groups-rings-fields

[velvet hull]

can someone help me real quick, is there an "easy" proof of showing that the tensor product of free modules is free, without invoking the fact that a free A-module M is always isomorphic to A^I for some set I (with that fact its easy but I don't wanna prove it, it's a pain in the ass)

[coral spindle]

What other definition of a free module is there?

[karmic moat]

i guess you could just write out the basis of the tensor product?

[coral spindle]

That invokes the fact wanted to be avoided

[karmic moat]

aw balls

[coral spindle]

I'm not really sure what you're looking for frankly

I don't think there could be any easier proof than what you're referring to and anamono was suggesting

[velvet hull]

I was trying to stitch together the UP diagrams somehow

and it doesn't look possible lol

[coral spindle]

OK

I think there might be a proof that says something like adjoints commute with colimits

but I couldn't put the pieces together for you

[karmic moat]

there's some acronym

[velvet hull]

yeah I don't think the diagram stitching is gonna work out

I'll just eat the bullet

[karmic moat]

right adjoints preserve limits

[coral spindle]

The diagram stitching isn't gonna work out I don't think, no

[karmic moat]

RAPL and i guess LAPC

LAPC sounds weird i dont like that one

[thorn jay]

That follows from some basic facts about the universal mapping property?

Like it's a basic fact in any variety that F_V(X) is the coproduct of F_V({*}) indexed by X

F_V(X) being the V-free algebra generated by X

[velvet hull]

any variety
AAAAAHHHH I cant escape

[glad osprey]

As Tariq said, you didn't quite prove what the proposition said. I assume the (aN)(bN) is defined as abN, ie. multiplication of the representatives? But in the proof you define it as pairwise multiplication of the elements of the two sets, so you only proved those two definitions are equal, assuming they're well defined

[delicate orchid]

lets try and not use terminology more advanced than is required

[thorn jay]

:3c

[delicate orchid]

what I would do is just prove that the simple tensors form a basis

I know you don't want that it's iso to A^I

but proving linear independence of the simple tensors can be done independently of that assumption

[karmic moat]

oh

[delicate orchid]

obviously this would follow as an immediate corollary

[minor gazelle]

You’re right 🥰🥰 I shall fix it now

[tardy hedge]

Tariq, the Loneliest Poet

[south patrol]

Chungus

[karmic moat]

real shit

[south patrol]

Ye

[coral spindle]

This is the most hilariously abstract way of saying that we can find a basis, come to think of it lol

Like dang

even the meow theory proof is just "we have a basis"

[south patrol]

Probably worth pointing out most people will not know about varieties of algebras lol

[elfin wraith]

“Basic fact”

[south patrol]

Ah ok nice

[narrow temple]

With that proposition, cycling 0 -> 1 -> infty -> 0 yields the transformation g(t)=t/(t-1) with the matrix ( 1, -1 \ 1 0) which fits the requirements for 4).
And for 5) it turned out that for any order 3 element i construct a element in PGL(2) such that g was conjugate to it, so I was basically done.
So I think I got the job done, but I dont have a good reason for myself as to why g(t)=t/(t-1) is such a "special" transformation and why I should expect the result in 5 to be true.
Maybe im missing facts from linear algebra, group actions or just intuition, so could someone help here if im missing a "deeper" reason?

[south patrol]

I should say like i did not intend any condescension, as I am unfamiliar with the definition of a variety of algebras myself lmao

[cyan skiff]

It has nothing to do with inclusion take aN=cN and bN=dN and try to prove that abN = cdN

[velvet hull]

well Im not familiar with them as well I just kind of understood the reply lol

[thorn jay]

That's what I'm all about baby

[thorn jay]

I cannot think in anything but tbh

[thorn jay]

I mean the proof is simple

Omgomgomgomgomg

[velvet hull]

studying was not the right word

[elfin wraith]

No no I’m quite sure it’s a basic fact of whatever the theory is lol, I was only joking because while I’m sure it’s basic to show I’d be surprised if many people know it

[dull ginkgo]

What’s the context here because how else would you define a free module

[velvet hull]

...using the UP?

