal?

Or you could say it's of degree 1.

Alternatively you could say that the extension is trivial, rather than the subfield.

[frosty pecan]

Sometimes I've seen inclusions K < L where K is strictly smaller called a proper subfield.

[next obsidian]

Lmfao

[coral spindle]

Uh and also

Q(zeta_4) =/= Q(zeta_8)

[celest cairn]

Sorry I meant Q(zeta_8) = Q(zeta_4), if it’s true.

[coral spindle]

But it's false

[celest cairn]

dang alr

[rotund aurora]

Q(zeta_n)=Q(zeta(2n)) is only true if n is odd

[coral spindle]

If you just picture zeta_4 and zeta_8 in the complex plane it's also quite clear imo

[next obsidian]

Lmfao

[coral spindle]

Ig it does require you to know that zeta_8 is an irrational point, oh well

Logically, a line inclined at 45 degrees and the rational circle never intersect!

[next obsidian]

It’s like definitely way way easier

Proof: look at it

[coral spindle]

Hey hey, it's intuition not a proof

[summer path]

Draw a picture

[next obsidian]

Hi tubu

Did u eated

[summer path]

Hi chmonkey

[rotund aurora]

I'm not sure why this is supposed to be helpful (to Saphire gaming), what if you picture zeta_3 and zeta6, what's the difference

[summer path]

No tubu got stuck in bed

[coral spindle]

Because they're obviously the negative of each other

[next obsidian]

Time to eaty

[coral spindle]

(For a particular zeta_6, but no matter)

[next obsidian]

What if someone called you butu?

Instead of tubu

[summer path]

[rotund aurora]

Q(zeta4)=Q(zeta8) is obviously false but Sapphire didn't notice

[coral spindle]

Your point being?

If you picture the roots, imo it becomes clearer.

This is why.

[rotund aurora]

that "just look at it" isn't a good hint, I think

[coral spindle]

OK

[next obsidian]

Q(zeta4)/Q(zeta8) = 4/8 = 1/2

[frank cosmos]

can i get a hint for this?

i got what when Z subset H we have |G:Z|=|G:H| * |H:Z| so then half the size

idk for the other case

the hint is 2nd iso thm

[cobalt heath]

Shouldn't you proceed similarly for that case as well?

[frank cosmos]

how? Z isnt a subgroup of H in this case

and the text suggets using 2nd iso

also, one more question:

this is a proof of cauchy

is this proof right: the number of elements with order p is -1=p-1 mod p thus the number of distinct subgroups is p-1/p-1=1 mod p?

[cobalt heath]

Well you may seek for another

[frank cosmos]

ok, the centralizer in H of A?

hm

[cobalt heath]

It is a but yea

[frank cosmos]

this dosent work though, since i need to know what G:Z is not G:Z_H

[cobalt heath]

Yeah, now you recall the hint?

[cobalt heath]

This

[frank cosmos]

applying it to H and Z?

dont see how that helps

[cobalt heath]

Why wouldn't it

[frank cosmos]

that just tells you HZ/H = Z/H cap Z

OH SHIT H cap Z=Z_H

[cobalt heath]

Are you in some trouble

[frank cosmos]

?

ok need to figure out the rest...

[cobalt heath]

This made me concerned a bit

[frank cosmos]

ok lol

[crystal vale]

If Z_36(written multiplicative) generated by x so for which integer a , the mapping from Z/48Z to Z_36 define by 1->x^(a) be group homomorphism?

[rocky cloak]

Does it work for a=1? Why/why not?

[crystal vale]

I think it's work because of laws of exponent

But if it's from Z_36 to Z_48 it will not work but I don't know how to state that for Z_48 to Z_36 it will work

[formal ermine]

i had a stroke trying to read this

[rocky cloak]

What is x^48, and what would it need to be for this to be a homomorphism?

[crystal vale]

x^(12)?

[rocky cloak]

Indeed, and what should f(48) be if f is a homomorphism from Z/48?

[crystal vale]

x is generater of Z_36 so take x =1 so it will be f(48) =12?

[rocky cloak]

Yes, but can there be a homomorphism that maps 48 in Z/48 to 12?

Like what do you know about the group Z/48?

[crystal vale]

For homomorphism 48 map to 0

[rocky cloak]

Yeah, and 12 is different from 0, so that can't work

[crystal vale]

Oh

For which a it will work?

[rocky cloak]

Well, how can we avoid the problem we just had, about 48a not being 0?

[crystal vale]

map 48 to 0

[rocky cloak]

Sure, but how

You choose a (where 1 is mapped)

[crystal vale]

1 map to 6

[rocky cloak]

6 would work yeah

[crystal vale]

1 map to 3n

[rocky cloak]

Yeah, you got it

[crystal vale]

Thank you

But 1 map to x^(a) and x is generater of 36 so x can not be multiple of 3 so there is no a ?

[rocky cloak]

I'm not sure what you mean

If 1 is mapped to x^a, then there clearly is an a involved

[crystal vale]

If I want to map 1 to 3 and x=1 then what my a?

[rocky cloak]

I'm not sure what you're saying. x is your notation for the generator in Z/36, so what does "x=1" mean? x isn't equal to anything that's just notation.

Like you're representating the elements as
x, x^2, x^3, ...
and you going that when 1 maps to x^a, then a must be a multiple of 3

[crystal vale]

1(mod36) is generater for Z_36

Okay

[rocky cloak]

If you're using additive notation, sure

[crystal vale]

Yes

[rapid junco]

Let R, S be domains and f : R → S be a homomorphism. Prove that either f (1) = 0 or f (1) = 1.

am I wrong by saying that you dont really care about if R is a domain here

or am I missing something obvious

[rustic crown]

homomorphisms that don't preserve 1?

[cobalt heath]

Rng homomorphism

[rocky cloak]

That's right, only the condition on S is relevant. And even then you can prove the same thing with much weaker assumptions than S being a domain.

[prime quail]

The quadratic formula fails for field with characteristic 2 because the bottom number is 2a=0.
Is there a quadratic formula for field with characteristic 2?

[wraith cargo]

there is but it's quite complicated

https://mathscinet.ams.org/mathscinet/relay-station?mr=680147

[velvet steeple]

Solved just after posting

[barren sierra]

no I posted in the wrong channel

I meant to post in #advanced-algebra so I deleted here and posted there

[hazy flax]

The preimage of a subgroup is a subgroup under a group homomorphism but why isn't the image of a subgroup also a subgroup?

I'm unable to see why the same argument doesn't work for the other case.

[rocky cloak]

The image of a subgroup is a subgroup

[coral spindle]

Who says it isn't?

[hazy flax]

Oh is it?

I thought you required surjectivity or something

[coral spindle]

No, you don't.

On the other hand, the image of a normal subgroup isn't necessarily normal

But surjectivity does guarantee this.

Is that perhaps the confusion?

[hazy flax]

Oh I see

[hazy flax]

yeah that might be it

thanks!

[rapid junco]

nice

[rapid junco]

the nilradical of Z/m is just 0 correct?

or would it be (p_1 * ... * p_n) where those are the primes without their p-adic valuations of m.

[coral spindle]

Well remember that the nilradical is the intersection of the prime ideals, and recall the prime ideals of Z/m

But a specific example may help with understanding the nilpotent elements

[rapid junco]

pZ/mZ

[coral spindle]

For example, what are the nilpotent elements of Z/8Z?

Think about how this generalises.

[rapid junco]

well the only thing im confused abt is that my proof works for both (m) being the nilradical and (p_1p_2p_3 ... )

so then these coincide?

[coral spindle]

Well again, what's the nilradical of Z/8Z

This should answer your question

If you want us to check your proof, then share it.

