#groups-rings-fields

406252 messages · Page 613 of 407

hidden haven
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So they won't generate isomorphic subgroups

signal remnant
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right

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like, when i first saw this exercise, i thought "this looks easy enough" but now here i am sooo confused.

hidden haven
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What I'm hinting towards is that if H_1 is a subgroup of G_1, H_2 of G_2, then H_1 ⊕ H_2 is a subgroup of G_1 ⊕ G_2

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So what you should try to see is that Zr is (isomorphic to) a subgroup of Zm, Zs (to) a subgroup of Zn

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And so Zr ⊕ Zs is (isomorphic to) a subgroup of Zm ⊕ Zn

signal remnant
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So in exercise 16 I showed that G_1 ≅ G_2 and H_1 ≅ H_2 implies that G_1 ⊕ H_1 ≅ G_2 ⊕ H_2.

hidden haven
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Nice that is being used here

hidden haven
signal remnant
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Yeah I can feel that

hidden haven
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Since here you don't actually have subgroups, but only isomorphic to subgroups

signal remnant
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sure

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hmmmm

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so say

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so say G is isomorphic to some subgroup H of some other group K. that is let G≅H for some H<K where K is a group. okay.

vestal snow
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Can someone give an example of a genus 1 (smooth, projective etc.) curve over F_5 or F_7?

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And a genus 2 one too

signal remnant
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how can you have a curve over a finite field? o.o

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i'm so close to throwing in the towel on this one. 😢

hidden haven
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Ok so {0,2} is a subgroup of Z4 isomorphic to Z2

signal remnant
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yes

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totally

hidden haven
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{0,2,4} in Z6 is isomorphic to Z3

hidden haven
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And you can try and see why I chose those as subgroups specifically

signal remnant
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i had an intuition that 4 and 6 were the interesting case as i was going to use those in a question before you proposed them. no i have a "feeling" for it sure

signal remnant
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{0,2,4} < Z6

hidden haven
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Yeah I chose them because they are small numbers which aren't prime, so have non trivial factors

signal remnant
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okay so {0,2,4} < Z6 and also {0,2,4} ≅ Z3 because 3 elements and elements in {0mod6,2mod6,4mod6} are all either order 3 or the identity

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is that what you're saying?

hidden haven
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You need more than that

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You can give an isomorphism here

signal remnant
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oh h

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oh uh

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ok one sec

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to show {0,2,4} ≅ Z3:
Define ϕ from {0,2,4} to Z3 such that for a in {0,2,4}, we have ϕ(a) in Z3 where ϕ(amod6) = (a%2)mod3 (aka a modulo 2 or even "the remainder of the quotient of a with 2"). this gives ϕ(0mod6) = 0mod3, ϕ(2mod6) = 1mod3, and ϕ(4mod6) = 2mod3.

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then show that the operation is bijective and operation-perserving

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would be next

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right?

hot lake
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4%2 = 2 ?

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just define phi by decreeing what the images of 0,2,4 are, there is no need to find a fancy formula

signal remnant
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oh lol whoops sorry

hot lake
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then you check that it is bijective and compatible with the group operations

signal remnant
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sorry brain went fart. i just mean a/2

hot lake
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oh lol

signal remnant
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ok so then I can go:
Assume ϕ(a) = ϕ(b) for a,b in {0,2,4}. ϕ(a) = ϕ(b) ⇒ ϕ(amod6) = ϕ(bmod6) ⇒ (a/2)mod3 = (b/2)mod3 ⇒ a = b. that gives injectivity.

south patrol
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You can also just read off an explicit inverse for example (ngl I would be tempted to say its just obviously bijective lol)

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but yeah, sure

signal remnant
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Surjectivity:
Let g in Z3. Then 2g is in {0,2,4} and ϕ(2g) = (2g/2)mod3. perfect.

hidden haven
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Right

vestal snow
novel parrot
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if m1/m1m2 is a module over A/m2 why is it a module over A

hot lake
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m1/m1m2 is an A/m2-module

signal remnant
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and finally a,b in {0,2,4} implies ab in {0,2,4} implies ϕ(ab) in Z3 and so ϕ(ab) = (ab/2)mod3 = (1/2)(amod3)(bmod3) = (1/2)ϕ(a)ϕ(b) WHAT

novel parrot
hot lake
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it's also an A-module I guess

novel parrot
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i imagine it being so too

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but i wanna be completey sure about it

hidden haven
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A/m modules are also A modules because A/m is an A-algebra

hot lake
signal remnant
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element-wise composition

hot lake
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just define a.(m mod M) = am mod M

signal remnant
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my coffee keeps getting cold because i take so long to type this 😢 lol

hidden haven
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So element wise doesn't make sense

novel parrot
hidden haven
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And you are talking about integers mod n, not functions so composition doesn't make sense

signal remnant
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so my ϕ is no good right?

hot lake
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you're using concrete groups so you want a concrete operation, not an abstract "it's just group composition lol"

hidden haven
signal remnant
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ϕ(amodb) = (a/2)mod(b/2)

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?

hidden haven
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There's an obvious A-module structure on M by composing these 2

hidden haven
signal remnant
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to get from Z3 to Z6

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mod3 to mod6

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oh

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err

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...

hidden haven
signal remnant
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lmao wait

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{0,2,4} is still mod6

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to get from mod6 to mod3

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and

hot lake
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so you have defined phi : {0;2;4} -> {0;1;2} by phi(x)=x/2 but then you are stuck you don't actually know what the operations are on Z6 and Z3 ?

signal remnant
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from {0,2,4} to {0,1,2}

novel parrot
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cowboyflonshed an A submodule is a A/x submodule just be restricting the scalars yeah?

hidden haven
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A submodule of A/x module?

signal remnant
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okay yeah so ϕ(amodb) = (a/2)mod(b/2) is fine and all except i am finding that ϕ in this case is not operation-preserving?

novel parrot
hot lake
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you don't actually know what the operations are on Z6 and Z3

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you check what the operations are

hidden haven
signal remnant
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o.o

hidden haven
signal remnant
hot lake
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by looking back at when you were told what groups Z3 and Z6 are

novel parrot
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um back to my question, m1/m1m2 is a A/m2-Module, how do we know its also an A-module