[thorn jay]

It's a basic fact that the fundamental class of the radical of a coherent condition is the prevariety generated by the fundamental class of that coherent condition

[dull ginkgo]

What UP are you using without referencing the indexing set of the basis

[velvet hull]

yeah, the UP with an indexing set

it just doesn't state the isomorphism explicitly, it's just some module F(I)

and I didnt wanna type the proof of isomorphism out

[dull ginkgo]

The humble adjoint

[velvet hull]

oh, trueee lmao

well that's easy

[south patrol]

Oh well then you can say that Hom(F(I) (x) F(J), M) = Hom(F(I), Hom(F(J), M) = Hom(I, Hom(F(J), M) = Hom(I, Hom(J, M)) = Hom(I x J, M) = Hom(F(I x J), M)

[velvet hull]

yes that's what I literally just typed out

[south patrol]

Lmao

[thorn jay]

😔

That's okay I'm used to no one caring about what I do /j

[south patrol]

Or observe there is a natural map F(I) (x) F(J) -> F(I x J) and each preserves colimits in each variable so you reduce to singletons

[coral spindle]

Oh nice

[south patrol]

So you just need to know F preserves the unit basically

[south patrol]

Says you

[velvet hull]

"wait, but isn't F1 x F2 also freely generated by S1 x S2" - me 1 minute ago

[south patrol]

Sorry dawg

[velvet hull]

its a tragedy

[somber sleet]

little question guys, in a UFD the decomposition of an element must always be finite, right?

[karmic moat]

right

[dull ginkgo]

Yeah. Any increasing chain of principal ideals must terminate. Apply that to the principal ideal of the element and consider elements dividing it generating larger ideals

[rocky cloak]

Yeah, infinite decompositions don't really make sense unless you have some notion of convergence

[karmic moat]

also i thought the definition of UFD says it should have a finite decomposition?

but I guess it depends on the source

[dull ginkgo]

GCD and having noetherian condition on principal ideals

[somber sleet]

I was asking because I didn't understand the first part of the proof of this statement in Algebraic Geometry of Gathmann,

[dull ginkgo]

because you can only like, divide an element so many times before the ideals generated stabilize, I.e they’re associates

Oh that’s literally the same proof lmfao

[karmic moat]

that's basically what mizalign said

[somber sleet]

jup hahaha, I think that's exactly what they are showing, it confused me because I didn't see what they were trying to show, but now it makes sense haha

[rocky cloak]

What part? That in a Noetherian ring, every element has a decomposition into irreducibles?

[dull ginkgo]

Yeah a divides b iff (b) is a subset of (a)

[somber sleet]

like I didn't see the need of showing finiteness

[somber sleet]

this @rocky cloak

[dull ginkgo]

In a way the inclusion ordering on principal ideals being pulled back through the principal ideal map is how you define divisibility

Alternatively :p

[somber sleet]

yeah it makes sense

[dull ginkgo]

Like most intro domain types depend on ideal structure

1. A domain is GCD (Greatest Common Denominator Domain) iff K(R) is closed under binary intersections.
2. A ring satisfies ACCP iff the inclusion relation restricted to K(R) is Noetherian, i.e there is no infinite strictly ascending chain of principal ideals
3. If a ring satisfies GCD and ACCP, then it is a UFD (Unique Factorization Domain)
4. A domain is Bezout iff K(R) is closed under addition of ideals
5. A ring is a PID (Principal Ideal Domain) if the principal ideal map is surjective, i.e every ideal is principal

Where K(x) is the principal ideal

So it kinda has to do with the structure of the sublattice of principal ideals

I miss algebra but the functional/numerical analysis mines call to me

[somber sleet]

algebra has my heart, why did you change sides? ahha

[dull ginkgo]

I liked functional analysis a lot because it mindmelds analytic structure and algebra but then got too lost in the suace

*sauce

I have an excuse to learn manifolds and lie theory for engineering mechanics tho

Sheaves seem like a powerful structure even in analytic settings so I'll likely take a stab at that somewhere

[delicate orchid]

oh my science....