[rapid junco]

sure i will do finishing touches and then that would be awesome

yeah the nilradicals should be (2).

ok well I will share both proofs and if you could clarify where I am going wrong for (m) that would be awesome.

I am having trouble showing r^a \in (p_1 p_2... p_n) implies r \in (p_1 p_2 ... p_n) because this ideal is not nessecarily prime.

Is there something I am missing?

r | r^a

[coral spindle]

Look at the integer factorisation in Z. You can lift this question about elements of Z/m to a question about elements of Z.

[rapid junco]

yeah so r | r^a in Z.

[coral spindle]

You're familiar with the p-adic valuation – use it

[rapid junco]

not overly, just wiki

oh im goofy

r | r^a so r | k * m

thus r is in the pi

[rapid junco]

what property were you suggesting

[coral spindle]

p_i | r^a so p_i | r.

I thought maybe if you saw this as a property of ordinary primes it would be more obvious.

You can also just see this as (p_1 p_2 ... p_n) = (p_1) (p_2) ... (p_n) in a more algebraic way.

[rapid junco]

ah I see

thanks for the help this is it for the most part.

So showing the inclusion for $(m) \subset N(Z/m)$ is trivial, but what goes wrong on the other hand?

[cloud walrus]

brayden letwin

[rapid junco]

If $r \in N(Z/m)$ then $r^n = 0 (m)$.

[coral spindle]

Considering (m) = (0), I should hope it's trivial!

[cloud walrus]

brayden letwin

[rapid junco]

$m \mid r^n$

[cloud walrus]

brayden letwin

[rapid junco]

$r^n = k m \implies r^n \in (m)$

[cloud walrus]

brayden letwin

[rapid junco]

ah I see but then we dont get any further

[coral spindle]

I really don't know what you're asking exactly. What goes wrong is it's false.

There are other nilpotent elements other than 0

[rapid junco]

right yeah I see now, was just confused cuz I thought I saw a way

but it was wrong

well thanks for the help

and clarification

[astral fractal]

Im working through Dummit and Foote and am confused as to how they came to the conclusion this group has order at most 6 just from the generators and relations presentation https://gyazo.com/974a28cb91654a0a5bdd3a6d8ae3985b

Is this supposed to be obvious?

Also a similar statement is made here without much explanation

https://gyazo.com/22b20ce33d68f2c35eff95b3464f7043

[fading field]

well, you have this commutation relation, which allows you to "separate" an arbitrary element (maybe you'll accumulate extra "powers" of x and y along the way), from that you get a process by which every element can be written as x^m y^n for some m and n, then you know that x^3 = y^2 = e, so there are at most 3 * 2 = 6 elements

[chilly radish]

Something here doesn't make sense or i'm going crazy, if x^3=1 for any n, doesn't it follow that n is either 1 or a multiple of 3?

[astral fractal]

how come you need the commutation relation in order to say that you can write any element in the form x^m y^n

the two dont seem to be related to me

[fading field]

well how are you going to switch them and still be able to say you have the same thing otherwise?

it's a free group

so $x^m y^n \neq y^r x^s$ without it

[cloud walrus]

smay

[fading field]

i mean it's a way of "physically moving" letters right/left

[astral fractal]

but let's say x^2 is an element in this group, can't we write it as x^2 y^0 without considering the commutation relation?

[fading field]

yes you can

[astral fractal]

that's what my confusion is as to why the commutation rule matters when writing an element in the group in terms of x^i y^k

[fading field]

because you might have like x^m y^n x^s

for example

and then you're hosed

[astral fractal]

oh

[fading field]

without it the order of the group is infinite!

[astral fractal]

so if we did not have the commutation relation, we would not be able to reduce this to be equal to some element already in the group?

[fading field]

yep

[astral fractal]

ah

[fading field]

well

"some element already in the group" meaning "something that we are familiar with" or "in a nice form"

x^m y^n x^s is already in the group

[astral fractal]

i see

something equvialent to it that is simpler

[astral fractal]

im still confused how x^3 = y^2 = e tells us anything about the order

it just seems that we found another way to write the identity

[fading field]

well what are the most elements possible

well you can't have like, x^4 for example

that's just x, which we would have already counted

[astral fractal]

right

and anything about x^4 will also be equal to something we already counted

x^5 = x^2

and so on

[fading field]

okay, so if we have a word, then we reduce it to our x^m y^n, and then that's equal to x^(m mod 3) y^(n mod 2)

[astral fractal]

i see

so we have x, x^2, y, 1, xy, x^2y

that would be true for any n

therefore the group will have order at most 6

[fading field]

yeah these are the only elements

[astral fractal]

interesting

i think it makes sense now

thanks for the help

[fading field]

well we didn't start with the assumption that x^3 = 1

[chilly radish]

Yea but you deduce it independent of n

[fading field]

yes

[chilly radish]

So it should apply for all n

But it can't

[fading field]

what do you mean that it can't?

[chilly radish]

Because it follows the order is at most 3, so either n is 1, 3 or greater than 3, in which case the order is 3, and since x^n=1 we have 3|n

The order of x

[fading field]

well they make a note that further collapsing can occur

the remark was that the order of the group was at most 6

[chilly radish]

Oh my bad

I misread it at the order of the group is 6

[fading field]

@astral fractal this is important actually

that you can get more collapsing

[crystal vale]

How can I show that additive group of rational number is not cyclic? If I assume if Q is generated by integer x then it probably not work and if I assume Q is generated by rational x then first I thought I will show contradiction by taking x^(a)=2 for positive integer a then it will contradict x is rational but what if a is negative? Any other approach

[long obsidian]

What if you took a generator then showed half of it can't be generated by the generator

[crystal vale]

Or by doing some case work if I show generater can not be in form of 1/x where x is integer then I will use when x=p/q then contradict

[long obsidian]

Is there an obvious way that the field of fractions of nested integral domains behaves?

Say $S \subset R$ is it still the case that $\hbox{Frac}(S)\subset \hbox{Frac}(R)$? This is obvious right?

[cloud walrus]

HausdorffT1

[uneven bobcat]

No, Frac(S) contains equivalence classes of pairs of elements of S, and Frac(R) contains equivalence classes of pairs of elements of R, so if S is strictly smaller, they will not be subsets of each other. Although there still is an injection Frac(S) --> Frac(R) that's easy to write down.

[crystal turtle]

All injections are subsets

Just like all isomorphisms are the identity sotrue

[topaz solar]

This one has caused many an L, like embedding things in themselves

[uneven bobcat]

Wait, so we're all working in HoTT now, and assuming there's no higher types?

[cobalt heath]

Categorical formulation is often quite clean

[long obsidian]

Is it possible to tell the degree of a field extension of something like $\mathbb{Q}(x)\subset \mathbb{Q}(x^{\frac{1}{2}})$ so the superset contains stuff like $\frac{x^{\frac{1}{2}}}{x^{6/2}+3x^{5/2}}$

Woops

[cloud walrus]

HausdorffT1

[uneven bobcat]

Yes. Observe that the polynomial t^2 - x in Q(x)[t] is irreducible, hence the extension Q(x)[t] / <t^2 - x> \cong Q (x^{1/2}) has degree deg (t^2 -x) = 2.

[long obsidian]

Ohh okay it's basically gonna be the same case for 1/3. Thank you!

[crystal vale]

Got it, thank you

[glossy crag]

Isn't the definition of a transcendence basis that F/K(x) be algebraic, why must it be finite?

