#groups-rings-fields

406252 messages · Page 567 of 407

unique juniper
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:/

rustic crown
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like U8 = {1, 3, 5, 7} and has subgroups {1, 3}; {1, 5}; {1,7} each pair intersect trivially, but clearly U8 doesn't have order 8

unique juniper
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yea

rustic crown
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this is the same example but ig written out completely

unique juniper
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but

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if we have sylow 2,3,5 then (sylow 2 * sylow 3) ^ sylow 5 still would be 1

rustic crown
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what is the group here?

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oh nvm, i didn't see the question

rustic crown
unique juniper
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but will G = its sylow groups multiplied?

rustic crown
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if we're assuming the sylow groups commute with each other then yep

unique juniper
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why?

rustic crown
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that's one hypothesis here

unique juniper
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i mean

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how do you know that G = sylow groups multipled given that they are normal & commute

rustic crown
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isn't this was the theorem says?

unique juniper
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its an assumption

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like if G = sylow groups multiplied then G is isomorphic to direct product

rustic crown
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for that we still have that H_1 ... H_n isomorphic to H_1 x ... x H_n

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cardinality of right right is |G|

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hence the subgroup H1... Hn = G

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did i get what you're asking right?

unique juniper
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maybe i can rephrase myself

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so if we want to use this

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We need the subgroups to commute, intersect with identity only and G = groups multiplied

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then we have result

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but my confusion is

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how can we be sure that G = subgroups multiplied

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in this case, G's sylow subgroups

rustic crown
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yea so for this instead of applying the theorem on G, apply it on the subgroup H = H1...Hn

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now the property H = H1...Hn is trivially satisfied by definition

unique juniper
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ok

rustic crown
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so we get |H| = |H1|*...*|Hn| = |G|

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since |H_i| = p_i^a_i and |G| = prod of p_i^a_i

unique juniper
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right

rustic crown
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this actually forces H = G

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since H is a subgroup with same size as G

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ie G = H1...Hn =~ H1 x ... x Hn

unique juniper
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i see

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thanks :D

rustic crown
unique juniper
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another Q :D

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if |G| = 36 then G is not simple

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number of sylow 2 is {1,3,9} and no of sylow 3 is {1,4}

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if we say no of sylow 3 is 4

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let S = sylow 3 groups

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Let G act on S

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Consider the corresponding permuation representation

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See that its not injective

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so kernal isnt just 1

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does this work?

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sure

rustic crown
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yea it works, but you need to say why the kernel isn't all of G, that's because by sylow 2nd, any two elements of S are conjugates. Hence the action is not trivial.

unique juniper
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yes

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ty

rustic crown
unique juniper
terse crystal
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Btw earlier there was a discussion about whether group of order of 48 is simple.Has it been solved yet?

unique juniper
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no but it can be solved the same way

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using group actions

terse crystal
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I see

unique juniper
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yup xD

terse crystal
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I proved directly by showing that there exists two subgroups of order 3 that has the same normalizer therefore impossible though.

terse crystal
unique juniper
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wait

terse crystal
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Yes since all sylow-3 groups are conjugate to each other therefore impossible to have the same normalizer

unique juniper
terse crystal
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Oh I made a mistake their normalizers are of order 12😂😂 then forget it

unique juniper
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with 2nd sylow theorem

terse crystal
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Index of N_G(H) equals to the number of subgroups conjugate to H

unique juniper
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whats a G-set lol

terse crystal
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Yes but I guess it’s a dead end

unique juniper
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oh

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i wasnt aware of it lol

golden pasture
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it does

unique juniper
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category :o

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i wish i knew what that was

golden pasture
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remain innocent

terse crystal
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Yes it has quite good properties like pullback kind of stuffs exist

unique juniper
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im gonna act like i understood what you said :D

terse crystal
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And any G set is isomorphic to a disjoint union of sets of the form of cosets G/H

unique juniper
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my school doesnt have a class about category i dont think

golden pasture
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r e m a i n

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i n n o c e n t

terse crystal
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My school doesn’t even teach algebra...😂

unique juniper
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this isnt undergrad stuff right.... ? categories

golden pasture
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my school teaches me

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how to expand (a+b)c or smt

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i cant rmb

unique juniper
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lol

golden pasture
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categories is high school stuff

unique juniper
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my highschool was emmmm calculus

golden pasture
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categories aren't useful unless you use it

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it's pretty redundant to learn it early on

chilly ocean
golden pasture
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ikr

chilly ocean
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applied category theory is a meme i think

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i don't think they've gotten any new result

golden pasture
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the only time i open text in category theory

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is when i realize i need it

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urs moment

terse crystal
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I think that theoretical physics almost is pure mathematics like I didn’t know that things like stability conditions of triangulated categories have physical backgrounds

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No, not just physical background I think it is only useful in the field of physics.

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I know a mathematics classmate who is studying stability conditions and he has no idea what it does since he doesn’t learn physics nor do I

unique juniper
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i did a physics class and hated it

terse crystal
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I don’t want to learn physics either I just want to know something about Noether theorem like how to prove it someday then don’t touch physics again.

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I searched it on wiki and the “ mathematical expression “ of it doesn’t look like mathematics 😂😂

unique juniper
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lol

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looks like differential equation stuff

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idk tho

terse crystal
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Yeah... I hope that there are some ways to learn mathematics extracted from physics but without learning physics for real.

unique juniper
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i need to study for exams :3

golden pasture
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noether is pretty neat

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you can somewhat fuzz the proof if you blackbox all the variational calc stuff and pretend it works

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which i dont actually know why it works yet

long beacon
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and by hint, i mean "thats all there is"

wind steeple
warped bane
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good evening guys

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anyone here who wanna talk about ZFC and how it's related to Zorn's lemma?

long beacon
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wrong channel bub

acoustic pine
long beacon
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but really its about dg cats

wind steeple
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mirror symmetry ? wtf

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I don't see the link between symmetry and triangulated categories x)

north widget
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hi

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im having smal abmout of trouble interpreting a question

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this is the context

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When it says f(x) has a factor the square of some irreducible polynomial in f. Does it mean
For f(x) = (x-a_1)(x-a_2)...(x-a_n) where a_i are in K, one of the (x-a_i) is equal to some x^2-b in F[x] with b in F?