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i dont really get what you said

hidden haven
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It's an A-module originally lol

hot lake
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you're doing things the wrong way what I would do is first show it's an A-module with the obvious definition then show it's an A/m2-module by checking m2 annihilates stuff

hidden haven
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It's a quotient of an A module by an A-submodule

novel parrot
hidden haven
novel parrot
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if m1/m1m2 was noetherian as A/m_2 module why also as A module?

hidden haven
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Being an A/I module means being an A module, but not the other way around (unless I is contained in the annihilator, first isomorphism theorem)

hidden haven
hot lake
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because when you are an A/m2-module, sub-A/m2-modules are the same as sub-A-modules

hidden haven
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Because m is contained in the annihilator of any submodule

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Sub-A-module

hot lake
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A-submodule ?

hidden haven
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All respect has been lost

hot lake
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R-sub-vector-space ?

hidden haven
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R-subspace

south patrol
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R-vector-space-sub

novel parrot
hidden haven
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R-vector-space-dom

hidden haven
novel parrot
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by applying the first iso theorem

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we can see that A/I is isomorphic to a submodule of A, (A- module)

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is that wt u meant

hidden haven
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No

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An A-module is exactly a homomorphism from A to End(M)

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Wait sorry gtg lol

novel parrot
signal remnant
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ty @hidden haven for your help

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i'm still a little stuck but i think i'll get there now

hidden haven
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I'm back 😌

hidden haven
novel parrot
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no

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but i think maybe you mistunderstood my question

hidden haven
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Yeah I was just saying how first isomorphism theorem is relevant

novel parrot
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like if we take any sub module of A-module

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but only let A/x act on this submodule

signal remnant
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Say I have a group G isomorphic to Z_r ⊕ Z_s. So g in G look like ( g_1 modr, g_2 mods), right?

novel parrot
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that submodule can now be a A/x module?

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no?

hidden haven
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See A/x is not a subring of A

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So it doesn't make sense to restrict scalars

hidden haven
novel parrot
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but A contains A/x no?

hidden haven
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How?

novel parrot
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isomorphically

hidden haven
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Z doesn't contain a copy of Z/2Z

signal remnant
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so (g_1 modr, g_2 mods) must be in Z_m ⊕ Z_n whenever r|m and s|n, right?

novel parrot
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of A/x

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but not the structure

hidden haven
novel parrot
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ok

hidden haven
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But that embedding will actually not give you a subgroup

signal remnant
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...uhhhhhhh

hidden haven
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Do you see why?

novel parrot
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so how do A/m1 module become A module?

signal remnant
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gimme a sec

hidden haven
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End(M), for an abelian group M, has a ring structure with pointwise addition and composition. With this, an A-module M is exactly an abelian group M, with a ring homomorphism A → End(M) (which defines the scalar multiplication)

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Try proving this equivalence

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It's extremely useful because now you can use stuff you know about ring homomorphisms to study module stuff

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It's similar to how an A-algebra B is exactly a ring B with a homomorphism A → B

novel parrot
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you mean the End(M) is a ring

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?

hidden haven
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Yes

novel parrot
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i dnt understand why this is needed

hidden haven
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Because once you have this, you have also a homomorphism from A to A/m. Then a homomorphism from A/m to End(M) allows composition to get a homomorphism A → A/m → End(M), which is an A module

signal remnant
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(g_1 modr,g_2 mods) is in Z_m ⊕ Z_n whyyyyyyyy

novel parrot
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definition of Zm and Zn

signal remnant
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it feels like (g_1 modr,g_2 mods) and (g_1 modm,g_2 modn) have nothing to do with each other.

novel parrot
novel parrot
signal remnant
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r and s are defined by the fact that r|m and s|n.

novel parrot
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so Zm contains Zr

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whats your question?

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u saying y they r equal?

signal remnant
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The exercise in question is "If r is a divisor of m and s is a divisor of n, find a subgroup of Z_m ⊕ Z_n that is isomorphic to Z_r ⊕ Z_s."

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but wait

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If r | m and s | n, then is it true that Z_r ⊕ Z_s < Z_m ⊕ Z_n?

novel parrot
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yah

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even if r <= m and s <= n

hidden haven
novel parrot
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start with a few small m,n and r,s and look to generalize it

signal remnant
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so all groups constructed in this way are automorphic?

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no we tried that 😦

hidden haven
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But you can embed Zr into Zm when r | m

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Recall the example of Z2 in Z4

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What was the embedding?

signal remnant
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ok so r | m implies Zr < Zm yes?

hidden haven
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And the one of Z3 in Z6

hidden haven
novel parrot
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just take the elements that have order r and s

signal remnant
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r | m implies Zr < Zm (and similarly for s,n)... so if Zr is automorphic, which it is, we're done? that seems like a copout.

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I'm sorry my thick head isn't getting this folks. I really appreciate all your help though.

hot lake
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where did you find that "automorphic" word

signal remnant
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automorphic means isomorphic to itself idk

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uh

hidden haven
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Zr isn't exactly a subgroup of Zm, it's isomorphic to one, and you have to identify which one

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Then every group is automorphic via identity lol

signal remnant
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right

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...am i being silly

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so Zr is NOT a subgroup of Zm?

signal remnant
novel parrot
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it is

hidden haven
novel parrot
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but u want a subgroup of Zm that is isomorphic to Zr

hidden haven
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We just say that it is

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But strictly speaking it isn't

signal remnant
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i. am. very. confused.