[teal vessel]

dumb thought. Suppose we take the set of all logarithmic functions, that is, the set L= {f: R+ -> R | f(xy)=f(x)+f(y)}. We know that these functions have an interesting relationship, specifically that they are all the same function offset by a constant, that is, we may arbitrarily choose some function ln such that log_a = ln/ln(a). I've been reading about quotient groups in this latest chapter I'm working on, and it spawned a question: does there exist some operation which groups this set L?

I suppose that would necessitate that there is some "identity element" in this set, though there is not an obvious choice for which element to identify it as beyond the "wouldn't it be nice if it was ln?" idea.

[cloud solar]

@rocky cloak I found out that the number of elements of the rings is divisible by 16. (For the problem with the finite commutative ring with at #idempotens=#nilpotents=#invertible elements if the number of elements >4 )

Is that right?

Even if one little ring in the ring product is isomorphic to F2

[rotund aurora]

are you sure they are the same function offset by a constant?

[teal vessel]

offset, scaled, same thing really (joking)

[rotund aurora]

It's not true

[teal vessel]

I should have said scaled. The change of base formula applies to arbitrary logarithmic bases (at least real valued ones), it doesn't need to be log_e

[rotund aurora]

There are many other functions besides logs that satisfy that functional equations

[teal vessel]

like....?

[rotund aurora]

you can put it in the form f(x+y)=f(x)+f(y) if you wanted

f(e^(log x+log y)=f(e^log x)+f(e^log y) so if you set g(x)=f(e^x) then g(log x+log y)=g(log x)+g(log y). Log is surjective anyway, so you just get g(x+y)=g(x)+g(y)

[teal vessel]

but that's not the same equation. I'm not talking about arbitrary addition and multiplication, I'm talking particularly about real number addition and multiplication

[slim kayak]

well, you do need continuity here

there are non-linear solutions to the cauchy functional equations, and given any such solution F you can construct f as F(log(x)), then f(xy)=f(x)+f(y) but f is not a logarhitm

[teal vessel]

I suppose, yeah. Otherwise choice can bite me in the ass

[slim kayak]

"which groups this set" ?

[teal vessel]

under which the set is a group

[slim kayak]

there are some natural choices

Mostly because you can associate the logarhitms with their bases

[teal vessel]

yeah, that's what I'm thinking

so I suppose if I could find some structure which holds for the bases as numbers I could then associate that to the logarithms through some homomorphism

[cobalt heath]

Perhaps it becomes finite extension with a suitable choice.. wait how does this work

[cloud solar]

Question

An example of a finite commutative ring with 2 idempotents, 4 nilpotents and 4 invertible elements?

[slim kayak]

isnt it being a finite extension part of the definition

[rotund aurora]

Yes, but it's not stated. That's Ocean's point I think

[rotund aurora]

If it has idempotents, that means it is a product of rings

[cloud solar]

It has only 2

[rotund aurora]

2 without counting the identity, right?

[cloud solar]

No

0 and 1 (unitary ring)

[rotund aurora]

why would you mention that then lol

[cloud solar]

I just wanted to know an example of finite commutative unitary ring with 2 idempotents, 4 nilpotents, 4 invertible elements.

[rotund aurora]

Z/8Z ?

[rocky cloak]

Yes, that's right. The number of idempotents is a power of 2. If there's only 2 idempotents you can show that the ring must have 4 elements.

If there are k >= 4 idempotents, then besides 0 and 1, idempotents are never units or nilpotent, so you will have at least 3k - 2 elements

[rocky cloak]

F2[x]/x^3, Z/8, F2[x, y]/(x^2, y^2, xy), Z/4[2x] are some examples.
I would guess this is all of them.

[jolly jewel]

Let G be a group such that for all a ∈ G we have a ∗ a = e. Show that G is abelian

Any ideas?

[cloud solar]

(ab)^2=e so ab=(ab)^-1=b^-1a^-1 but a^-1=a and b^-1=b so ab=ba

[jolly jewel]

ok I got it! Thanks

[jolly jewel]

I actually had it done another way so I just wanted to check...
(a * b) * (a * b) = e
Implying (a * b) = (a * b)^-1
(a * b) = b^-1 * a^-1
(a * b) = (b * a)

[hallow knoll]

How can you check if this is associative? There has to be a better way than to check every possible option right?

[formal ermine]

no

well you can take some shortcuts

but at the end of the day its just checking every combination

[teal vessel]

I'd start by noticing that C seems to do nothing, so you can redo the rows and columns by putting C first in both rows and columns. This gets your identity element out of the way, then you can check all non-identity elements for associativity

[rocky cloak]

I guess you can recognize that this is isomorphic to C2xC2, which is a group

[crystal vale]

What background needed for Category?

[rocky cloak]

For category theory? Not a lot of background is strictly needed, but it depends a little on your motivation for learning.

Knowing some basic algebra and algebraic topology is a good source of examples. If you want to actually use the tools you're learning for something you might want to know some representation theory, homological algebra, algebraic geometry or algebraic topology.

[crystal vale]

My background involve basic Algebra groups, rings , field so now on which topic I should focus on?

[rocky cloak]

Well, what do you want to learn?

[crystal vale]

Not specific but I want good background in algebra

[cloud walrus]

Mike Hawk

[rocky cloak]

Well you may want to learn some Galois theory if you don't know that. You could also learn the basics of modules and representation theory, maybe even learn about characters of finite groups.

Once you know a little bit about modules you could learn some homological algebra, at which point you might also want to pick up some category theory. After that maybe some commutative algebra.

[crystal vale]

Okay thank you I will do Modules first

[untold basalt]

actually also deg(q)=n

[glossy crag]

Let y be a root of q in an algebraic closure of Z/pZ, then K=Z/pZ(y) is an extension of Z/pZ of degree n (if you don't know what the notation Z/pZ(y) or "algebraic closure" mean, then just let K be the quotient ring Z/pZ[x]/(q), this is a field because q is irreducible and q has a root y in it). The field K has p^n elements (as an n-dimensional vector space over a field with p elements), which implies its every element satisfies a^p^n=a (the size of the multiplicative group is p^n-1, so a^{p^n-1}=1 for all non-zero a by Lagrange's theorem, multiply both sides by a). Thus y is a root of x^p^n-x, so q divides it (as the minimal polynomial of y).

[untold basalt]

thanks

[slim kayak]

If f and g are non-zero divisors and they are contained in each others radical ideals, does it follow that f and g are associated?

[crystal turtle]

No. Take f = p and g = p^2 in Z for a prime p.

[south patrol]

Real

[slim kayak]

Right, weird

(V(p) and V(p^2) both give an effective cartier divisor of spec Z here or am I trippin? I just need one affine cover and taking the trivial cover works fine)

[formal ermine]

nlab differentiates between a norm on a field and an absolute value, by saying that an absolute value is a multiplicative norm, and that a norm does not need to fulfill |fg| = |f| |g|

other texts define norm == absolute value tho

so why does nlab make the distinction

[rocky cloak]

The operator norm is a good example of a norm that isn't multiplicative. If you don't have a distinction you would need to come up with a new word for what the operator norm is.

[south patrol]

I think this is pretty standard terminology from nlab

though you can also call the absolute value a multiplicative valuation and other things ig

[barren sierra]

how do I use the universal property of free groups for this?

[barren sierra]

I mean ok say i is the inclusion S -> F(S) where S = {a, b}

so then apply the universal property the function S -> F(S) taking a to a and b to ba

the universal property says there is a unique homomorphism F(S) -> F(S) such that a maps to a and b maps to ba

actually I think I got it I was way over thinking it

[crystal vale]

How many pair of elements of D8 generate D8 .... is it 8?