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or like

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one of the (x-a_i)^2 is an irreducible in F[x]?

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Im thinking the last one.

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Because it says: f(x) has as a factor, meaning one of the (x-a_i), the square of some irreducible polynomial in F[x], p(x) in F[x] irreducible such that (x-a_i)^2 = p(x)?

golden pasture
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it means like

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f(x) = g(x)h(x)^2

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for some g,h in F[x]

north widget
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ok

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so not one of the linear factors, but one of factors in general

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well wait

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it says f(x) is composed of linear factors in F[x]

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K[x]*

magic owl
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i cant find this from googling for 10 seconds so i will ask here does anyone (maybe @latent anvil) know what the locally finite condition means in the context of a hopf algebra?

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Graded hopf algebra if it matters

golden pasture
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if it's all linear factors it's immediate

north widget
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ok

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seen that word twice today, ngl a little triggered

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"immediate"

golden pasture
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ooft

north widget
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oih

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ok

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lol

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because a_i and a_j will be the same

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and thats 0

golden pasture
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yesh

north widget
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ok so ig thats it for one side of the iff

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the => way might be a little different

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wait no

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if its 0 that means one of the factors appears twice 🤩

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ok

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ig it wasnt that hard

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ty 🙂

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roots* instead of factors

magic owl
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nvm sham i figured it out

golden pasture
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but now f may not be a product of linear factors

north widget
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wot

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no i think it is defined as such miss ariana

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in the hypothesis it is defined as being that way in K[x]

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or am i missing smol detaille

golden pasture
north widget
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ah

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it does ask about that.

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it still seems somewhat doable

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if two of the linear factors in K[x] are the same

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hmm

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is the factor squared in F[x] instead of K[x] 🤔

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can't know i assume

golden pasture
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but the linear factors may not be in F[x]

north widget
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but if we assume like you said f(x)=h(x)g(x)^2 in F[x]

golden pasture
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consider for instance (x^2+1)^2 over Q[x]

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it has disc 0

north widget
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btw

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quick question

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how do u know determinant so quickly

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oh

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reapeated roots

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?

golden pasture
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ye

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it factors to (x+i)^2(x-i)^2

north widget
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ya

golden pasture
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but x+i isnt in Q[x]

north widget
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but x+i isnt in F[x]

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but i was falsely thinking that since we know it appears twice , (x+a_i)^2 MIGHT BE IN Q[X]

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BUT THAT ISNT NECESSARILY TRUE

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caps

golden pasture
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ye but it is in K[x] (here K=Q[i] probably)

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yea unfortunately not that direct

north widget
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i think you gave me a really clean lead 🙂

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with the f(x)=h(x)g(x)^2

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but then again

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g(x) in K[x] might not have g(x)^2 in F[x]

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but my goal ig is to show that f(x)=h(x)g(x)^2 using the fact that it has repeated roots in K[x]

golden pasture
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yesh

north widget
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hmm

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so like

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we know for a_i in K f(x)=(x-a_1)...(x-a_i)...(x-a_n)

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or since del f(x) = 0 => f(x)=(x-a_1)...(x-a_i)^2 ...(x-a_n)

golden pasture
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maybe a better place to look at things is consider the ||gcd|| as well as the ||derivative||

north widget
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wanting to go some stupid cheaty way, I would decide to regroup f(x)=h(x)g(x)^2 with h(x)=(x-a_1)...(excluding x-a_i^2)(x-a_n)

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ok

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so

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i def wont look at those spoilers

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but

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tell me if it has to do with galois groups

golden pasture
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nop

north widget
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ok

golden pasture
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nothing too fancy is really needed

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it's just kinda hard to motivate

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(at least the solution im thinking off)

north widget
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ok

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so

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ill look at spoilers in an hour if i dont find any leads

golden pasture
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kool :D

north widget
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ok

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so

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we also know f(x) is monic

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but like

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idk if that helps tbh

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prolly not

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oo

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maybe something to do with basis of splitting field

golden pasture
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it's a field so monic doesnt rlly matter

north widget
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or degree

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hmm

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they do mention degree of polynomial

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and the extension K is of degree n

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so like

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K:F = n

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but no intermediate fields

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unless i decide to take some L as the splitting field of (x-a_i)^2 over F

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wait nvm

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it cant be a splitting field

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im pretty sure

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its either in K or in F

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ok ariana I have another idea of thinking about K as a F linear basis

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we know basis of K is {1,a_i,a_i^2,...,a_i^(n-1)}

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and f(x) is composed of a product of (x-a_i^k) for k<n

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if we know one of these factors is repeated then (x-a_p)^2 with p being one of the a_i^k ends up turning into another a_i^q?

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i cant phrase it that well but im pretty much trying to say that knowing that one of the roots is repeated and is in K means that it gets moved to another basis because it gets squared

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nvm im thinking its simpliler than this

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bruh

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is K just F adjoined some elements in K because K is splitting field over F

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so like

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K = F(a_1,a_2,...a_n)

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and each of these elements can be represented by irreducible monic polynomials in F

golden pasture
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quite tricky to think along these lines

north widget
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i might look at ur hints because even thinking along these lines is gonna lead me to dead end it feels

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i was thinking that since one of the a_i is repeated in f(x)

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either a_i becomes one of the other a_k or it goes into F

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but if it goes into a_k that means something

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can't really think what though

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oh it isnt irreducible

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well wait nvm

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its (x-a_i)^2 not x-a_i

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yeah wait

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if (x-a_i)^2 = (x-a_k) that tells us something but isnt clear to me atm

golden pasture
north widget
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ariana, each a_i is a root to some irreducible polynomial in F[x] right

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solike

golden pasture
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yesh

north widget
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can you replace a_i with f_{a_I}(x) to represent that?