[tardy hedge]

^Ya bro was cooking with that one

[delicate orchid]

that is kind of cool

[tardy hedge]

What is K(R)?

Set of principal ideals of R?

Uh

[delicate orchid]

I hope so

[karmic moat]

fraction field

[delicate orchid]

I was more thinking like

K_1 or K_0

but that literally wouldn't make sense it has to be the set of prinicpal ideals

or the poset of them, rather

[thorn jay]

Grothendieck group type shit

[tardy hedge]

What is that

[karmic moat]

fg modules modulo some relation

[thorn jay]

Type shit

[tardy hedge]

Type shi

[karmic moat]

relation is like some short exact sequence relation type shit

[tardy hedge]

Relations 🔥

[karmic moat]

okay iirc it's like

[thorn jay]

Congruences my beloved

[karmic moat]

K_0(R) is the group of isomorphism classes of fg R-modules

modulo the relation that if you have three fg R modules that fit into an SES 0 -> A -> B -> C -> 0, then you set [B] - [A] - [C] = 0

[delicate orchid]

define it using the universal property or I'm blocking you

too late!

[karmic moat]

bro </3 this is the one i learned in class

[delicate orchid]

this is the better definition anyway

[karmic moat]

can you put a ring structure on the grothendieck group?

[thorn jay]

Tensor product

?

[karmic moat]

oh nice

[thorn jay]

Idk just a guess but probs

Though I'm not sure it works well cuz tensor products don't preserve SES

[rocky cloak]

If you have some exact monoidal product, like for example over the group ring or a hopf algebra, it induces a product on the Grothendieck group

[karmic moat]

hmm

yeah i guess there's an issue with tensor product of modules then

[thorn jay]

Okh yeah of course

[thorn jay]

Ext and Tor ❤️

Over semisimple algebras that does form a ring

[karmic moat]

i wonder if there's anything cool that happens if you replace "fg R-module" with "flat R-module" in the definition of K_0

they form a category and the tensor product is exact so you get a grothendieck ring

[thorn jay]

Is there always a set-valued amount of flat modules?

[karmic moat]

wdym set-valued?

[thorn jay]

Do the collection of their isomorphism classes form a set or a proper class

[karmic moat]

hmm

tbh when defining K_0 i didn't even think about that for fg modules

the only definition i have is K_0(R) = \oplus Z[M]/~ where M are fg modules and ~ is that relation

[delicate orchid]

worrying about size issues is for nerds icl

[karmic moat]

but yeah i guess i didn't think about it being set-valued

[delicate orchid]

K_0 is a functor so like w/e

[karmic moat]

so it still works for the category of flat modules then?

[delicate orchid]

if they're closed under extension, sure

that shit should be an abelian category

only thing that worries me is kernels and cokernels

bad news gang - it isn't always abelian

however I have stopped caring about short exact sequences. Just quotient out by [M_1]+[M_2]-[M_1 (+) M_2]

[karmic moat]

should be, you can just look at the long exact sequence of tors right?

[karmic moat]

fawk 💔

[rocky cloak]

fg projective modules might be better

[karmic moat]

hmmm i see

sometimes by K_0 people mean fg projective modules right?

eg stacks https://stacks.math.columbia.edu/tag/00JC

[delicate orchid]

I've seen this yeah

[karmic moat]

algebraists should just agree on one thing

pmo </3

[thorn jay]

There are many different definitions of the word ALGEBRA

[karmic moat]

true

[south patrol]

Isnt this the normal thing lol

[karmic moat]

my prof just used fg modules

[south patrol]

No

Just consider free modules

[thorn jay]

Oh yeah ofc

[vast verge]

Can someone explai'n where the green part comes from?