[sonic coral]

i’m not sure what you mean by pairs, but D_n is always generated by two elements, r and s

where s has order 2, r has order n, and rs=sr^{-1}

[summer path]

a "spinny" and a "flippy"

[sonic coral]

one thing i’ll keep ruin is be careful with what you read on dihedral groups. it’s not really a convention one way or the other to use D_n or D_2n

[crystal vale]

Yes but can it generates by r^(3) and s ?

[sonic coral]

well i jus told you it can be generated by r, an element of order 4, and s, so see if you can figure out how to get an r from those and you’ll have your answer

[crystal vale]

Yes if I can write sr^(3)=rs then I will get r

Question number 4 ....what question want?

[rocky cloak]

In the lattice, the subgroup generated by two other subgroups is their lcm in the lattice. So if you just look at the subgroups generated by a single element you can see which pairs generated D4. (Also remember that (r) = (r^3))

[jovial prawn]

If we have a field extension K -> L such that [L : K] = n is finite, does it follow automatically that L = K(a_1,...,a_n) where these a_i's are the roots of some minimal polynomial? That is is every finite field extension given by adjoining roots of a minimal polynomial of degree n in L?

[mighty kiln]

Yes, you just keep adjoining a new element until everything is included

[jovial prawn]

Does the minimality of the polynomial play a role here?

[mighty kiln]

Oh you meant like using a single polynomial for all ai?

[crystal vale]

So is there total 8 pair?

[rocky cloak]

I count 12

[crystal vale]

(r,s) (r3,s)
(r,sr) (r3,sr)
(r,sr2) (r3, sr2)
(r,sr3) (r3,sr2)

[rocky cloak]

Now you're just missing those pairs that don't contain r or r^3

[crystal vale]

(s,R2)?

[rocky cloak]

No, that don't give you all of D4

[crystal vale]

Yes

[rocky cloak]

You have 4 elements that are not rotations

Maybe try combining some of them

[crystal vale]

Yes got it

(sr, sr2)

(sr3,sr2)
(s,sr3)
(s,sr)

Thank you

[jovial prawn]

Yeah I meant like an n degree polynomial with roots a_1,...,a_n

[rocky cloak]

I'm not entirely sure what you're asking, but if you have an irreducible polynomial of degree n, and you adjoin all the roots to K, you won't necessarily get a degree n extension.

For example adjoining all roots of x^3 - 2 to Q gives a degree 6 extension.

If you however just adjoin one root, you do get a degree n extension.

It is however not true that any finite extension can be described as adjoining a single root of a polynomial. You may look into the primitive element theorem.

[jovial prawn]

I'm trying to understand what is meant by saying that all finite extensions are algebraic. Can you elaborate on "If you however just adjoin one root, you do get a degree n extension."? If I have a field extension K -> L with [L : K] = n, the can I always find a polynomial (possibly minimal) such that adjoining one of the roots say a of this polynomial in L I'll get L = K(a)?

[mighty kiln]

No

For instance adjoining a cube root of two to Q doesn't get you all the roots of x^3-2

Any finite simple (adjoining by one element) extension is of the form K[x]/P(x) for some irreducible P, which is the minimal polynomial

But not all finite extensions are simple

An algebraic extension is one where for any element x in the extension, there is a finite simple extension K → K(x), i.e. x has a minimal polynomial

But it may not be possible to get all the elements using just one such extension

[rocky cloak]

In most cases you come across L can be described as adjoining a single root of an irreducible polynomial, but there are a few exception.

The simplest exception would probably be to have K be the field of fractions F2(x, y) over F2 with two variables and have L be F2(sqrt(x), sqrt(y)).

Then L has degree 4, but any element in L generates an extension of degree 2 (or 1)

[cobalt heath]

Hm that gives me some curiosity, is there a statement about cardinality of algebraic closure?

[mighty kiln]

The algebraic closure of an infinite field has the same cardinality

And the algebraic closure of a finite field is countable

By just enumerating the polynomials

[cobalt heath]

Is there no case where the closure is finite?

[cobalt heath]

I wonder why is this.

[celest furnace]

No and you get this by showing an algebraic closure would need to contain F_p^n (i.e. the splitting field of x^p^n - 1) for every positive n

[cobalt heath]

Ah, I see

A bit underwhelming that there is not much structure here

[celest furnace]

Wdym

For finite fields there is so much structure that every finite field is F_p and every finite extension is F_p^n

[cobalt heath]

I mean cardinality of algebraic closure being so simple

[celest furnace]

Ah

[mighty kiln]

Wait are you be in model theory reading group

[cobalt heath]

Well, I failed to participate regularly there

Laziness got me

[mighty kiln]

It's probably mentioned there somewhere

[pine crow]

If M is a module and U is a submodule of M. It exists an isomorphism from M to U. Can you conclude U=M? Does this differ if it isn't a module but a group, ring, vector field?

[cobalt heath]

2Z is a submodule of Z

[pine crow]

That's what I thought as well. In this proof I am trying to understand they conclude somehow from M and U being isomorph that they are the same.

the premise being a bit more complicated though. not sure if this is a the right channel for it. K is a field and K[t] is the polynom ring. M is a finitely generated K[t]-module and and M_tor the torsion-module of M. To proof is that dim(M)< infinite being equivilant to M=M_tor the proof basically gets to the point that we know M is isomorph to M_tor and they conclude M=M_tor

In the proof the they just casually say M=M_tor. I feel like I am missing something crucial that is super obvious but I just dont see it

[mighty kiln]

Every element of M_tor is torsion, so if M is isomorphic to M_tor then every element of M is also torsion

[cinder hill]

How is the hint supposed to help? Can't you just show that the number of Sylow 13-subgroups is 105 and through a few more computations show that G would contain more elements than it does if it were simple?

Honestly proving the hint was way harder than the problem

[glossy crag]

I know next to nothing about transcendental extensions, anyone got a good reference for their basics (stuff that gets used in number theory/AG all the time)? One thing I'm interested in learning in are separating transcendence bases, e.g. I've read that if L/K has trdeg 1, then x trans can be chosen so that L/K(x) is separable.

[hidden wind]

anyone know of a abstract algebra textbook in the manner of Viro’s Elementary Topology?— in that it sets up everything for you to work out all the proofs yourself, mwahaha

[balmy belfry]

Does anyone know how I am meant to prove that if the order of a in a group G is n, then the order of a^k is n/(n,k)

Or rather

How am I meant to show that it is the smallest

Number that behaves like the order

I proved the first part but I can’t prove it is the smallest

Wait can I just say nk/(n,k) is the smallest multiple of n and k?

Thay doesn’t feel like it really proves it tho

If a^x = e then n|x right so then n should divide the order of a^k

And nk/(n,k) is the smallest multiple for that is true right

By definition of GCD

[daring nova]

If you squint hard enough, it does

[balmy belfry]

K thx

[slim kayak]

If m is the order of a^k then a^km=e, so km must be a multiple of n. Furthermore if p is a multiple of n trivially a^p=e. From here it should be clear, no?

[balmy belfry]

🐸🐸🐸🐸🐸

[rocky cloak]

Yeah, just assuming that none of the sylow subgroups are normal and counting the number of elements you get should overshoot the number of elements in G.

[pine crow]

yep that was ... kinda obvious. I always miss those conclusions. Thanks a lot

[rotund aurora]

This sounds false? Like consider A=K(x^p, y^p,....) below B=A(z,w,...)/(z^p-x^p,w^p-y^p,....) and B(T)/A where T is an indeterminate

And K has char p

[deft pecan]

Elements of Abstract Algebra by Clark is exactly this

[rotund aurora]

There is an equivalence between varieties of some kind of dimension n over k and finitely generated field extensions of transcendence degree n over k. So the study of such transcendental extensions is the content of algebraic geometry. I know books that focus on n=1 from the field extension perspective (Rosen is fine), but idk for higher dimensions. But I guess the translations are most of the time quite explicit.