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so like

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(x-f_ai(x))^2

golden pasture
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ehhh

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very shady

north widget
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ig not right

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its more like (x-?)^2 = f_ai

golden pasture
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im not rlly sure how this would work out :p

north widget
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and i would say that (x-?)^2 = x^2-?(depends on characteristic of field)?

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but the x^2 part would still undoubtly be there

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it should also stay irreducible in F[x] though, retracting this statement

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x-? isnt in F[x]

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so x^2-? might not be

golden pasture
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(it wouldn't be)

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unless char 2

north widget
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ok 15 minutes and ill look at hint

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ok

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so

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going along with the line of reasoning that (x_a_i)^2 = x-a_k for some k<n, this implies that (x-a_k)^2 = x-a_l for some l <n

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this would keep repeating though

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but it has to terminate somewhere qq

golden pasture
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(x-a_i)^2 is quadratic in x

north widget
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yes

golden pasture
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it doesnt make sense to say it is x-a_k

north widget
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no i mean like

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it might be the root of the irreduicible polynomial of a_k

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so like

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irr(F,a_i)^2 = irr(F,a_k)

golden pasture
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ehhh

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wat

north widget
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ok not necessarily

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ig my wonder is

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(x-a_i)^2 cant really be in K[x]

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or nvm

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it has to be

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but i mean like

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(x-a_i)^2 might be an irreducible in F[x]

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or it has to be

golden pasture
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it would generally not be in F[x]

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consider (x-i)^2 not in Q(x)

north widget
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yea that makes sense

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as a counter example

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ig the differnece here

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is that

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well not really

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i was thinking we can take an irr(Q,isqrt2) as an example

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and the splitting field would be F(sqrt2,i)

golden pasture
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what is irr

north widget
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irreducible polynomial in Q that has root isqrt2

golden pasture
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ok so x^2+2

north widget
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well fuck

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that example doesnt work

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but yea that would be it

golden pasture
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if you jus sprinkle in a bit of galois in it it would work

north widget
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which part though, degree of extensions?

golden pasture
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the gap is that all your galois conjugates must appear as roots

north widget
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yea 1/a_i

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right?

golden pasture
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so you take a product of Π_σ (x-σa_i) and that gives you a polynomial in F(x)

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where σ goes over the embeddings of F(a_i) to some algebraic closure

north widget
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ok i forget what embeddings are

golden pasture
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but it isnt really a trivial thing to work out

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(intuitively the different roots of the minimal polynomial of a_i)

golden pasture
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this would be a more galois theory based approach

north widget
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hold up

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i dont understand notation of your product

golden pasture
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soo essentially it takes a product over all the F-embeddings from F(a_i) to K

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the rough idea is

north widget
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wait

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is K a seperable field extension?

golden pasture
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minimal polynomial of a_i must divide into f(x)

north widget
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it is right?

golden pasture
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maybe lemme go back a few steps

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so the main idea in this is

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if we have some polynomial f(x) in F(x)

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and it's splitting field is K

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then if a_i is a root of f(x) in K

north widget
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ok

golden pasture
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then the min poly of a_i must divide f(x) in F(x)

north widget
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ok

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it divides because

golden pasture
north widget
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seperable?

golden pasture
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divides cuz minimal polynomial magic

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(essentially if it doesnt, then you can argue by bezout that funny things will happen)

north widget
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idk what bezout is

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im only undergraduate man qq

golden pasture
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bezout is like

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if a,b are coprime

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i can find some c,d such that ac+bd=1

north widget
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oh

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that old trick

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that i used a bunch of times for things i barely remember

golden pasture
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so if a is your min poly and b is f, some linear combination of them gives 1 but a_i is still a root to both of them

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so a_i is a root to 1

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which is very illogical

north widget
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so it isnt coprime

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ok

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that seems unintuitive though

golden pasture
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slight hint for a much easier solution is to use this bezout/gcd thing to your advantage

north widget
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wait

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so like first we assume a = min(F,a_i) and b = f(x)

golden pasture
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yea essentially we have a super nice property that if f(a_i)=0 and g(x) is the minpoly of a_i, then g(x) actually divides into f(x)

north widget
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and a,b are coprime

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so an+bm=1

golden pasture
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yesh

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now plug in x=a_i

north widget
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that gives that 0 = 1

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and that is bollocks

golden pasture
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yup

north widget
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ok

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so then a_i is a factor of f(x)

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but wait

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we knew that

golden pasture
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(minpoly of a_i)

north widget
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we are taking f in a_i

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ok

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ooo

golden pasture
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and your min poly is in F(x)

north widget
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alright

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but this is independent of repeated facotrs

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oh wait

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using the fact that del F = 0

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we have that min a_i ^2 is also in F[x]

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ooookk

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so if min a_i divides f(x) once, and (x-a_i)^2 is a factor of f(x) in K[x]

golden pasture
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the reason we can actually say this is due to the separable part

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separability ensures your minimal polynomial has no repeated roots

north widget
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ok

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so then minimal polynomial is most likely degree 2

golden pasture
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minpoly can have any degree

north widget
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of a_i

golden pasture
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yea it can have any degree

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say like

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f=(x^4+1)^2

north widget
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ok

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oh

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the square of irreducible part though

golden pasture
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ok now the example works

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yea

north widget
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x^3-1 irr in Q

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wait

golden pasture
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(actually it isnt thats why i needed to edit haha)

north widget
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isnt there a forumla

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like

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x-1(x^2-2x+1

chilly ocean
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I think -1 is in Q

north widget
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or something

golden pasture
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x^3+1 and x^3-1 has roots -1 and 1 respectively