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with that

hot lake
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I still don't know what is Zn and what is the group operation on Zn

hidden haven
signal remnant
hidden haven
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Like before working with the direct sum

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Try to just list all subgroups of Zn

hot lake
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that's kinda difficult for an abstract n

hidden haven
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Yeah but do it for small n

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You'll see a pattern

signal remnant
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o_o um ok let's see

hidden haven
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And then you can say what exactly the pattern is

hot lake
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but for real, jadeejade I remember you being confused at what were the group operations on Z3 and Z6, and unless I missed something during lunch, I haven't seen you dig out the definitions and write what the operations are

signal remnant
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Z1 < Z1, okay. Z1 < Z2 and Z2 < Z2, okay. Z1 < Z3 and Z3 < Z3, ...right?

hidden haven
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Yes catThin4K

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But how exactly

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Like what is Z1 inside Z3 as a subset

signal remnant
hidden haven
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Yes

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Zef asked you before and you said pointwise composition

signal remnant
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oh i thought they were asking about the ⊕

novel parrot
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how do you do the direct sum notation symbol btw?

hot lake
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we weren't doing the direct sum of groups problem back then I think

signal remnant
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it's a unicode character idk

hot lake
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do you have a definition of Zn anywhere

signal remnant
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this is the only exercise i've been talking about this whole time as far as i'm aware

hot lake
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where did you even find those exercises

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ah

signal remnant
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this is exercise 19 in chapter 8 of Gallian's textbook Contemporary Abstract Algebra

novel parrot
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Z3 is probably just regular multiplcation mod 3

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so like 2 times 2 = 4 = 1

signal remnant
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oh in the book the definition of Zn is

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The set Zn = {0, 1, . . . , n - 1} for n >= 1 is a cyclic group under addition modulo n.

hot lake
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did you read this part

signal remnant
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a million times yes

hot lake
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there it is

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Zn

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addition mod n

signal remnant
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right

hot lake
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that's the group operation

signal remnant
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right

novel parrot
hot lake
signal remnant
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... the wrong operation for ab?

hot lake
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well either you read it as multiplication and gave up, either you didn't remember that composition was addition mod n and gave up

signal remnant
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uhh one sec

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wait so are you saying that my logic wrt the operation-preserving property of ϕ was marred by mixing up addition with multiplication?

hot lake
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well you never looked at the operation that was supposed to be preserved

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what is 4 composed with 4 in Z6 ?

signal remnant
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(4)(4) = 2mod6

hot lake
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and we prefer to write that as 4+4 = 2 mod 6

signal remnant
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sure

hot lake
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and is phi(4+4 mod 6) = phi(4) + phi(4) mod 3 ?

signal remnant
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yes

hot lake
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then it's good

hidden haven
hot lake
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you can check 0+0, 0+2, 0+4, 2+2, 2+4 if you want

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you will find it does preserve the group operations

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so your phi is an isomorphism from ({0;2;4} , addition mod 6) to ({0;1;2}, addition mod 3) = Z3

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lemme go have a look on the subgroups definitions

hidden haven
signal remnant
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and ϕ(2+4 mod 6) = ϕ(0 mod 6) = 0 mod 3 with ϕ(2 mod 6) + ϕ(4 mod 6) = 2 mod 3 + 1 mod 3 = 0 mod 3? cool!

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OKAY AWESOME so my ϕ does work in general?

hot lake
hidden haven
hot lake
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so here, {0;2;4} is a subgroup of Z6

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it is a subset of Z6 satisfying blablabla

novel parrot
hot lake
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and it is isomorphic to Z3

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but !

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when we say "Z3 is a subgroup of Z6"

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there is potential for confusion

signal remnant
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well yeah i'm confused right now lmao

hot lake
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after all, Z3 = {0;1;2} with some other operation

signal remnant
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not as much as when I got here though!

hot lake
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and {0;1;2} can be understood as a subset of Z6

signal remnant
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sure

hot lake
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but we don't want to say that {0;1;2} is a subgroup of Z6 because that's false

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so when we say Z3 is a subgroup of Z6

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we always mean

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Z3 is isomorphic to a subgroup of Z6

signal remnant
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...waaaiiit say that one more time.

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so what you're saying is

hot lake
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your phi is an isomorphism from Z3 to {0;2;4} with addition mod 6

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{0;2;4} is a subset of Z6 that makes it a subgroup of Z6

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so Z3 is isomorphic to a subgroup of Z6

signal remnant
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so what you're saying is

hot lake
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and that has less potential for confusion than "Z3 is a subgroup of Z6" that can be interpreted as "{0;1;2} is a subgroup of Z6"

signal remnant
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Z3 < Z6 can only be resolved through this ϕ

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that is otherwise it would necessarily just be a subset

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like the elements of Zm can't be understood modulo n without invoking this structure

hot lake
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yeah

signal remnant
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but do any of the elements of Zm ever "change" if you change the modulo/modulation?

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or map anywhere i guess you'd say

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sorry if i'm being vague

hot lake
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well in this book the elements of Zn are defined as the numbers 0 to n-1

signal remnant
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sure

hot lake
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other books would define Zn in another way

signal remnant
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hmm

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what i mean is

hot lake
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the element would be equivalence classes of numbers

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and then you would have Zn is actually disjoint from Zm if you look at them as sets

signal remnant
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how much does changing the modulo/modulation change the structure of the residues?

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or how can i quantity/qualify that

hot lake
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I'm not sure what that's asking

signal remnant
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oh by residue i mean equivalence class

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sorry

hot lake
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so then the elements "0" "1" and "2" in Z3

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are not the same as the "0" "1" and "2" in Z6

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so there is less potential confusion when saying Z3 is a subgroup of Z6

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it automatically means Z3 is isomorphic to a subgroup of Z6

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because Z3 wasn't a subset of Z6 to begin with

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technically

signal remnant
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and this ϕ is unique in constructing that bridge, and ⊕ inherits that structure?

hot lake
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in what chapter is the ⊕ exercise

signal remnant
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chapter 8. the whole chapter is just on ⊕ lol

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exercise 19

hot lake
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well google won't let me read that far

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so when we say A is a subgroup of B, most of the time we are talking about some isomorphism

signal remnant
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the whole pdf for edition 8 is online somewhere, which is the edition i'm using

hot lake
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and not necessarily some actual inclusion

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also sometimes there can be several isomorphisms possible

signal remnant
hot lake
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so it's always nicer to know which one we are talking about

signal remnant
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that's interesting

hot lake
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you could have had psi(0) = 0 psi(2) = 2 and psi(4) = 1

signal remnant
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oh so ϕ is not unique?

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in general?

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it would just be messier to write it out?

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sometimes*

hot lake
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so in fact, when someone says "remember that Zn is a subgroup of Zkn, blablabla", you should always write "remember how we have phi : Zn -> Zkn defined by phi(x mod n) = kx mod kn, is an injective group morphism, blablabla"

signal remnant
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yes

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i am convinced that this ϕ is an isomorphism yes

hot lake
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so there can be lots of stuff said implicitly

signal remnant
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sure. ok so i have my group that is isomorphic to some external direct product of residues of Z modulo whatever.