[south patrol]

E.g. Lagrange applied to G/H

[candid patrol]

If [G:H] = n, then for every x € G, we have x^n = 0 in G/H, but 0 = H, so (in G) x^n € H

[arctic trail]

This is rather simple

[vast verge]

I know I'm very sorry

[arctic trail]

Noooo sorry

[vast verge]

It slipped my mind

[arctic trail]

I didn't mean it like that

I misunderstood the question

I thought it was finite order**

[white oxide]

Let M_i be a family of A-modules, and suppose that each M_i is flat. Can I have a hint for showing that the direct sum M of the M_i is also flat?

[next obsidian]

Show that tensor product distributes over direct sums

[white oxide]

Do you mean that $A \otimes \bigoplus_{i \in I} B_i \cong \bigoplus_{i \in I} A \otimes B_i$?

[cloud walrus]

okeyokay

[karmic moat]

yeah

[white oxide]

I see, thanks

[white oxide]

Can somebody please explain to me why $\mathfrak{p}^{ec} = \mathfrak{p}$ for a prime ideal $\mathfrak{p}$ which doesn't meet $S$ as in iv?

[cloud walrus]

okeyokay

[velvet hull]

by (ii), p^ec = bigcup_{s in S} (p : s). But p is a prime ideal, and p does not meet S, so if x is in (p:s) such that xs is in p then x has to be in p. Hence p^ec is inside p, the other containment is easy

[rocky cloak]

Well, the elements in p^e look like x/s for x in p. Then elements of p^ec will look like when x/s is in A. I.e. y with x = sy in p.

Then apply p being prime to sy

[covert cliff]

In the semi direct product on the last line is the operation just multiplication?

(A, a)(B, b) = (aAB, ab)

Or what is it?

So by first isomorphism GL(n,F)/SL(n,F) = Fx

[dull ginkgo]

(Internal) semidirect product is of the form

AaBb = AaBa^-1 ab = (A aBa^-1) (ab)

[minor gazelle]

I fixed some mistakes for proving that multiplication between cosets are well defined.. is this proof now better or still need to be fixed? And the proof showing the quotient group is a group are both methods valid? Proposition 1.35 is a proof showing a ~ b iff a^{-1} b in H which is how the left cosets are defined. Then I argue that multiplication for the normal group is well-defined, and then manipulate the system of representative directly

[regal mango]

Hello everyone if anyone wants to study group theory and practice its question from herstein in the month of August can fill out this form

https://forms.gle/NJthoYupPc4Hjepe6

[cyan skiff]

The problem is the way you replace a single element of N with N. What you are basically assuming is that $a^{-1} a' $ commutes with every element of G which is a stronger condition than normality. Normality basically comes down to $ [
ah_1 = h_2a \quad \text{for } h_1, h_2 \in N
]
$

[cloud walrus]

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

[cyan skiff]

So use this for the proof

[cyan skiff]

Plus, You haven't shown that the quotient group is closed and I don't get your inversion proof. You can't claim that if aN = bN then a(-1) = b(-1).

[minor gazelle]

I shall be fixing it.. I found the concept a bit abstract 🫣🫣

[minor gazelle]

How’s now

I felt my idea of showing it that way might be correct? But was unclear? Maybe? So this time I made it more explicit.. or my idea was wrong completely 🫣 then I shall revise the concept completely

The commutative like operation is granted by normality

[tough raven]

They haven't specified the monomorphism but supposing it is t ↦ diag(t, 1, ..., 1), we have (A, a) ↦ A diag(a, 1, ..., 1) as the bijection, so (A, a) (B, b) ⇒ A diag(a, 1, ..., 1) B diag(b, 1, ..., 1) = A diag(a, 1, ..., 1) B diag(a, 1, ..., 1)^{-1} diag(ab, 1, ..., 1) ⇐ (AB', ab), where B' = diag(a, 1, ..., 1) B diag(a, 1, ..., 1)^{-1} is the matrix obtained from B by multiplying it first row by a and dividing its first column by a.