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generally this argument is kinda long and meh, a bunch nicer way is to look at the ||derivative||

north widget
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ok

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ill look at spolilers now

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bruh

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derivative

golden pasture
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haha

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yesh

north widget
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what did earlier spoilers say

golden pasture
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it appears

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derivative and gcd

north widget
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bruhh

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ok

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so using seperable mostly

golden pasture
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differentiating p(x)^2 gives 2p(x)p'(x) so gcd gives us what we need

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much faster

north widget
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wait hold up

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so we have that K = F(a_1,a_2,...,a_n)

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and seperable means that derivative of min a_i is not 0

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right?

golden pasture
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yea

north widget
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how tf this translate to no repeated roots?

golden pasture
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the idea is that if f(x) has a repeated root

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gcd of f(x) and f'(x) is *not one

north widget
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ok

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brb

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using intuition

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so like

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gcd = 0 iff f'(x) = 0

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?

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wtf

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gcd is greatest common divisor

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how tf one of them 0

golden pasture
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oops gcd is not 1*

north widget
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ok

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oh

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so like in x^2

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gcd(x^2 ,2x)= x

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no?

golden pasture
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0 can occur for finite characteristic - this can actually detect multiple roots for nonseparable case

golden pasture
north widget
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so x^2 is seperable

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wait what

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ik there are seperable polynomials

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but if we are reffering to field extensions

golden pasture
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separability is a thing about field extensions

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in general case we have
gcd(f(x),f'(x)) is not 1 <-> f has a repeated root in splitting field

oblique river
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Separability is also a thing about polynomials

north widget
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its the same concept though?

oblique river
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Yes

north widget
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ok

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so

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so is a seperable field extension, a field extension over a seperable polynomial?

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or in other words

oblique river
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Idk what a field extension “over” a polynomial means

north widget
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?

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isnt that the terminology

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like

oblique river
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No

north widget
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you are given a field

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you make a polynomial

golden pasture
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(splitting field)

north widget
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the roots of it make another field

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ohh

golden pasture
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that isnt quite the definition of separable field extension tho

oblique river
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Yeah so if you do that process to a a separable poly you get a separable extrensiob

golden pasture
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separable extension demands min poly of every element in the field extension to be separable

north widget
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so if K is the splitting field over a seperable polynomial in F[x] then K is a seperable field extension?

oblique river
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“Splitting field of” not “splitting field over”

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And yes

north widget
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ok

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i need to get my language exact

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ah ok

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so like

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since K is seperable in our example we have that gcd(f'(x),f(x)) ~=(not equal) 1

golden pasture
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(!= usually not equal in text context)

north widget
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ok

oblique river
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Sorry idk what “our example” is

north widget
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um

golden pasture
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it isnt 1 because it isnt 1 in K(x)

north widget
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so gcd depends on field?

golden pasture
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gcd is independent of field actually

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which is super nice

north widget
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ok

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wait

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it isnt 1 in K(x) because a root is repeated

golden pasture
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You can compute the gcd in F(x) and compute it in K(x) and should get the same result if you expand everything

oblique river
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Im confused, K is a separable field extension

golden pasture
oblique river
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And the polynomial f(x) is also separable

golden pasture
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it's as nice as it gets lol

north widget
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ok

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im trying to follow implicitations

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ok

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so

golden pasture
#

so a rough tldr theres like a few ways you can "detect" repeated roots

discriminant = 0 tells you theres a repeated linear factor in the splitting field

a repeated linear factor tells you gcd(f,f') is not 1

gcd(f,f') is some polynomial in F[x], say p, and p^2 must divide f purely by properties of f' and gcd

north widget
#

wait

#

gcd(x^2-1,2x) is also not 1

#

it isnt repeated though

#

its not an iff?

oblique river
#

Are you sure about that?

north widget
#

yeah

#

wait

#

right

oblique river
#

What do you think that gcd is

north widget
#

um

#

bruh

#

you cant divide by x

#

or 1

#

wait

#

only 1

#

fuck

oblique river
#

Bruh what lol

north widget
#

x^2-1 you divide by x-1 or x+1

#

for 2x you only got x and 2

#

so gcd is 1

#

ok

#

so it doesnt have repeated

golden pasture
north widget
#

wait

#

we dont need disc = 0 to tell us that gcd(f,f')!=1 since we get that f is seperable

#

wait

golden pasture
#

we do actually

north widget
#

seperable means derivative not equal to 0

golden pasture
#

yes

#

it essentially takes care of cases like in char p, the polynomial x^p-k

oblique river
#

Why are we involving separability here?

#

That not prt of the problem at all

north widget
#

its a given in the problem though

golden pasture
#

(we don't actually need separability)

oblique river
#

We are given that f is separable

#

And therefore so is K

#

All this means is that if f has a double root then each of those roots come from a different irreducible factor

north widget
#

oh ok

#

seperable doesnt imply no double roots

#

wait it does though

#

we just went over this

oblique river
#

Separable and irreducible implies no double roots

#

The whole point is that f(x) in this problem could be the square of some irreducuble poly

golden pasture
#

separability tells you that your derivative doesnt vanish, this isn't sufficient as you can have finite characteristic

oblique river
#

In which case its discriminant is 0

#

And it has repeated factors

north widget
#

buncho, thats one side of the iff and i understand that part

oblique river
#

Ok so actually there seem to be two definitions of separable out there

#

One of them is “all irred factors have no repeated roots” and the other is “has no repeated roots at all”

#

I was using the former

#

Which side are you trying to understand now?

north widget
#

disc = 0 => f(x)=h(x)g(x)^2 for both in F[x]

#

i think miss ariana has given me the key

#

even though she gave me nice recap ill try and restate once more

#

Lets assume that f(x) isnt divisible by minimal polynomial of a_i in F[x].