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i know that so long as this isn't the trivial case that this group exists and is not trivial

hot lake
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I got a pdf for 7th edition

signal remnant
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probably good enough.

hot lake
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looks like a nice book with tons of exercises

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ah you said exercise 19

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17

signal remnant
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a maths professor of mine who specialized in Galois groups and something else that he connected them to told me to do every exercise in this book years ago so i am and yeah lol

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yeah, 17 in yours

hot lake
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is there an exercise saying that if r divides m then find a subgroup of Zm isomorphic to Zr ?

signal remnant
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no

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or

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...no, I don't think so

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lemme check

chilly ocean
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gallian is so bad imo

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maybe just for exercises is okay tho

signal remnant
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the book is pretty much a book of exercises but yeah the reading material is rather sparse

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ch.4 ex.47 "Let G be a cyclic group of order n and let H be the subgroup of order d. Show that H = {x in G | |x| divides d}."

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i've already done Ch.4 Ex.47. I'm quoting it as a result we can use.

novel parrot
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just read dummit and footes algebra book

hidden haven
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Have you seen the problems that narwhal posts here

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From d&f

novel parrot
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lol

hidden haven
hot lake
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as in classify all simple groups of order <= 10000 ?

novel parrot
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lmfao

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you can always skip a few problems

hidden haven
novel parrot
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idk gallian has too many boring exercises

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and d&f has more contnet

hot lake
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and yeah lol it doesn't say ANYTHING except the definition of direct sum and the chinese remainder theorem though it doesn't say it's chinese remainder theorem

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the book is only for the exercises

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it doesn't even say "if H is a subgroup of G then H + K is a subgroup of G + K"

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which is occasionally useful

signal remnant
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We also have Ch.6 Ex.53: "Let a belong to a group G and let |a| be finite. Let ϕa be the automorphism of G given by ϕa(x) = axa^-1. Show that |ϕa| divides |a|. Exhibit an element a from a group for which 1 < ϕa| < |a|."

hot lake
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I'm not sure that's going to be relevant

signal remnant
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well what i mean is the converse of lagrange's theorem is not true

hot lake
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so yeah if I were you I would prove "if H is a subgroup of G then H + K is a subgroup of G + K"

signal remnant
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at least, before this point in the book

hot lake
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forgive me for not having a fancy (+) key on my keyboard

signal remnant
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is fine

hidden haven
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Imagine not being able to type ⊕

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"Honorables" in the math server

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😌

signal remnant
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well i mean I have what I think is a very useful result that G ≅ G ⊕ {e_H}...

novel parrot
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what does topological abelian group mean?

signal remnant
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for both directions

hot lake
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it does one example

signal remnant
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that's a later example

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it even quotes the exercise in question

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oh wait is that in ch. 8?

hot lake
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it's in the middle of chapter 8

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yeah

hidden haven
hot lake
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and it also assumes that " if H is a subgroup of G then H + K is a subgroup of G + K" has been already magically absorbed by the reader

signal remnant
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for each divisor, eh?1

novel parrot
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or like

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hm

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whats inversion

hidden haven
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x ↦ x⁻¹

hot lake
hidden haven
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G → G

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And multiplication is G x G → G

novel parrot
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but

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whats continous

hidden haven
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G x G with the product topology

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A function is continuous if the inverse image of every open set is open

novel parrot
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so G satisfies the topology axioms

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and the group axioms

hidden haven
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Yes

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And the continuity of the group operations with respect to that topology is a compatibility condition

signal remnant
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Z_m ⊕ Z_n having a subgroup isomorphic to Z_r ⊕ Z_s if r|m and s|n is like saying Z_m ⊕ Z_n has a subgroup isomorphic to Z_r ⊕ Z_s if <m/r> ⊕ <n/s> < Z_m ⊕ Z_n???

novel parrot
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im not really familiar with topology catThin4K

hidden haven
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Compatibility between the 2 structures you've put on G

hot lake
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wtf your book does morphisms after isomorphisms ??

hidden haven
oblique river
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I think that's actually somewhat standard

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isomorphism is more of a fundamental concept imo

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like, two things being equivalent

signal remnant
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am i right?? o.o

hot lake
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hmm maybe

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I'm not sure what " <m/r> ⊕ <n/s> < Z_m ⊕ Z_n" means and if it's equivalent with "r divides m and s divides n"

next obsidian
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That seems so backwards dude…

oblique river
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how is "two things are the same" not fundamental

next obsidian
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Like you have to explain what it means to be a group hom to get an isomorphism yeah? It’s not just an invertible map

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So why not introduce the maps which respect the structure and then say if it’s invertible it’s an iso

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Like idk, even showing the inverse of a hom is always a hom is nontrivial and you need to know what a hom is to show that

oblique river
#

yes, it's part of the definition of an isomorphism

#

an isomorphism is a function which satisfies .....

#

you just dont label the first part as a "morphism" until later

hidden haven
#

An arrow with 1 arrow head is simpler than an arrow with 2 arrow heads smugsmug

next obsidian
novel parrot
#

so what are we taking the closed sets as?

oblique river
#

let me guess, you also think that abstract algebra classes should define magmas, loops, and semigroups before groups

next obsidian
#

God no

oblique river
#

because those structures have a subset of axioms of a group

next obsidian
#

Those are trash

signal remnant
#

If r|m and s|n, then <m/r> ⊕ <n/s> < Z_m ⊕ Z_n and <m/r> ⊕ <n/s> ≅ Z_r ⊕ Z_s so we are done.

hot lake
#

is it cyclic ?

next obsidian
#

I just think talking about what maps exist first makes more sense

hidden haven
hot lake
#

wait it says order d already

next obsidian
signal remnant
#

isomorphisms between groups are operation-preserving bijective maps

novel parrot
next obsidian
#

Like, it isn’t “we take these to be the closed sets”

#

It’s the choice of group structure with a choice of topology

#

And they have to play nice

hidden haven
#

A better definition is operation preserving invertible maps catThin4K

novel parrot
#

ok

next obsidian
#

Example, give any group the discrete topology it’s now a topological group

#

But that’s not very interesting

novel parrot
#

idk what discrete topology is lmao

next obsidian
#

Every set is open

hidden haven
#

Everything is open

next obsidian
#

It makes every map from it continuous

hidden haven
#

And therefore everything is closed roopopcorn

#

You can also take the indiscrete topology

#

Then everything that can be not open is not open

hot lake
#

the noisy topology

signal remnant
next obsidian
#

Anyway let’s hop to topology to talk about topological groups more

hot lake
#

that was super hard to read but yeah

hidden haven
signal remnant
hidden haven
#

Looks good catthumbsup

signal remnant
#

HOORAH

#

i feel like Erdos

#

omg

#

yaaay

hot lake
#

I kept thinking there was a > missing

#

my brain trying to match brackets

#

and I was like "no wait this is the subgroup generated by stuff thing and this one just says subgroup"

#

"it actually makes sense"

signal remnant
#

Good enough for me.