[tribal moss]

I like https://gowers.wordpress.com/2008/08/12/how-to-use-zorns-lemma/

[languid trellis]

I'm struggling to see why being a solution to congruence (1) implies you solve congruence (2). Help pls

[rocky cloak]

Raise both sides to the power of q_i

[languid trellis]

oh

That's obvious then lol

Thanks haha

[karmic moat]

yeah

basically you can have multiple maximal elements

key word is maximal vs maximum

[thorn jay]

Consider a poset where no elements are related

[karmic moat]

the word "maximum" implies unique

[south patrol]

m is a maximum means m >= x for all x, which implies uniaueness

m maximal means that it is impossible to have m < x

They say maximal, not "maximum"

Different things

Yeah i mean it is true

You can only have one maximum, but can have multiple maximal elements

Maximal + every element is comparable is enough

Easy exercise

[karmic moat]

take a subset of N

[south patrol]

Or i can do if need be but i suggsst u should try

[karmic moat]

for example

[south patrol]

Wdym lol

[karmic moat]

sorry i should say upper bounded subset of N

[south patrol]

Sure ye

[karmic moat]

finite subset

[south patrol]

But then there are no maximals

Yes

Great

[karmic moat]

wouldnt the maximal element just be the largest element in your finite subset of N lol

it satisfies the definition of maximal

[south patrol]

Ons thing that woyld work is "subsets of N of cardinality <= n"

[south patrol]

Oh lol

Sure but then that is a maximum

I wasnt sure what the point of your comment was

[karmic moat]

yeah i mean as an example that maximal + every element comparable implies maximum

[south patrol]

Okie sure

Yes exactly

[crystal vale]

How (yx-1)a = 0?

[dull ginkgo]

Tim’s false

*it’s false

Ignore it

[south patrol]

You can visualise his nicely by drawing like a tree

And ordering by height

There need not be a tallest leaf

[karmic moat]

ewwwww

graph theory ewwwwwww

[crystal vale]

I am thinking for commutative ring with unity

[south patrol]

This is a good question

[tribal moss]

Jagr's example works there.

[crystal vale]

Sorry I don't know which one

[south patrol]

k[x,y]/(y(x^2 -1)) and y, xy

[tribal moss]

There's also Z[x,y,z]/(zyx-x) with a=x, b=yx, which I abstracted further from Jagr's one.

[crystal vale]

Yes someone gave me this hint but I didn't make any contradiction here

[crystal vale]

Yes Ra = Rb

[tribal moss]

You just need to show that there's no unit u with ux=yx.

[south patrol]

Z[x]/(x - x^3) presumably also is good lol

x^2 R = xR

[crystal vale]

How?

[south patrol]

Uhh

[crystal vale]

I mean I tried

[south patrol]

No nvm i think thdy differ by a unit

[tribal moss]

No, I found a unit that turns x into x² there.

[south patrol]

Yeah as a ring this is just Z x Z x Z

And so it should not be possible

[crystal vale]

Is that tricky ?

[rocky cloak]

It's Z x Z[x]/(x^2 - 1) at least

[crystal vale]

Why not Jagr ?

[south patrol]

Ah my bad

[rocky cloak]

That's just my name

[south patrol]

Cause (x+1, x-1) is not (1) lol

[crystal vale]

Not Jagr 😅

[tribal moss]

Hmm, I didn't completely finish all the details, but I think it's just term chasing.

[rocky cloak]

.

[crystal vale]

Okay I will try again

[rocky cloak]

The ideal generated by a set is a particular ideal yeah

[thorn jay]

There is only one ideal being generated by a particular set yeah

[tribal moss]

I think you can probably prove that the only units in that ring are 1 and -1.

[dull ginkgo]

Is there like an actual commutative example

[crystal vale]

Oh

[rocky cloak]

There are lots of ideals that contain something, but only one of them is the smallest ideal containing something

(which is what generated means)

[thorn jay]

Definitions are hard

[rocky cloak]

Are the given examples not commutative enough?

Or what does that mean "actual commutative"?