#

So f(x) and min(F,a_i) are coprime => f(x) * m+min(F,a_i) * n = 1

#

but this is a contradiction since a_i is a root of both f(x) and min(F,a_i) which implies 0=1. So, f(x) is divisible by minimal polynomial of a_i in F[x]

#

Next we have that (x-a_i)^2 is a factor of f(x) in K[x] since Discriminant of f(x) = 0

latent anvil
north widget
#

ok

#

and gcd(f,f')!=1 because Discriminant of f(x) = 0 implies repeated roots

#

so there is a p(x) that divides both

#

ok

#

and if we have p(x)^2

#

why is it true that 2p(x)p'(x) divides f.

#

yeah why does p^2(x) divide f

#

ariana said purely on properties of f' and gcd

oblique river
#

sorry i'm not sure why that's relevant here? If disc = 0 then you have a repeated root alpha which has a minimal polynomial p(x). f(x) is divisible by p(x). But. f(x)/p(x) still has alpha as a root, since the multiplicity of alpha as a root of f(x) is at least 2, but the multiplicity of alpha as a root of p(x) is 1. therefore f(x)/p(x) is also divisible by p(x), which you can rewrite as saying that f(x) is divisible by p(x)^2

#

if you want to go about it your way, suppose that f(x) = p(x)g(x) and f'(x) = p(x)h(x). Differentiating f(x) with the chain rule, we get f'(x) = p'(x)g(x) + p(x)g'(x), which is equal to p(x)h(x). therefore, the whole left hand side p'(x)g(x) + p(x)g'(x) is divisible by p(x). Since the right term is divisible by p(x) this means the left hand term is as well

north widget
#

oh ok

oblique river
#

but notice that p(x) and p'(x) have a gcd of 1

#

so "p(x) divides p'(x)g(x)" implies that p(x) divides g(x)

#

so back to our original f(x) = p(x)g(x), we see that f(x) is in fact divisible by p(x)^2

north widget
#

wait

#

p(x),p'(x) has gcd one because there is no repeated root

oblique river
#

yes that is the hypothesis of the problem

north widget
#

since p(x) is minimal polynomial

#

right?

oblique river
#

p(x) is an irreducible factor of f(x)

#

and in the problem it says

#

all irreducible factors of f(x) are separable

north widget
#

which means they have distinct roots

#

or gcd(p,p')!=1

oblique river
#

I think you mean =

#

but yes

north widget
#

hold up

#

distinct roots implies gcd =1

#

repeated implies not

#

ok

#

seperable implies distinct roots implies gcd=1

#

ok

#

you cleared a lot up for me

#

really appreciate ya

#

i feel bad though becaause it feels as if ive been given the answer

oblique river
#

np and gl 👍

north widget
#

even though i understand it now

#

how do you overcome this guilt buncho

oblique river
#

you have to be shown some answers

#

or else you won't know how to do it at all

#

if youve never seen a technique before, how are you supposed to know how to use it?

#

i just showed you a new technique

magic owl
#

(related: there are always more problems no matter how many you are walked through)

north widget
#

true

#

but i just feel guilty whenever i ask for help idk

#

even if i thought through it a little bit

oblique river
#

that's not good

north widget
#

or a lot

oblique river
#

there's nothing wrong with asking for help

north widget
#

the way i see it is if i ask for help doesnt that mean im not ready to work on the problem by myself

#

so i should just relearn previous classes because i am not knowledgeble enough

oblique river
#

math is supposed to be hard and if every single problem you do is obvious/easy it means you aren't in the right class

#

idk what else to say other than i think your attitude about asking for help is wrong

north widget
#

perhaps

#

maybe my mindset will change 😔

oblique river
#

the only way you're going to get better at math

#

is learning from people who came before you

#

work on problems with other people

chilly ocean
#

You can also derive the theory yourself I guess

oblique river
#

if you do everything in isolation you're only handicapping yourself

#

yes, and you will take the same length of time it took them to do it originally

#

i.e. 50+ years

north widget
#

oh i guess thats a perspective i havent really thought of

#

isnt every undergraduate math class an accumulation of decades of work in a sense?

oblique river
#

yes

#

more than decades lol

#

just because you learn something in a month doesn't mean it was invented in a month

#

the theorems you learn were full-on research papers at one point in history

#

so like you are free to try to rederive everything yourself but don't be surprised when it takes you a decade to learn field theory while it took your peers 2 months

#

because they asked questions and used the resources available

#

i didnt suggest googling answers lol

chilly ocean
#

I think it's sort of okay to look up answers as long as you understand why it works, understand why the approach was used (what led the person to try it?) And stuff like that

oblique river
#

i gotta go now

#

gl with your work

north widget
#

inah

#

i like asking ppl

#

i enjoyed asking here because i am given opportunity to understand answer

#

id say google overall sucks

#

but best experience ive had is looking up stack exchange answers for an analysis proof

#

but all i got were hints

#

so i followed the hints and arrived at answer myself

#

i like hints.

warped bane
#

Let $E$ a vector space and $ F=Vect(e_1 ,e_2 ,e_3)$ and $ G =Vect(w_1 ,w_2) $two vector sub-spaces of $E$. What is the difference between $F+G$ and $FUG$?

cloud walrusBOT
chilly ocean
#

If I understand correctly e1 + w1 may not be in F \cup G.

#

You can see it by looking at R^2, taking F = lin(0,1), E = lin(1,0), F \cup G should be OX and OY axis, but there are vectors such as (1,1) in F + G.

warped bane
#

I'm afraid I don't get your point

chilly ocean
#

Oh, you are asking how you define those sets?

#

$F + G := { e + w : e \in F, w \in G }$

cloud walrusBOT
warped bane
#

well the two notation left me puzzled

chilly ocean
#

F u G is just a set union. (At least that's how I interpret it)

warped bane
#

I see

#

so F u G is not equal to Vect(e1,e2,e3,w1,w2)?

chilly ocean
#

No, if that's a normal set union.