[rotund aurora]

This is true in general if you replace separable with algebraic. If you look at the property that a subset of L/K be algebraic (i.e., dependent) or transcendental (i.e., independent) you get a matroid, which basically formalizes the notion of independence of a vector space. So you can show that transcendental bases exist and they all have the same size (if infinite, this requires AC), and the proof is the same as in the vector space case

I'm sure you already knew this, but just in case

[languid sapphire]

why tau to the power 25 equal to the other thing? I know the order of the part they added is 5, and so it should be something divisble by 5, but i dont understand why specifically 25?

also sigma is equal to (3 11 7 6)(8 2 5)(1 4)

[hidden wind]

wonderful, thank you

[thin bough]

y^2 = (1)^3+7(mod 37)
y^2 = 8 mod 37

how do I go about solving this ?

[boreal inlet]

Uhm.. aren't both relations completely equivalent

Also if you just gave y^2 = 8 mod 37, I don't think any solutions exist

[thin bough]

yeah i'm trying to understand why not solution exists

[boreal inlet]

y^2 - 8 is irreducible in F_37 [y]

That means it cannot have a root in F_37

I think that is enough to see it

[chilly ocean]

in general you can use eulers criterion to check if something is a quadratic residue mod p

[boreal inlet]

That's also there yes

I completely forgot about that, too engrossed in Rings and Fields i guess 💀

[chilly ocean]

before the edit it said mod 27 and i was so confused and had to double check that 27 isnt prime 🤣

😭

[thin bough]

yeah sorry

[boreal inlet]

I mean you can work with non primes but as far as my experience with elementary number theory goes it's pain

[chilly ocean]

the only other criterion i know only works for N=pq

is there another one?

[boreal inlet]

I think you can extend it to cases of Z/p^nZ

Like here 3^3 = 27

It has to be prime powers

I'm not sure how exactly it works though, I have to work it out

[thin bough]

thanks for the help

[boreal inlet]

You're welcome

That brings me to a question

But it's about Field theory

Let C[t] be the polynomial ring over C on a single variable, and let C(t) = Frac(C[t]).

Let G be a subgroup of the group of C-algebra automorphisms of C(t), such that it is generated by t -> 1/(1-t), and t -> (t-1)/t. I have to find Fixed field of G wrf C(t).

If the first automorphism is sigma and the latter one tau, we see tau = sigma^2 wrt composition, and hence this group is just the cyclic group of order 3.

But after that I'm completely confused on what to do.

Fixed field of G wrt C(t) = C(t)^G = {k in C(t) | sigma(k) = k for all sigma in G}

[thin bough]

so you just have to find 2 roots x,y that are in F_37 to show that a solution exists ?

[boreal inlet]

Yes. If you can factor the polynomial into linear ones, the constant terms in each factor is your root

[thin bough]

for x=3, it shows that 3^3+7(mod 37) has y value of 16 but i don't get why

oh i get it

it's because 16^2 mod 37 = 34

[elder wave]

so real for that

[slim kayak]

if 27 was prime, what would it be divislbe by? Certainly not 3, since then 57 = 27 + 30 would be divisible by 3 too

[rocky cloak]

Notice that if f is any rational function, then
r(f) = f + tau(f) + tau^2(f)
Is in C(t)^G.

Then just checking that t has degree 3 over C(r(t)) gives you that C(t)^G = C(r(t)).

[flint crater]

So given $e_1 \wedge e_2 + e_3 \wedge e_4 \in \bigwedge^2 \mathbb{R}^4$. How can I show that this is not elementary. I am a bit confused how to proceed at all

[cloud walrus]

ℝ

[flint crater]

The bigwedge thing doesn't quite look right but idk how to properly do that so sorry

So I know that if it is elementary I could write it as $v_1 \wedge v_2 = e_1 \wedge e_2 + e_3 \wedge e_4$

[cloud walrus]

ℝ

[flint crater]

(if I am not mistaken)

[celest furnace]

Not sure what channel this is for but it’s not this one

Maybe try #diff-geo-diff-top

[balmy belfry]

HI RIKU

🐟🐟🐟

[boreal inlet]

Amazing. I'd have never thought of this.

Checking degree 3 I can do by a certain exercise. This element r(f) is not in C but in C(t), and definitely can be written in a form g(t)/h(t) wherw both h(t) and g(t) are non zero non constant polynomials where gcd (g,h) = some unit in C.

I have proved a result before which shows that the degree of C(t) over C(r(t)) is maximum of degree of g and h, which I can manually calculate to be 3

The first part is easy to show, I just show image under sigma and tau

After this, I see that

C(r(t)) \subset C(t)^G \subset C(t)

So

[C(t) : C(r)] = [C(t) : C(t)^G] [C(t)^G : C(r)]

As 3 is prime

[C(t)^G : C(r)] is either 1 or 3

[boreal inlet]

Actually wait now I'm not sure how to proceed. Do we know for sure C(t)^G can't be same as C(t)? For that we need to show an element which is not fixed. Oh wait that's trivial lmao

Yeah we are done

Thank you very much

[cobalt heath]

I wonder how it happens that permuting roots are this hard

[lime junco]

my condolences

[stuck fiber]

Let $R$ be a ring and $I$ a left ideal of $R.$ Let $A$ be a left $R$-module and $S$ a non empty subset of $A.$ Define $IS={\sum_{i=1}^{n}r_{i}a_{i}: a_{i}\in S, r_{i}\in I, n\in\matthb{N}}.$ I'm trying to prove that $IS$ is a submodule of $A,$ but I'm missing closure under addition

[cloud walrus]

Kenshin
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[south patrol]

Closure under addition is the more immediate thing, since I is closed under addition

[stuck fiber]

Really I thought everything else seemed obvious

[crystal vale]

Is ring of continuous function on the interval [0,1] integral domain?

[summer path]

It is not, try to find a counterexample

[crystal vale]

Got it

If I remove continuous part is it remain ring?

[crimson notch]

hey guys, stupid question: wouldn't that "subfield" in theorem 27.19 rather be a "subring" or I am loosing something there? (btw the book is a first course in abstract algebra, by fraleigh)

[stuck fiber]

Subfield seems fine since Z_n is a field when n is prime

[crystal vale]

Zp is field , when you map T: Z to F then Zp is isomorphic to image of T and because Zp is field so image will be field

[crimson notch]

oooooooh I knew it was something basic that my mind wasn't taking into account lmaoo

thanks!!

[rocky cloak]

IS is by definition a sum of stuff. If you have one sum of some stuff, then add another sum of some stuff you get a sum of some more stuff.

But yeah, everything is pretty immediate here.

[boreal inlet]

Let G be a subgroup of C-algebra automorphisms of C(t), isomorphic to Z. What should be the fixed field of G?

This G is generated by the map which takes t to t + 1

One thing I'm sure is that C is contained in C(t)^G

[cobalt heath]

So it sends f(t) to f(t+1)?

[boreal inlet]

Yes

[cobalt heath]

Yeah, solve f(t) = f(t+1)

[boreal inlet]

You mean determine coefficients of f(t) where this happens?

[cobalt heath]

Yeah, maybe hard to show directly but doable imo

Another way to see this is that f(t) - f(t+1) = 0.

[boreal inlet]

Will keep this in mind.

Before that I reached that if some f(t)/g(t) is in the fixed field, then

f(t+k)g(t) = f(t)g(t+k) for all natural numbers k

[cobalt heath]

Well i thought this kind of shows it obviously but well, still need to see it is zero polynomial

Meanwhile I can indeed handwave it like, since f(t) = f(t+1) as a polynomial, f(0) = f(1) = f(2) = ... (by applying the evaluation map)

Can this happen?