[dull ginkgo]

I mean for the

Whole xR = yR but x and y are not associative

*associate

What

[rocky cloak]

k[x, y]/(y(x^2 - 1)) and Z[x, y, z]/(xyz - x) are both pretty commutative to me

[south patrol]

Lol i mean jagr gave commutative examples

[tribal moss]

That's what the task is.

[dull ginkgo]

Oh shit yeah

[thorn jay]

It's not

[south patrol]

In general

[thorn jay]

It's isomorphic to the poset of ideals of R/I though

[dull ginkgo]

I wish I didn’t suck at algebra lo

[south patrol]

I mean many posets in nature will have maximals which arent a maximum

[rocky cloak]

I don't who this jagr guy is, but they sound cool

[thorn jay]

They're probably just a fake

[south patrol]

Idk jagr is pretty cringe

But u seem ok

[dull ginkgo]

The set of all ideals containing a set is well-founded

[crystal vale]

That means you quotient it again with y

[thorn jay]

A cheap clone of you, after all, not not jagr2808 is equal to jagr2808, so they're just copying you and putting a "not" in front of it

[ashen heron]

new group chat beef damn

[crystal vale]

Why must the x-terms of unit be zero? I am not sure about the x-terms of the unit, you mean x will be vanish

Not vanish, you mean all x-terms will be zero

[rocky cloak]

Like a polynomial will look like
a0 + a1x + a2x^2 + ... + b0y + b1yx + ...
so if you quotient out by (y) you get
a0 + a1x + a2x^2 + ... in k[x].

This is only a unit of a1=a2=...=0

[south patrol]

Fun exercise

[dull ginkgo]

Ring theory moment

[crystal vale]

I got upper part but what does it mean this is the only a unit of a1=a2=...=0

[rocky cloak]

Well, we know what the units of k[x] are

It's the constant polynomials

[crystal vale]

Yes

Got it

[knotty badger]

-# nonzero ofc

[tribal moss]

Instead of "quotient out y" it may be easier to think of: if you have any ring homomorphism out of R, then units must map to units.
In particular, for R=k[x,y]/(yx²-y), this is true for the evaluation homomorphism R > k[x] that sends x to x, y to 0.
This kills all tems in a polynomial that contain at least one y, and we know if we start with a unit that should still produce a unit -- that is, just a constant term.

So if u is a unit in k[x,y]/(yx²-y) it must be of the form c + (terms all involving y)

[crystal vale]

Doesn't contain, is that typo?

[tribal moss]

Whoops, fixed.

[crystal vale]

Okay

[crystal vale]

I see

I got jagr's example

I mean is that natural for you to think about all these? I mean I tried but I didn't think in that way to take the quotient again and find the forms of units

[tribal moss]

I'd say, not super natural for me -- I have to learn as I go each time I involve myself with one of these exercises ...

[crystal vale]

Okay can I take K[x,y]/<x-xy> in replace of K[x,y]/< y(x^2-1) >, I don't think it change anything

[tribal moss]

I think that preserves the property that the only units are constants, but what are your a and b then?

[crystal vale]

No, now I am not thinking about a and b, I am thinking about the form of unit change or not

So it doesn't change, right?

[tribal moss]

I think that's correct. But (to be honest) I haven't been fully grasped the last part of Jagr's argument yet.

[crystal vale]

You mean the only unit element is of the form a + f(y)y + g(y)yx

[tribal moss]

Well here, you get unit elements have the form a + f(x)x, since you interchanged x and y when you went from y(x²-1) to x(1-y). And then you don't need a separate g(x)xy term because x=xy.

[crystal vale]

Sorry, from y(x^2-1) to x(1-y), I don't get it

I got it f(y)yx^2 = f(y)y

[tribal moss]

Jagr said K[x,y]/(y(x²-1)), you said K[x,y]/(x(1-y)) -- it's a different variable that factors out of the generator.

[crystal vale]

Oh

[crystal vale]

Oh I quotient it by y, i think it doesn't matter, right?

Then it will be a + f(y)y

Oh no

[tribal moss]

I don't think you can straightforward "quotient it by y", because the ideal <y> does not contain <x(1-y)>.