#

here is an example, consider E=R^2, and F=the x-axis and G=the y-axis. then the union is just a cross looking shape

#

Lol, look up 8da. cocatThink

#

ah lol

warped bane
#

is there anyone relationship between FuG and F+G

chilly ocean
#

I mean, there are special cases where those are equal.

warped bane
#

I assume that FuG C F+G

chilly ocean
#

i guess you can say F+G is generated by FUG, but this is sort of trivial

warped bane
#

i think I'm having it all wrong

#

satisfying enough

#

I appreciate the help

north widget
#

oh ok

#

just want a hint

#

for part b it says show discriminant of f(x) is in F

#

now tell me if this is right path to think of

#

i should write out the products and then make a generalized polynomial that satisfies it as a root and show that it is reducible into linear factors in F[x]?

#

ok

#

not really sure how to "make a generalized polynomial that satisfies it"

chilly ocean
#

maybe that is possible, although this is not the way i would do it

#

well, i don't have the details figured out, and maybe what i'm thinking of doesn't even work, but i think one way to do it could be to consider the fact that ||(\delta f)^2 is fixed under all permutations of the roots||

north widget
#

ok

#

@chilly ocean is the spoiler a hint?

#

or does it forfiet answer

#

forfeit*

chilly ocean
#

i guess it forfeits the answer if it works, but my algebra is not fresh enough in my head if the details work out

north widget
#

Ill look but before that

#

would my way really work or is it fallacious

#

because say you write out a generalized polynomial that satisfies it as a root

#

the generalized polynomial will be in terms of a_i's which are in K

#

how would I make any conclusions about its place in F

#

general speaking ofc

chilly ocean
#

but eg (\delta f)^2 is apparently symmetric

#

this sounds like a good thing

#

and i'm not exactly sure what it means to find a polynomial that satisfies it and reduce it into linear factors, if we wish to show that (\delta f)^2 is in F, then i think this would mean that the polynomial has to be x - (\delta f)^2

#

so it is basically equivalent to showing that (\delta f)^2 is in F

north widget
#

wait

#

idk if del f ^2 expands like that

#

because we dont know the characteristic

#

also wdym its apparently symmetric

chilly ocean
#

i'm not sure what you mean, regardless of the characteristic it should. maybe 2alpha1 alpha2 goes to 0 if we are in characteristic 2

#

you know, (\delta f)^2 is invariant under swapping alpha1 and alpha2

north widget
#

oh ok

chilly ocean
#

i hope i am not taking you down the wrong path, i could be saying irrelevant shit 🙂

north widget
#

the thing is ive heard this

#

no no

#

our chapter goes over symmetric functions

#

i dont actually understand it too well because it wasnt discussed in lecture afaik

#

like when is a polynomial not symmetric?

#

arent roots always interchangeable?

chilly ocean
#

yeah, if you have a function in the form (x-r1)...(x-rn), then roots are always interchangeable and the function stays the same. but we are talking about an expression involving r1,...,rn (alpha1,...,alphan).

north widget
#

ok

#

so when factors are linear roots are always interchangeable

#

actually i cant think of an example when they arent for polynomials in 1 variable

chilly ocean
#

linear or not shouldn't matter, eg, you can work in a field extension in which the polynomial splits, then swap the roots, then come back to the original field, and it will be again the same

#

i'm also not even sure what it means to swap roots if we are not already working in a field extension where the roots exist

north widget
#

so is there an example of a polynomial in Q(sqrt2) that isnt symmetric

chilly ocean
#

i'm not sure what you mean

north widget
#

um

#

Im trying to understand what you mean when you say a polynomial is symmetric by getting an example of when it isnt.

chilly ocean
#

sorry, i just meant like we have (\delta f)^2 which is a formula involving alpha1,...,alphan, and we can take any permutation of these alphas and (\delta f)^2 remains unchanged

north widget
#

is this true though

#

so if we have (a_1-a_2)(a_2-a_3)(a_3-a_4) and then swap to get (a_1-a_2)(a_2-a_4)(a_4-a_3)

#

every instance of a_3 becomes a_4 and a_4 becomes a_3

#

its the same?

chilly ocean
#

you are talking about n=4 case? but (a_1-a_2)(a_2-a_3)(a_3-a_4) is not equal to \delta f

north widget
#

yeah n4 case

chilly ocean
#

\delta f should be (a1-a2)(a1-a3)(a1-a4)(a2-a3)(a2-a4)(a3-a4)

north widget
#

oh

#

i didnt know thats what it would look like

#

so it contians every combination of roots

chilly ocean
#

yeah

north widget
#

multiplied together

#

or permutation is more correct?

chilly ocean
#

something like that, whatever it is, just every pair of roots is in there

north widget
#

ok

#

ill look at your spoiler now

#

oh you say delf^2 is fixed

#

thats because f is symmetric you are saying

chilly ocean
#

erm, because delf^2 is symmetric

north widget
#

yea sorry thats what i meant

chilly ocean
#

or whatever the terminology is, i could be goofing it up

#

have you seen the fundamental theorem of galois theory?

north widget
#

yeah they are a bunch of theorems that relate fields to their galois group which is the group of automorphisms which permute roots of the splitting field

#

some have to do with degree

#

i remember the ones dealing with degree mostly

#

also random question

#

do we know if our field F is char 0 or not?