[boreal inlet]

Holy shit I completely forgot about this one

[cobalt heath]

It is quite difficult to recall & remind this

Esp., as you see, since polynomial we are dealing with now is abstract structure

[boreal inlet]

By the way, wait, does this suffice? I want to determine if something larger than C is the fixed field here

[cobalt heath]

You can do the exact same with rational functions here.

That is, f(t) / g(t)

(C(t) is made of rational polynomials, right)

[boreal inlet]

Yes

[boreal inlet]

Which reduces to this case. This doesn't happen for all polynomials btw, because x and x^2 directly contradict it.

I think iff the degree of f and g are equal this should hold.

Atleast for degree 2 it holds

Have to check if it's true for other degrees

[cobalt heath]

Evaluation should still work for rational functions as well

You can regard this as an classical algebraic geometry exercise btw

[boreal inlet]

Starting with f(t+k) = f(k) gives me this

[cobalt heath]

I mean

f(0)/g(0) = f(1)/g(1) = f(2)/g(2) = ...

[cinder hill]

You can just use third iso to prove that G'/G''' is the direct product of the other two quotients and hence abelian right?

Does that work?

[rare grotto]

hello everyone

can anyone recommend me a channel for absrtract algebra

would really appreciate it

[ashen heron]

what do you mean recommend you a channel

this is the channel for abstract algebra lol

#❓how-to-get-help to get a help channel

[rare grotto]

oh oh sorry

youtube channel

[ashen heron]

[rare grotto]

haha

[ashen heron]

https://www.youtube.com/@elliotnicholson5117 and https://www.youtube.com/@MichaelPennMath

[rare grotto]

thanks matye

are they good?

also what about discrete math

[ashen heron]

i barely follow math youtube, ive just seen other people recommend these channels here before and i don't know about discrete math. try asking in #discrete-math

[rare grotto]

alirght thanks!

[wraith cargo]

no
third iso just says that G'/G'' is the quotient of the two quotients
It needn't be that it's the direct product

[boreal inlet]

I am unable to see what to do after this.

[cobalt heath]

How many zeros does f(x) / g(x) = f(0) / g(0) have, in this case?

[boreal inlet]

I mean, is f(x)/g(x) a polynomial? I don't understand what do you mean by zeroes of f(x)/ g(x).

[dim widget]

So suppose a rational function was invariant under t \to t + 1. What can you say about the set of roots?

[boreal inlet]

If by rational functions you mean the elements of form f(t)/g(t), then the roots of h(t) = f(t)/g(t) is precisely the roots of f(t)

And in that case f(t) and f(t+1) have the same set of roots

[rare grotto]

where do i study abstract algebra from

[boreal inlet]

#book-recommendations message

[rare grotto]

is there something like videos

[boreal inlet]

You might try out Richard Borcherds' lectures

[cobalt heath]

Any functions can have zeros

Oh I worded it confusingly

I mean zeros of "f(x) / g(x) - f(0) / g(0)"

[boreal inlet]

Oh okay

[rare grotto]

are there other options

i am in despreate help of someone who teaches it

[cobalt heath]

So you can say something about its set of roots
(Or, zero sets)

[elfin wraith]

This might be quite a vague question so sorry if it’s hard to answer, but my course notes have just defined the evaluation of a polynomial as a mapping
$$R[X]\to\text{Maps}(R,R)$$
For a commutative ring R. I’m perfectly happy with that, but below it is an exercise to show that mapping isn’t injective if $R=\mathbb{F}_p$for prime p, again fine with this

But as a note to the question the notes mention “this means that in general, even over fields, a polynomial is not just a special type of function!”

Could anyone elaborate on that? Are they just trying to emphasise the POV that polynomials should be viewed as elements of a ring which can be evaluate rather than a function in their own right?

[cloud walrus]

Nope

[cobalt heath]

But.. but one can refute functional extensionality

[elfin wraith]

What is functional extensionality? Never heard of that before

[cobalt heath]

Well I was just talking in terms of constructivist math

I think it basically means that equality for polynomial is not equality of it as a function, so you cannot regard polynomials as a subset of 'space of functions'

About extensionality: https://en.m.wikipedia.org/wiki/Extensionality

[boreal inlet]

That I got.. but what do I get from it?

Only thing I can think of is this the following:

[cobalt heath]

The following?

[boreal inlet]

(\prod_{i = 1}^n (t - \alpha_i) = \prod_{i = 1}^n (t + 1 - \beta_i))

[cloud walrus]

Riku

[boreal inlet]

Which would say each root beta_i is (alpha_j) - 1 for some j

This I can write because C is algebraically closed and hence both f(t) and f(t+1) splits over C

[cobalt heath]

Hmm, what is alpha and beta?

[boreal inlet]

Oh uh alpha_i are the roots of f(t), and beta_i are the roots of f(t+1) if you consider t+1 to be a variable

As it's the same polynomial, then the set of roots are unique upto permutations

[cobalt heath]

Yes

Let's keep it rather simple. Say, a is a root of f

Then can you trace what should also be a root of it?

[boreal inlet]

a - 1?

Or a+1?

[cobalt heath]

Is it either-or, or both?

[boreal inlet]

It can be both

Don't know if both always

Because let's say there can be repeated roots

a occurs maybe twice

a is also a root of f(t+1), hence, a - 1 is a root

but as a is a root, there should also be a root a+1 somewhete

[cobalt heath]

Wouldn't that mean it is both?

[boreal inlet]

...oh right

Yeah my bad

I'm awfully slow today 😔

[cobalt heath]

Nah it's fine, I'm often slower

[boreal inlet]

But wait doesn't that mean there are infinitely many roots

[cobalt heath]

Yes

[boreal inlet]

.... what the.

That's absurd no

How can a polynomial have infinite roots

[cobalt heath]

One polynomial over C has so many roots that you cannot count

[boreal inlet]

no way, is it the zero polynomial

[cobalt heath]

Ye

[boreal inlet]

lmao

ok so something is definitely wrong

[cobalt heath]

Now we are nearly done

[cobalt heath]

Wdym

[boreal inlet]

No I mean by this, i can't start with a non-zero polynomial

[cobalt heath]

Do you recall where we began as an assumption

[boreal inlet]

...oh, right, we didn't say f(t) has to be nonzero. We only needed g(t) to be non zero

So this works

Wait does this mean C(t)^G is trivial?

[cobalt heath]

Well btw

Not every polynomial in C[t] has a zero!

[boreal inlet]

That is true

Non-vanishing polynomials

Which include the constant polynomials

[cobalt heath]

Indeed

[boreal inlet]

Wait is it only the non-zero constant polynomials?

[cobalt heath]

On C[t], every nonconstant polynomial has a zero

[boreal inlet]

oh my god how did I forgot about that.

Yeah so

I think this is good enough evidence to conclude C(t)^G has to be C

Or am I missing something

[cobalt heath]

Exactly.

[boreal inlet]

FINALLY

thank you so much

[cobalt heath]

You’re welcome!

[cinder hill]

No you must have misread it I typed G triple prime

[boreal inlet]

Honestly this blew my mind

[wraith cargo]

yeah that's what I meant sorry

[cobalt heath]

This problem has Algebraic geometry feeling, so def check it out!

[boreal inlet]

Will do, thanks

[boreal inlet]

I still can't get over this solution why and how is it so good

[dim widget]

It’s good because it’s geometric!