#

Thanks everyone. 😄

#

how did y'all get to be so good at this stuff?

#

😛

hot lake
#

lots of tears

hidden haven
#

Speak for yourself 😎

next obsidian
#

Practice

signal remnant
#

I really appreciate this help. I don't really have anyone in my daily life that has much clue about this advanced stuff. Even those that are supposedly good at math ha

hidden haven
#

Remote learning or self study? catThin4K

signal remnant
#

This is self-study.

#

I have a mathematics minor from university.

#

but i consider mathematics to be essential for personal development

#

engineer by trade

hot lake
#

that book tells you so little in the chapters

signal remnant
#

haha right?

#

still

#

a very smart person challenged me long ago

novel parrot
#

use dummit and foote instead 🙂

hot lake
#

and then "here, have 80 exercises"

signal remnant
#

noo i'm already too far into this one

#

can't stop now

#

lmao

hidden haven
novel parrot
#

your only about 50 pages in dummit and foote

signal remnant
#

I enjoy the problem-solving process. That is the most important part for me. The subject matter is a bonus hah.

#

I promise I'll take a look at it.

next obsidian
#

Wut

novel parrot
#

you will enjoy dummit and foote alot more and learn more!!

signal remnant
#

I'm excited that I'm almost through the Part on Groups though

hot lake
#

I messed up

signal remnant
#

gah paywalls

#

ok maybe later

novel parrot
#

dummit and foote is the best algebra book

#

even with all the extra words

#

still the best

hidden haven
#

Are the tensors in physics just elements of a tensor product? catFone I've seen talk of rank n tensors though so I'm not sure

novel parrot
#

arnt tensors in physics just n dimensional vectors

next obsidian
#

Don’t think so

#

Lol

novel parrot
#

like how in python u can put a list inside of a list n times lol

#

idk but in programming they call those tensors i think

next obsidian
south patrol
next obsidian
#

That’s actually like…

#

Really close to the idea of the tensor power

hidden haven
#

Yeah

hot lake
#

they are sections of tensor products of the tangent and cotangent bundles ?

hidden haven
#

I see

south patrol
#

This professor was a string theorist so his stuff is a bit more normal lol

#

Well mathematical

hidden haven
#

String theorists are normal?

next obsidian
#

Yeah haven’t u ever seen nLab

#

Urs is very normal

south patrol
#

Well, compared to the maths taught by less mathematical physicists and looking at it from a maths perspective lol

next obsidian
#

maths

#

BRITISH

south patrol
#

Yes

next obsidian
#

BRITISH

south patrol
#

I am a brit oof

next obsidian
#

In constant awe the English manage to get their own language wrong

novel parrot
south patrol
#

I agree math makes more sense lol

next obsidian
#

zed

hidden haven
south patrol
#

😔

next obsidian
#

Yeah they’re based on the BRITISH ENGLISH

south patrol
next obsidian
#

Which is WRONG

#

Jesus potato

#

Do they actually say that

hidden haven
#

then why did you say based?????

south patrol
#

Lmao

#

No

#

That was a meme

next obsidian
#

Ok lol

south patrol
#

sometimes I call it that ironically though

next obsidian
#

Slim

#

Aren’t you a stupid American

#

Where were they from?

south patrol
#

Oof

next obsidian
#

O K

#

If it was like

#

Some dumb European country

#

Then smh

#

No place to speak

#

European country tierlist:

#

America > everything else

south patrol
#

ok

next obsidian
#

🇺🇸

hidden haven
#

Haha America isn't in Europe stupid American

next obsidian
#

That’s because Europe doesn’t matter enough

hidden haven
next obsidian
#

I think Australia gets a pass to be cool

#

They’re kind of like the Australian America

south patrol
#

Australia is the Australian America

#

You heard it here folks

next obsidian
#

Hahahaha they said heard it hear

#

British moment

#

😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂🤣🤣

south patrol
#

someone else did that last time smh

hidden haven
#

Can't even speak the language they claimed to have invented 😎

next obsidian
#

Curious!

#

Okay okay okay so hear me out ok

#

If the English speak the right thing English

#

Why does every other country learn English like an American

#

OWNED

#

If this isn’t true then stfu

south patrol
#

Not even true

next obsidian
#

Those countries learning British English aren’t even learning English

#

Wtf

#

I just got a notification from Adobe acrobat on my phone

south patrol
#

stupid american

next obsidian
#

:(

paper flint
#

Half of the world learnt American English because MS Word defaults to American English.

hidden haven
#

British English is called that because it's the British version, but the American version is called English because it's the original

signal remnant
#

tensors are a generalization of scalars/vectors/matrices. they are basically just rank n transformations on elements of nm-space kinda

next obsidian
south patrol
next obsidian
south patrol
#

Oh so you admit saying america > Britain is a shitpost? owned

hidden haven
next obsidian
hidden haven
#

Anything about Britain is a shitpost

next obsidian
south patrol
#

Gotem

next obsidian
#

Some claim that Rudin was just an elaborate shitpost

south patrol
#

it was

#

The Weil conjectures were shitposts that happened to be true

#

and were predicted by Nostradamus

signal remnant
#

.>

south patrol
#

Huh

signal remnant
#

if anyone has a problem with this feel free to meet me outside of the next AMS conference on the west coast lmao

hidden haven
#

This feels nothing like the math tensor lol

south patrol
#

so the space of all tensors is basically just R^n up to isomorphism

#

Nice

signal remnant
#

this is inspired by what i learned from Dr. Rabin, a high-dimensional manifolds type dude

south patrol
#

Also on the topic of tensors, when would mathematicians talk about tensors having to satisfy certain transformation rules - is that in a diff geo context?

signal remnant
#

gosh more than just diff geo i'd imagine

hidden haven
#

In diff geo you have exterior products which are multilinear maps but alternating catThin4K

south patrol
#

yeah

#

I just find it weird physicists imposing certain rules, even just like how vectors transform, sometimes

hidden haven
#

But idk outside of that I've only seen tensors as the tensor products, not individual tensors as maps

south patrol
#

without further clarification / context

#

glad to be doing maths now lol

#

*math

potent briar
#

all 1-dimensional vector spaces have only 2 subspaces right?