#

or is there no way to tell from our givens

#

ok

#

so my textbook is giving big leads

#

they didnt describe symmetric functions too well

#

but they brought up a field extension by symmetric functions (tbh best way I can describe it) and one of the galois theorems

#

not too much context in one picture );

chilly ocean
#

if you are talking about whether K is a separable extension of F in problem 12, then i think this is probably true, since K is the splitting field of f and it says the irreducible factors of f are separable

#

but as to whether the field F is char 0, i would guess it is possible to construct examples satisfying the givens in problem 12 such that it is not char 0

north widget
#

yeah ig

#

wait a second

#

ig the way it wants me to do this is to make an intermediate field, say L, using the symmetric function (being the discriminant) and show that its index compared to base field is 1 or [F:L]=1 using the fact that [K:F] = n.

maybe K:L = n also?

#

because the way im seeing it is that the discriminant contains every root and every permutation of it

#

and the galois group of K over F should also contain every permutation of roots

chilly ocean
#

do we know that [K:F]=n?

north widget
#

yea

#

in the problem we are given that K is a splitting field of F over polynomial f(x) which is of degree n

#

oh wait

#

it can be greater than n nvm

#

*K is splitting field of f(x) over F

steady belfry
#

abstract help pls

chilly ocean
#

Ok, so for example, let's take the left diagram

#

What happens if you start with an element (g, h) and follow the left path?

steady belfry
#

umm what is a projection map

chilly ocean
#

You know, like an element (g, h) in GxH maps to g in G, that would be the projection onto the first coordinate (pi_1)

steady belfry
#

so u wanna see if pi_1 of g = pi_2 of h right

#

wait no other way around

#

of g of pi_1 = h of pi_2

chilly ocean
#

Erm, I'm not sure I follow

steady belfry
#

just for the commutative part

#

idrk what im doing

#

lol

#

so if you put g,h on the left side, what happens?

#

how does it work for all of H

#

because you dont know what H is

chilly ocean
#

So I'm saying like, if you start with (g, h), and you go through the first map on the left, you get g, right?

#

What happens when you go through the second map on the left?

steady belfry
#

h?

chilly ocean
#

How do you get h? It should be an element of GxH

steady belfry
#

uhh idrk what im doing

chilly ocean
#

So if you go along the left path, and you start with an element (g, h), it should be like (g, h) -> g -> (g, e)

steady belfry
#

right so the right path would be (g, h) -> h - >(e,h)

chilly ocean
#

Yeah

steady belfry
#

which means its not commutative?

chilly ocean
#

Now the question is, does (g, e) always equal (e, h)?

steady belfry
#

oh

chilly ocean
#

Yeah

steady belfry
#

which is no

#

so

#

not the left one

chilly ocean
#

Yeah

#

(Unless something stupid like G=H=the trivial group)

steady belfry
#

so the right diagram

#

you wanna put in (g,h) again right

#

left path just gets you (g,h) -> (g,h)

#

and right path is (g,h) -> (g,e) -> g

chilly ocean
#

There are two paths in the right diagram, G -> G and G -> GxH -> G (both starting and ending at G). So we should start with an element in G, ie just g

steady belfry
#

oh

#

so it is commutative

#

g->g,e->g

chilly ocean
#

Yeah

steady belfry
#

ok for the first one

#

how does g->g,e map to GxH

snow flint
#

e is in H

steady belfry
#

oh

#

ah

#

i see

#

e is in G too then right

#

@snow flint @chilly ocean

snow flint
#

yes, although e is just used to represent the identity of whichever group ur talking about

steady belfry
#

so for any g in G, g will map to the pair (g,e) where it creates all the elemnts of GxH

#

right

snow flint
#

unless the only member of H is its identity, it wont create all elements of GxH

steady belfry
#

wait so why isn't the left diagram commutative

snow flint
#

because the pairs (g,e) and (e,h) are not always equal

#

you should be able to see that they are only equal when g=h=e

steady belfry
#

yea

#

ok and right side is just g and g are

#

always equal

#

yes

#

ok this makes sense

snow flint
#

yup

steady belfry
#

like a coordinate point

#

pi_1 = g and pi_2 = h in (g,h)

snow flint
#

yes

golden pasture
steady belfry
# snow flint yes

last thing, for the one on the right, when g-> (g, e), it doesn't map to all of GxH

#

right

#

so does the middle step matter or uhh

snow flint
#

do you mean the pi functions

steady belfry
#

no just the g-> (g,e) part

snow flint
#

i mean, it wont be a map from GxH to GxH then

steady belfry
#

ur mapping G to GxH first right

#

does that matter at all lol

#

wait does mapping mean that its mapping to codomain or range

#

right diagram

snow flint
#

oh sorry

#

oh, so you're asking if the first step on the right path is pointless?

#

because we revert it right after?

steady belfry
#

idk i was just doing reverse logic to see if the left one could be right because (e,h) and (g,e) are subsets of GxH since GxH are codomain

#

and for the right diagram, when g -> (g,e), it doesnt map to all of GxH

#

u still have to think abt the intermediate step to see if it works?i think

snow flint
#

wait, why are you concerned about it mapping to all of GxH

steady belfry
#

uhh idk

#

lol

snow flint
#

uh

#

i cant rly help u then lol. not sure what u mean

steady belfry
#

idk im just so confused lol

#

im just trying to help a friend with abs alg but im 15 ;-;

snow flint
#

hav u tried getting older

steady belfry
#

true

#

based

snow flint
#

mmm okay, so originally you just wanted to find out which of the two were commutative, right?

steady belfry
#

ya

snow flint
#

which you now understand?

steady belfry
#

p sure

#

yea

#

first diagram maps to two unequal things

#

second diagram maps to two equal things

#

therefore proving commutativity

snow flint
#

okay yup

steady belfry
#

if thats a word

snow flint
#

wait i said communicative lol

#

okay but yes ur right

#

so now whats ur question

steady belfry
#

idk my friend is tripping me up

#

is the second one right 100%

snow flint
#

tell ur friend to ask here

#

yes

#

the pi (projection) map undoes the addition of the "e" component from before

steady belfry
snow flint
#

uh

steady belfry
#

thats not the point of commutativity though right

snow flint
#

yea

steady belfry
#

she said thats why g,e and e,h dont matter

#

uhh

snow flint
#

,w commutative diagram

cloud walrusBOT
snow flint
#

they definitely do matter

steady belfry
#

uhhh

#

"but g,e is just a subset of GxH and not all of GxH

#

how do you know the maps dont have to be onto"

snow flint
#

uhh

#

what is even the issue

#

i dont see how onto/not onto is leading to the misunderstanding

steady belfry
#

idk tbh

snow flint
#

they dont have to be onto cuz, it doesnt matter? it just has to be a mapping?