[elfin wraith]

Just bumping this because i still can’t really work out what he was referring to with that comment

[glossy crag]

That's exactly what they're emphasising, you have a procedure that for every "formal" polynomial in R[x] (an infinite tuple of elements (a_0,a_1,a_2,...)) gives you a function R->R (map c\in R to a_0+a_1c+a_2c^2+...), but this procedure does not always output different functions for different polynomials, i.e. is not always injective. It is injective e.g. over infinite fields (if formal polynomials f and g take on the same value at every point, then they are equal as elements of K[x], i.e. as infinite tuples (a_0,a_1,a_2,...)), but over finite fields this is not true: over F_p, the formal polynomials 0 and x^p-x take on the same value at every point of F_p (because a^p\equiv a mod p for all a\in Z). So polynomials as functions R->R and as formal tuples (a_0,a_1,a_2,...) are not exactly the same (they are the same e.g. for infinite integral domains, in particular infinite fields).

[elfin wraith]

Ah ok thanks that does make sense, I worked out the function wouldn’t have been injective by FLiT but I wasn’t 100% certain that this is what he was getting at

Really wish this wasn’t an LA course the ring theory section has been by far the best lol

[dim widget]

I think thinking about basic ring theory through the lens of linear algebra is a very good thing

many people will go through the whole undergrad curriculum and not learn these things

[elfin wraith]

I actually think we’re doing the opposite and approaching LA via ring theory, but yeah it’s not terrible but the full course is basically building up to the proof of the Jordan normal form which I guess just doesn’t feel like all that exciting of a result

But who knows maybe I’ll actually find it amazing when I get there lol

[south patrol]

Well there is a nice proof of jordan normal form using ring/module theory which is cool

And more conceptual imo

[static glen]

Okay you view a vector space as a K[X]-module and fire some decomposition theorem on it, but i think it is also nice to work out the jordan canonical form using methods of linear algebra

[glossy crag]

JCF via LA is perfectly conceptual imo, it's the rational CF that's a bit awkward (at least the proof I've seen).

[severe linden]

(D&F p71) What would be the intended way of proving this here? "Just doing" it, or is there some clever way with lattices (as this is the chapter)?

[hidden wind]

i love how abstract algebra is bombarding me with excuses to draw figures

mwahahaha

also how i can compute some of these symmetries just by looking at the object the symmetries act on haha

just casually folding a pentagon in the study hall

*insert jens duck dance*

possibly meaningless figures but fun fun

[glossy crag]

There's really not much to it, at most a line of calculation.

[severe linden]

Yeah, just proving that s² = (r²)² = (sr²)² = 1 is enough right?

[dim widget]

yes

rotation by 180 degrees commutes with reflection

[severe linden]

Yeah

I was just wondering if there was some amazing way using lattices it wanted me to do, but okay!

[dim widget]

there are lots of ways to do it because it's such a small group

[severe linden]

Yeah that makes sense

In another question, I was asked to find the 12 pairs of elements which generate D8, given this lattice. I quickly found (s,rs), (s, r³s), (r²s, rs), (r²s, r³s) and also (r,s), (r,r²s), (r, rs), (r, r³s) because their only common supergroup (is that the term)? is D8 as can be seen from the lattice

Could someone give me a hint on how to find the remaining 4 ones?

[coral spindle]

I'll point out <r^4> for free

Oh wait, is this D_8 of order 8 or 16?

[severe linden]

The order of r is 4

[dim widget]

it looks like it's order 8

[coral spindle]

Right, I thought this was an incomplete lattice lol

[severe linden]

Mhmm, could it be that there are in fact only 8 and that the question is wrong?

From what I see, whenever you pair up two other elements, you always get something which isn't D8

[coral spindle]

I can't seem to see what the other four would be. The only common supergroup of the cyclic subgroups being the whole of D_8 is precisely the condition needed (as you have realised), and unless I'm missing something very silly there are indeed only 8 such pairs.

[severe linden]

Quick check that I didn't misinterpret the question?

[coral spindle]

Not that I see.

[severe linden]

Interesting, well, thanks for your time anyways!

[coral spindle]

I imagine there is a fun approach to showing this is sufficient using the action of D_8 on these pairs of generators, but I can't see how it would work without sifting through some details.

[hidden wind]

i hate this ambiguity, lol

[mighty kiln]

r³ perhaps?

[wooden fulcrum]

Hello

[coral spindle]

Yeah, whenever I use D_n I do make sure to state it's of order n or 2n, it's a horrid place to be in but I guess this is the bed we've made :(

[wooden fulcrum]

Say I am to write 1/[polynomial with some variable) without using anything but rational numbers in the denominator

if I know something about another polynomial of the same variable

Apparently it's to do with field extensions but the parts explaining that are missing from our lecture videos so I've no clue what's going on

[formal ermine]

then...?

[wooden fulcrum]

(concrete example write 1/(α^2 + α +1) without using anything but rational numbers in the enumerator when I know that α^3 +2α^2 = -2)

[wooden fulcrum]

what do I read to understand how to solve this

[formal ermine]

how to solve what

[wooden fulcrum]

a question such as this

[formal ermine]

theres no question

[dim widget]

it's true that this is illegible

but i think the question is something like

[formal ermine]

thank god im not going insane

[dim widget]

Let K = Q[x]/f(x) with f(x) monic irreducible

Then let $\alpha \in K$

[cloud walrus]

kålrot

[wooden fulcrum]

well a task

[dim widget]

write \alpha^{-1} in the basis given by x, using only the expression of \alpha = \sum_i a_i x^i, and the coefficients of f

[wooden fulcrum]

to write 1/(α^2 + α +1) without using anything but rational numbers in the enumerator when I know that α^3 +2α^2 = -2

[formal ermine]

i keep forgetting what is top and what is bottom in a fraction

[wooden fulcrum]

enumerator is bototm

this is one of the ways to solve it that we were shown

[formal ermine]

denominator is bottom

numerator is top

[wooden fulcrum]

I guess I can repeat it to get solutions for other exercises of this type but idk what I'm doing lol

[formal ermine]

enumerator doesnt exist

[wooden fulcrum]

ah wtf sorry

im on 45 minutes sleep

I googled twice how to call the bottom thing and forgot it both times

[formal ermine]

is what tteg said what ur looking for?

[wooden fulcrum]

idk we didn't get the field extension videolectures

[formal ermine]

but ur using it here

[wooden fulcrum]

that's the prof solving the question on video

he referred to it as if it's something that we know but for some reason the field extension videolectures weren't uploaded so it made no sense to me

I looked up field extensions on wikipedia and that looked all fine but I didn't find anything relating to this

[dim widget]

$\alpha*(\alpha^2 + \alpha + 1) = \alpha^3 + 2\alpha^2 + \alpha - \alpha^2 = -2 - \alpha^2 + \alpha$. If we add $\alpha^2 + \alpha + 1$ we get $2\alpha - 1$, so we see that $(\alpha + 1)*(\alpha^2 + \alpha + 1) = 2\alpha - 1$

[cloud walrus]

kålrot

[dim widget]

Now we want to find the inverse of 2\alpha - 1 which is simpler

[wooden fulcrum]

so the question is literally "write 1/[polynomial with some variable alpha] without using anything but rationals in the denominator if you know [different polynomial with variable alpha equals zero]" sorry for the bad phrasing before

and like I jsut don't get wtf it is I'm doing

how can I even flip a polynomial like that

[dim widget]

Okay here is an example:

[wooden fulcrum]

or a non-polynomial into a polynomial I mean

[dim widget]

suppose x^2 + x + 1 = 0

then we want to find 1/x

but expressed as a polynomial in x with rational coefficients

well x^2 + x = -1 thus x(x + 1) = -1 so we see that -(x + 1) = 1/x

so that's how we flip the polynomial 1/x

In your problem if x^3 + 2x^2 + 2 = 0

then you want to find 1/(x^2 + x + 1)

[wooden fulcrum]

so uh it's

we find an inverse of the 1/something and an inverse of that that is a polynomial?