#

the trivial ones

hidden haven
#

Yes

potent briar
#

thank you

south patrol
#

If it has a non-zero vector, that'll span the entire space

potent briar
#

yeah

#

i am having trouble understanding the dual space of a vector space

#

i know the definition

#

but am i supposed to be able to visualize all the linear functionals somehow?

novel parrot
#

depends on the vector space

signal remnant
#

say we have an element x of a vector space X of rank k. then we can associate any tensor T with dimensions n₀ × ... × nᵢ × k to the left of x to get an element y with dimensions n₀ × ... × nᵢ × 1.

novel parrot
#

latex looks better than unicode KEK

signal remnant
#

Tx = y

hidden haven
#

Because they are linear transformations from V ~ F^n to F once you fix a basis of V

#

The infinite dimensional case is also similar. You're just choosing an element in F for each basis element

potent briar
#

ok ok but wait i haven't gotten to transformations on my book

hidden haven
#

Wait monkaS

#

How do you have functionals before transformations

signal remnant
#

whoa

potent briar
#

ask Halmos

hidden haven
#

Do you know what a basis is?

potent briar
#

yes

signal remnant
#

ok i gotta get back to real work now. anyway thanks everyone.

potent briar
#

a linear funcitonal is a homogenous function from the space to the field

#

that's it isnt it

hidden haven
#

Ok so a functional on V is the same as just a function from a basis B of V to F

#

Because if you give me a function from B to F, it can be "linearly extended", uniquely, to get a functional on V

#

Not sure what homogeneous means here

hidden haven
#

These 2 are inverse bijections

#

And so it's sometimes easier to just think of a functional as a function from a basis to F

#

But I think it will be clearer once you read about linear maps

#

Because they're just special cases of linear maps

#

And knowing about linear maps will give you vocabulary to talk about this easily

potent briar
#

ok thank you

#

homogenous means T(av + bw) = aT(v) + bT(w)

#

a, b scalars, v, w vectors

hidden haven
#

Right that's what linear is too

#

Except linear maps go from vector space to vector space instead of vector space to field

#

The right side still makes sense in a vector space

potent briar
#

we did define already an isomorphism

#

which is a map from V to W that is linear

#

linear maps go from V to V?

hidden haven
#

Isomorphism is not just a linear map from V to W

#

It's an invertible linear map

#

Or a bijective linear map

potent briar
#

yeah ofc

unreal portal
#

For part (a), isn't every finite field isomorphic to some F_p?

unknown.png
celest mantle
#

maybe it suggests p prime so you could look at F_p^2 or something like that

hidden haven
#

Finite fields are 𝔽_{pⁿ}, not just 𝔽ₚ

barren sierra
#

what is the different between the bracketed red part and part c

unknown.png
#

isn't this subgroup mentioned in C literally Cayley's theorem?

novel parrot
#

yes

#

u prove it with part a and b

#

and first isomorphism theorem

#

how does A_o become a subring of A

#

A_0A_i < A_i for everyy i so it can "handle" multiplication from every A_i

#

but then what

barren sierra
gritty sparrow
novel parrot
#

ook

gritty sparrow
#

So you can multiply elements and it stays in A_0

novel parrot
#

ok

#

i see

#

ok

novel parrot
#

so in ur case, kernel is 0

#

so G is isomorphic to the image on S_G

#

and images are subgroups

barren sierra
#

what is G / Ker(f)

novel parrot
#

in your question

#

kernel is just 0

#

so G/kernel = G

barren sierra
#

yea identity maps to identity and nothing else maps to identity

#

I see

novel parrot
#

i skipped the section on topology

#

hopefully i wont be needing it

hidden haven
#

In fact the stuff in red brackets is more general because (c) restricts it to finite groups for some reason

novel parrot
#

is there any other example of graded rings except polynomial rings?

#

is there an obvious one that i dont know

#

thats just summing everything except A_0 ya?

fossil shuttle
#

you can associate it to a topological space

#

maybe an easier example is the exterior algebra generated by a vector space.

novel parrot
hidden haven
#

Or tensor algebra too I suppose? hmmCat

fossil shuttle
#

yeah.

novel parrot
#

i dont know anythign about any of those

fossil shuttle
#

so the tensor algebra is like the polynomial ring except variables don't commute.

hidden haven
#

Tensor algebra should be easy since you know what a tensor product is

novel parrot
#

but what are the subgroups

fossil shuttle
#

so you have like, "polynomials" of the form xyzyzyx

upper pivot
#

In number theory the space of all modular forms is a graded ring.

fossil shuttle
#

they'd still be the homogeneous "polynomials" of degree n, like in the case of polynomial rings

#

the exterior algebra over k in variables x1... xn is also like the polynomial ring except that instead of multiplication being commutative, it's anticommutative in the sense that x_i * x_j = -x_j * x_i

#

this could be slightly wrong

novel parrot
#

$A = \bigoplus_0^{\infty} A_i$

cloud walrusBOT
#

ActiveChapter

695704832979239002.png
novel parrot
#

so any x in A is actually an ordered pair yeah?

#

i dont really understand homogenous

fossil shuttle
#

yeah, sure, you can think of it that way. or it's a formal linear combination of ordered pairs

#

every element is a finite sum (n_1,x_1) + (n_2,x_2) .... + (n_k,x_k) where n_1 < n_2 < ... n_k and x_k \in A_k

#

something like that

#

and (n,x)+ (n,y) = (n,x+y)

#

except that we impose the law (n,0)=0 for n, where i guess 0 here denotes the empty sum.

#

you might not see it written as an ordered pair but you can always give an isomorphism to another module where the elements are actually pairs

novel parrot
#

ok ...

fossil shuttle
#

a homogeneous element is a sum with only one term, so just (n,x)

novel parrot
#

why are we allowing multiplication?