prisma ibex
#

why are we talking about onto here? This isn't relevant to why the left diagram doesn't commute

#

the left diagram doesn't commute simply because you get different answers in general for the two morphisms GxH->GxH in this diagram

steady belfry
#

,w projection map

cloud walrusBOT
steady belfry
#

uhh

prisma ibex
#

projection maps are the obvious ones out of products here

steady belfry
#

is there a definitoin for projection maps

#

i can't find one online

prisma ibex
#

yes it's literally just

#

if you have a product

steady belfry
#

first coordinate for pi_1 and second coordinate for pi_2

prisma ibex
#

the projection maps are the ones that project out individual factors of the product

#

yes

#

they are the coordinate projections if you will

steady belfry
#

lolol

#

is this the same thing

prisma ibex
#

not quite

#

I guess in the example you're thinking of with groups

#

the projection maps from products are a special case of this

#

but products make sense in many more contexts and projection maps out of projects always make sense whenever products make sense

steady belfry
#

i see

#

ty all for your help

chilly ocean
#

anyone?

rustic crown
#

e1 --> 0, e2 --> e1

#

oooh someone made me blue

rustic crown
# rustic crown e1 --> 0, e2 --> e1

if you want the reasoning, by rank nullity both rank and nullity are 1, so pick a basis for kernel say {u} and so u will die. extend this to a basis {u, v}. Now we want image to be <u>, easiest thing to do is map v to u.

chilly ocean
#

im still processing all this 😐

rustic crown
#

oh F i see a [2] at the end... don't say this is from a test >.<

chilly ocean
#

no test is tomorrow

#

this is a previous year question paper

rustic crown
#

(okie i'll trust you on that)

#

another way to think this would be to see what is T^2

#

since image of T = kernel of T, we can say T^2=0

chilly ocean
#

i'll tell you what i was thinking

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i read somewhere that any linear transformation from R2 to R2 can be written as (ax+by, cx+dy)

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so i was trying to use that information

rustic crown
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yea this would do, only thing is that you get 4 variables so gets kinda tedious

rustic crown
chilly ocean
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so first i calculated the null space

chilly ocean
rustic crown
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what i said earlier meant (1,0) --> (0,0) and (0,1) --> (1,0)

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so basically i gave the matrix for the transform

[0 1]
[0 0]
chilly ocean
rustic crown
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if you have studied about eigenvalues, you would know that T^2 = 0 reduces the calculations by a lot!

chilly ocean
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btw thanks!

rustic crown
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that equation just says that both eigenvalues would be 0

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(and if you want an overkill for the problem, use the jordan normal form)

chilly ocean
chilly ocean
rustic crown
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seems like you would know the words "rank", "nullity" and "basis"

chilly ocean
rustic crown
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rank + nullity = dim R2 = 2, but the condition says that rank = nullity

chilly ocean
rustic crown
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so the kernel is 1 dimensional, and hence i can pick a basis for it, say {u}

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picking anything in R2 - <u> will be linearly independent to u, so lets pick v

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{u,v} then is a basis for R2

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and its enough to tell you the image of basis elements to define the linear transformation!

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since u was in the kernel, it must go to 0.

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but we have the condition that im T = ker T

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so v must land inside the kernel, that is a multiple of u

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T(u) = 0 and T(v) = c*u

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kernel of this map is indeed <u> and if c is not 0, then image is also <u>

chilly ocean
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i kind of get it now

rigid cave
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Why does the conjugacy class of 5 cycles in A5 split and why doesn't this happen to for example the 3 cycles?

carmine fossil
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Because you can't find a even permutation x such that x(1,2,3,4,5)x^-1=(1,2,3,5,4)

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That would imply
(x(1),x(2),x(3),x(4),x(5))=(1,2,3,5,4)

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And whatever you choose for x(1),you end up with an odd perm

rigid cave
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Yeah that is a great explanation!

carmine fossil
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For 3 cycles,you can choose a x in such cases such that x is even

rigid cave
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Okay, but is there any "criterium" for this?

carmine fossil
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idts

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Are you sure 3 cycles don't split?

rigid cave
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I think so, in a5 at least

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in A4 they split

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Okay so let g be a odd permutation and let's look at g(12)(34)g^(-1). Let g'=g(12), which is even. Then g'(12)(34)g'^(-1) = g(12)(34)g^(-1) which shows that the double transpositions do not split. Could you somehow "generalise" this? So for example, let's look at tau (in A5) and it's conjugacy class. If tau consists of disjoint cycles such that they commute with each other, then the conjugacy class don't split. Is this right?

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But then the contraposition of this would be "conjugacy class splits => the above does not hold" which is not that interesting since we are looking for criteriums for which it splits and not it's consequences.

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Wait, do the composition of transpositions commute if and only if those transpositions are disjoint?

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So to show that the 3 cycles, (123), don't split in A5 we could use a similar argument. Let g be odd and let g'=g(45). Then g'(123)g'^(-1) is g(123)g^(-1) so we could always "transform" the odd g to an even permutation. This "trick" and the above do not work for (123) in A4 which shows that they split there (if there are no more "trick" to be used). It seems that the criterium here is that if tau commutes with an odd permutation then the conjugacy class of tau don't split. However, I am not sure.

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And in fact, this works! You could prove this using the exact same method as above