[dim widget]

well this means that you want a polynomial p(x) = a + bx + cx^2 such that p(x) * (x^2 + x + 1) is a polynomial divisible by x^3 + 2x^2 + 2

to do this is an annoying linear algebra problem but it is always solvable as long as the original polynomial is irreducible (which is the case for your polynomial)

in your specific case it's easier because the degrees are small

so the system of equations you're supposed to solve is not terrible

[wooden fulcrum]

ah I think I'm starting to get it a bit

I kept thinking of a general polynomial as a function and how we were supposedto flip it

[dim widget]

yes, it's only possible to flip it once you impose the condition that x^3 + 2x^2 + 2 = 0

then you can flip any polynomial which is not divisible by x^3 + 2x^2 + 2

[wooden fulcrum]

so I have 1/polynomial1
an I want to find a polynomial2 such that polynomial1 * polynomial2 = 1 ?

[dim widget]

sort of

you want polynomial1 * polynomial2 = 1 + polynomial3 * (x^3 + 2x^2 + 2)

here there will typically be many choices for polynomial3

[wooden fulcrum]

[dim widget]

here's a simpler example

do you know about modular arithmetic?

[wooden fulcrum]

yes

[dim widget]

okay, suppose we have a prime number p (say p = 7) this is the analogue of an irreducible polynomial

[wooden fulcrum]

we had lectures on groups and rings/fields and stuff just this was left out so it came very out of the blank

[dim widget]

then i can ask you find the inverse of 35 modulo 7

[wooden fulcrum]

is this modular arithmetic with polynomials

[dim widget]

1/35 mod 7 is a number n such that $n35 = 1 + m7$ for some $m$

[cloud walrus]

kålrot

[dim widget]

yes this question about inverses is basically about modular arithmetic with polynomials

[dim widget]

anyway continuing with this

Now we want to find 1/(2\alpha - 1)

[wooden fulcrum]

but the last part is 0

[dim widget]

it's 0 modulo x^3 + 2x^2 + 2 yes

[wooden fulcrum]

having in mind that such a polynomial is zero we're just finding polynomial2 such that polynomial1 * polynomial2 = 1 no?

well since if the zero polynomial comes up we can just poof it?

I guess that's just a rephrasing of what youd said

[dim widget]

yep

my way just makes it easier to apply an algorithm

I feel

anyway the way to do with is to do polynomial division

using Euclid's algorithm

[wooden fulcrum]

that's the thing the prof showed us

then plug into bezout

[dim widget]

yes

[wooden fulcrum]

pog I think I get what is wanted of me now

so if the zero polynomial wasnt irreducible we'd get a correct but a bit too big of a solution right?

[dim widget]

err

if the two polynomials are not coprime then there will not be a solution

just like p does not have an inverse mod p

since 0 never has an inverse

[wooden fulcrum]

ah

and if they're coprime then the solution will be too big but correct no?

but modulo its zero divisor it would equal the right smallest solution?

[dim widget]

i don't know what you mean by too big

but if they are coprime it doesnt matter if one is irreducible or not

however if one is irreducible then it is easier to check that they are coprime

[wooden fulcrum]

a power of the actual solution I guess

like we'd get a polynomial2 for which polynomial1 * polynomial2 = 1 mod polynomial 3 holds

but there would exist a polynomial2' | polynomial2 for which it also holds or something

[dim widget]

here why dont you try this example

what is the inverse of x + 1 modulo x^3 + 3x + 6 = 0

[wooden fulcrum]

like same as say with mod 3 it would be true that 16 = 1 mod 3 but 4 = 1 mod 3 also

ok will do thnaks

[rocky cloak]

Think carefully about the subgroup generated by r

[coral spindle]

Oh lmao indeed it was something silly

@severe linden let it be put this way: <r> = <?>

[severe linden]

<r³>?

[coral spindle]

U got it

[severe linden]

Yeah, okay haha

Damn

Thank you all

[wooden fulcrum]

OHHH

did it 2 different ways and got same results with both pog

it's starting to make sense nice, thank you so much

ig I still don't know the whole story regarding the field extension part of it or whatever but I will come back to that later

[rotund aurora]

is there an easy description of the ring of real valued continuous functions as a ring?

[dim widget]

no

[rotund aurora]

ok

but isn't the field of fractions like R(continuum many variables) ?

[dim widget]

i doubt it

some functions have square roots

there are a lot of algebraic relations

i don't think it's just generated by a bunch of algebraically independent elements

[rotund aurora]

Ah I see what you mean

[dim widget]

yeah I think it's quite tough, probably writing down some description would run into logical issues (maybe you need choice?) it's not reasonable to expect I think.

[rotund aurora]

You probably need choice just to invoke a transcendental basis

If you write down a polynomial in many variables and look at its zero locus in R^n then every connected component can be parametrized by continuous functions lmao, so yeah very complicated

sry maybe not all of the connected component, you have some restriction because it ought to be a function. But some part at least

[hidden wind]

is there a group homomorphism for any m > n from Zm into Zn ?

no that can't be i must be very confused

there is no homomorphism from Z3 into Z2

[rotund aurora]

Z/mZ is larger than Z/nZ, so idk what you mean by "into"

[hidden wind]

onto

sry

[rotund aurora]

There are many examples

Suppose m is divisible by n. Then there is a canonical homomorphism Z/mZ-->Z/nZ

[hidden wind]

yes but not for any m > n, right

[rotund aurora]

For any no, because m and n can be coprime

[hidden wind]

excellent, thanku

[coral spindle]

If we have a group homomorphism f : G → H and g in G, then |f(g)| divides |g| — the order of the elements.

So if you want a surjective homomorphism f : Z/m → Z/n you know something in particular

Firstly, that the image of 1 in Z/m generates the whole of Z/n — maybe this isn't totally clear so you should think of this.

Secondly, the order of 1 is m and the order of f(1) is n

So in particular, you must have that n divides m. This is a necessary condition for an 'onto' map to exist.

And you can probably see quite easily why this is also sufficient.

@hidden wind that should clear up any lingering doubt

If you didn't mean for the map to be onto (i.e. surjective, although I dislike this 'onto' terminology) then there are trivial maps that work.

[hidden wind]

thank thank i've embarassed myself plenty for good motivation to figure this out

*insert jens duck pray*

my continued confusion in elementary number theory is making abstract algebra much more FUN, mwahaha

[spiral wolf]

I have the minimal polynomials found but I’m not sure how to proceed

If I could get a nudge in the right direction I would appreciate it

[celest furnace]

,rotate

[spiral wolf]

Thank you

[celest furnace]

You need to find a polynomial such that f(2^1/4+i) - i =0

[spiral wolf]

Yes

[celest furnace]

Maybe divide the minimal polynomial by like 1+x^2?

[spiral wolf]

Possibly? I did an easier version of the problem for practice: express sqrt(6) as a polynomial of sqrt(2)+sqrt(3)

I got 1/2(sqrt2+sqrt3)^2 - 5/2 just by brute forcing it

But this polynomial did not seem related to the minimal polynomials x^2-6 and x^4-10x^2+1 in any obvious way

Not by polynomial division for one

[celest furnace]

I mean you could probably do something similar

Take this to the 4th power to get rid of the 2^1/4 term then use the minimal polynomial relation

[spiral wolf]

Let me keep at it then, that’s what i was trying originally

[celest furnace]

Feels like there should be a better way though

Maybe not though this is like related to the proof of the primitive element theorem and that is hugely non constructivr

[spiral wolf]

This was helpful thank you