#

A_i are groups?

fossil shuttle
#

it's a ring, so there's an addition and a multiplication law

#

the multiplication law is required to satisfy

#

(n, x) * (m,y) = (n+m,z )

#

for some z depending on x and y

novel parrot
#

the actual elements of A are just (x1, x2, ... , xn, ...) where x_i is in A_i and multipliying two of these ordered pairs we get another pair that has its points in A_i

#

its like that right?

#

but errr multiplication isnt defined on the subgroups?

somber marsh
#

where do i have measure theory channel

#

well if this works for measure theory what is the K set there

unknown.png
novel parrot
#

you gotta show 1 = 0

#

so multiplication = addition

gritty sparrow
#

A group is binary operation which is associative with an identity and such that every element has an inverse, they want you to show that the only ring where the multiplication has these properties is the zero ring

small bison
#

in a non-trivial ring, the best you can ask from multiplication is that the non-zero elements form a group

#

ie you have a division ring/field

#

that problem is asking for something slightly stronger: you want everything (including 0) to be invertible

primal pier
#

Broke: field
Woke: division ring

#

in algebraic topology there's something called the "cohomology ring"
you can associate it to a topological space

if you're not referring to the poincare connection idk

#

what you're referring to

latent anvil
#

the cup product makes the direct sum of groups $H^\ast(X; R) = \bigoplus_{i=0}^\infty H^i(X;R)$ into a (graded-commutative) ring

cloud walrusBOT
#

goblin shamrock

228562950703546368.png
latent anvil
#

for R a commutative ring

#

or alternatively, the wedge product of differential forms makes the direct sum of de rham cohomology groups of a manifold into a ring

#

they're isomorphic so it doesn't really matter

primal pier
#

ye and the isomorphism is poincare lemma or theorem or whatever

latent anvil
#

It's de rham's theorem

#

The poincare lemma is that the de rham cohomology of a contractible manifold is zero

barren sierra
#

ok sorry I'm still stuck on this

#

I do not get how normality / kernals would help me answer this

#

I have an answer

#

G = the identity permutation, A_3, or S_3

#

but I just said that and gave examples of valid homomorphisms

#

not sure how I can leverage normality or the kernal of phi

obsidian sleet
#

the image is isomorphic to the quotient group G/ker(phi)

barren sierra
#

haven't learned isomorphism theorems

obsidian sleet
barren sierra
#

yea

#

no

obsidian sleet
#

stare what did they expect from u monkaS

barren sierra
#

I've learned about quotient groups in my intro to proofs class

#

but we never covered it in this class

#

uh

#

like what isomorphisms and homomorphisms are

#

I've learned what a normal subgroup is

#

and I know what a kernal is

terse crystal
# barren sierra

Just choose any three elements of order 2 of G (not necessarily different) then map (12),(23),(13) to those three elements respectively

barren sierra
#

the identity, the alternating group, and S_3 itself

barren sierra
#

never learned order

obsidian sleet
#

tf is ur class doing

barren sierra
#

I mean I presume these things will come soon

#

yea

terse crystal
barren sierra
#

well if Ker(phi) = {id} then we only have that phi(id) = id and so everything else must map to a non-identity element

#

and vise versa if Ker(phi) is S_3 then for all s in S_3 we have phi(s) = id

#

ok

obsidian sleet
#

i will take it from here

barren sierra
#

ty 😭

obsidian sleet
#

yes you are correct

#

so have you shown that if the kernel is trivial group the map is injective

barren sierra
#

then we have isomorphism?

obsidian sleet
#

yes

#

since finite cardinality

barren sierra
#

that intuitively makes sense

obsidian sleet
#

6 items map to 6 items injectively so it must be surjective etc.

#

now in reverse when the kernel is the whole group the image is the identity, which u got yes

#

that ends two cases

#

now to deal with A_n

#

wait

upper pivot
#

I mean we are mapping into the image, which makes this always surjective

#

We don’t need a finiteness condition

obsidian sleet
#

right im dumb

#

yes

barren sierra
#

I mean I answered the question but here's how I answered it

#

I know trivially mapping phi(s) to the identity is a homomorphism

obsidian sleet
#

i can't read i thought it was from S_3 to S_3 but kekw

barren sierra
#

the identity function is a homomorphism

obsidian sleet
#

what john said is good

barren sierra
#

actually wait

terse crystal
#

Oh my bad, I thought you needed to classify all φ but it’s φ(S_3) not φ… spamakin is right just calculate all normal subgroups of S_3 and take quotient. The only non-trivial one is A_3 I think

barren sierra
#

does this help?

#

I did this problem

#

and I showed this

obsidian sleet
#

uh

barren sierra
#

can I leverage these results?

#

or no

obsidian sleet
#

at first glance im not seeing it immediately

barren sierra
#

fair enough

#

ok

obsidian sleet
#

im still kind of thinking about how to do A_3

upper pivot
#

Just do what@you were

#

Find all the quotients

barren sierra
#

uh

#

how?

obsidian sleet
#

so if all of the even permutations map to the identity

#

what is left

barren sierra
#

the odd permutations map to themselves

#

?

obsidian sleet
#

well the map is from S_3 to some group G

#

idk what G has but

barren sierra
#

I sent the full question lol

#

G is a group

#

that's all I got

upper pivot
#

There’s no need to think so hard about this, just think about what the size of The quotient

obsidian sleet
#

he doesn't have first iso

barren sierra
obsidian sleet
#

so we have to go through and investigate the image

upper pivot
#

So do you know what quotient groups are yet?

barren sierra
#

from my intro to proofs class but hasn't been covered in this class

upper pivot
#

I see

barren sierra
#

so in theory I am to solve this without that

upper pivot
#

Yeah sorry I’ll let metal guide then

obsidian sleet
#

wait 😭 john don't abandon me im still thinking too

#

im lowkey working out the problem with spam tbh

#

but i have suspicions

upper pivot
#

Oof ig look at the cardinality of the image

#

When kernel is A3

obsidian sleet
#

product of two odd permutations is even right

barren sierra
#

yes

obsidian sleet
#

you might be able to use this fact

upper pivot
obsidian sleet
#

so what will end up happening is that

#

id = phi(odd*odd) = phi(odd)*phi(odd)

#

uh

#

is this right

barren sierra
#

I may skip algo lecture tomorrow and just go to office hours

upper pivot
#

That’s correct metal.

barren sierra
#

so one homomorphism is

phi(a) = a if a in A_3 and phi(a) = id otherwise?

upper pivot
#

That would not be a homomorphism, no, you meant the other way around

barren sierra
#

what

upper pivot
#

A3 is all even, including the identity

#

It’s the kernel you want

barren sierra
#

right

#

hm ok

upper pivot
#

S3 is a pretty small group, honestly just check what the homomorphism has to be

barren sierra
#

Wait wait so find normal subgroups, set those as kernals, identity transformation the rest?

upper pivot
#

By testing it out on all elements

barren sierra
#

I am confused and just doing this solves my problem but still leaves me confused

upper pivot
#

I mean you learn nothing from this problem in general IMO cause you have no tools

#

Like it would give you insight if you could use first iso or whatever

#

But as it is right now it’s kind of just a computation problem

barren sierra
#

I mean ok the only think interesting about isomorphisms and kernals that I have learned

upper pivot
barren sierra
#

is something similar to lin alg and solutions

#

so if we have phi(x) = g

#

the set of all solutions x to this is 1 particular solution + any element in ker(phi)

#

is that what first the isomorphism theorem is?

upper pivot
#

Hmm not exactly

barren sierra
#

hm ok

#

nvm then

#

so the idea is find all normal subgroups

#

set those as kernals

upper pivot
#

Yeah that’s the main insight, everything after that is computations

barren sierra
#

and define the rest of the function as just mapping an element to itself if it isn't in the kernal?

upper pivot
#

I mean it’s not a map to itself so that doesn’t exactly make sense right

barren sierra
#

lemme word it better

#

the process is

find some normal subgroup N of S_3

define phi: S_3 to S_3 such that

phi(n) = id if n in N
phi(n) = n if n not in N
?????

upper pivot
#

This is not a homomorphism then.

barren sierra
#

hm

upper pivot
#

What I’d say is, force your kernel to be something and see what falls out

#

And the@map is not@in general going into S3

barren sierra
#

right but it's up to isomorphism right?

#

so arbitrarily I can pick G as subsets of S_3

upper pivot
#

What I’ll say about A3 is that every odd permutation differs by an even permutation from each other

#

No.

barren sierra
#

and everything else is just isomorphic?

terse crystal
# barren sierra

I was careless. You need to divide the problem in several cases. Case 1 G has no element of order 2, then the answer is {1}. Case 2 G has elements of order 2 and there doesn’t exist two elements of order 2 of G such that their multiplication has order 3 then the answer is {1},Z/2Z. Case 3 there exists two elements of order 2 of G whose multiplication has order 3 then the answer is {1}, Z/2Z and S_3

upper pivot
#

You wont have the image@be a subgroup necessarily

#

You are annihilating everything@in the@kernel yes

#

But elements in the kernel also interacts with everything else right

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For example take a map from Z whose kernel is 2Z

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You cannot describe this image as a subgroup of Z

barren sierra
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right

upper pivot
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(The image is called the quotient)

terse crystal
upper pivot
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But yeah so as I said

terse crystal
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He said “up to isomorphism”

upper pivot
terse crystal
upper pivot
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It’s obvious from context I mean it’s not a subgroup of itself…

terse crystal
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OIC

signal remnant
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@hidden haven just made this as a thank you gift 🙂

tensors_meme.png
next obsidian
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This is pretty good lol

primal pier
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Explain Cech co/homology to me like I know de Rham co/homology

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just kidding dont

hidden haven
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Nice KEK don't understand most of it though catThin4K

primal pier
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There's just too much to learn. OhNo_cat

dire bramble
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is this just asking me to show that the images of Z under each ring's initial homomorphism are isomorphic?

vast quiver
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is it referring to this sorta definition? "The characteristic is the natural number n such that nZ is the kernel of the unique ring homomorphism from Z to R"

dire bramble
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I'm not sure; I wasn't given any definition in class

vast quiver
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mhm

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well that's a definition that might be useful to use lol

dire bramble
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is that equivalent to what I asked?

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I'm leaning yes

vast quiver
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I think that's the definition that the problem is referring to. Use this definition to show that characteristics of isomorphic rings are the same

dire bramble
barren sierra
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It's true that if there is an injective homomorphism between finite groups

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then that homomorphism is an isomorphism?

thorn delta
barren sierra
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endomorphism is from group to itself?

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hm ok

thorn delta
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Ye

vast quiver
# dire bramble the things are indeed equivalent, since Z/ker(i) = Z/nZ = im(i), so if im(i) = i...

another way to do it is to use the uniqueness of the initial homomorphism (which is pretty much what your argument uses when you say Z/nZ=Z/mZ implies nZ=mZ). If R,S are your two rings, and phi:R->S the isomorphism, then you could have said phi∘i is a ring hom Z->S, which by uniqueness of the initial hom implies j=phi∘i. This gives us ker(j)=ker(phi∘i), and ker(phi∘i)=ker(i) because phi is an isom.

barren sierra
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imma have to go through and prove and disprove those

thorn delta
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Also, a counter example to what you asked would be the unique map {e} -> G for any G != {e}

vestal snow
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How/why is the curve y^2=x^3+1 a projective curve?
Doesn't a projective curve mean something that arises as roots of a homogeneous polynomial?

prisma ibex
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Yes you need to homogenize it first

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y^2=x^3+1 is an affine curve

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The corresponding projective curve is y^2z=x^3+z^3

vestal snow
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Isn't that a surface?

prisma ibex
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No

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It’s a curve in P^2

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I’m taking P^2 to have homogeneous coordinates [x:y:z]

vestal snow
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Oh wait

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I get it

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My bad

prisma ibex
vestal snow
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I was thinking of A^3

prisma ibex
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Yee

vestal snow
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Arigato

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@prisma ibex Won't homegenouzing the polynomial also introduce one extra rational point?

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(0:1:0)

prisma ibex
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It will yes

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The affine elliptic curve is the projective elliptic curve minus the point at infinity (which is the identity for the group law on the elliptic curve)

vestal snow
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So when we say that we need to consider the rational points on the curve y^2=x^3+1

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Do we include the point at infinity?

prisma ibex
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It depends