So if you try that, you collapse the entire ring to K -- which is valid but doesn't tell you anything useful.

[crystal vale]

Can I map homomorphism such that K[x,y]/<x-xy> to K[x] ?

Such that x goes to x and y goes to 0

[tribal moss]

No, because then f(0) = f(x-xy) = f(x)-f(x)f(y) = x-x·0 = x.

[crystal vale]

My bad

[tribal moss]

In order to make an evaluation morphism from a quotient of a polynomial ring, you need to make sure that your chosen images of the variables satisfy the relations in the ideal you quotient by.

That's why you could map x to x and y to 0 in K[x,y]/<yx²-y>.

Because mapping y=0 makes sure that everything in <yx²-y> will map to 0 when you make the same assignments in a map from the raw K[x,y].

[crystal vale]

Got it

[crystal vale]

So which part?

[tribal moss]

The one where he got from "every unit has the form a + f(y)y + g(y)yx" to "every unit has the form a".

(And I was sorta depending on that when I claimed the only units in Z[x,y,z]/<xyz-x> are ±1).

(But now I'm not sure his argument for k[xy]/<yx²-y> actually does generalize to Z[x,y,z]/<xyz-x>).

[covert cliff]

Why not just define it to be the product rather than the semi direct product?

[crystal vale]

When he said this?

[tribal moss]

Here.

[crystal vale]

So you mean you are doing again map K[x,y]/< y(x^2-1) > to K[y] such that y -> y and x-1 maps to 0

[dull ginkgo]

Because the setwise product of two subgroups need not be a subgroup. It is when one is normal.

When they have trivial intersection you get a semidirect product

It’s just standard terminology and highlights how one is normal :p

[covert cliff]

Oh and ig in this case SL(n, F) is normal and these have trivial intersection so its just the "same" notation

[tribal moss]

Yeah, I think I get that now. More elementarily "x maps to 1" and then "x maps to -1" tells us that
a + (f+g)y and a + (f-y)y are both units in k[y] means that f+g=f-g=0 in k[y], which is easily solved to get f=0 and g=0.

[covert cliff]

and the isomorphism $SL(n, F) \rtimes F^\times \to GL(, F)$ is just given by $(A, a) \mapsto aA$?

[tribal moss]

(And that doesn't generalize neatly to Z[x,y,z]/<xyz-x> so my universal example wasn't as good as I thought).

[cloud walrus]

Khush

[crystal vale]

a + (f+g)y and a + (f-y)y how it is unit in K[y]

I got it

[tribal moss]

We know the only units in K[y] are the nonzero constants.

[crystal vale]

Yeah great

How this long idea comes to mind man

But I have a doubt, how I can show that x ≠ kxy for any unit k in K

[tribal moss]

If x = kxy holds in K[x,y]/<y(x²-1)>, then it still hold after we map to K by f(x)=1, f(y)=1 (this is an allowed mapping because it satisfies y(x²-1)=0) -- so 1 = k.
But it also still holds after we map to K by f(x)=1, f(y)=2, so 1 = 2k too.

Oh, wait, Jagr's example has a = y, b = xy, so it's not x ≠ kxy you need to show but y ≠ kxy.

[crystal vale]

After all this is commutative

[tribal moss]

Yes, but x and y have quite different roles in the ideal we quotient out, and that influences which mappings we can use to argue with.

x=kxy can be refuted in one go just by mapping f(x)=1, f(y)=0, which gives 1=0 no matter what k is.

To refute y=kxy, I think we need two mappings, first f(x)=1, f(y)=1 to get 1=k -- and then f(x)=-1, f(y)=1 to get 1=-k.

[crystal vale]

But how can I say y ≠ xy in K[x,y]/<y(x^2-1) >, I think map, x -> -1 and y to 1 then it will give 1 = -1, which is not possible, right?

[tribal moss]

Yeah, that would be a way.
A more intuitive, but also more technically roundabout way would be to show that every coset in K[X,Y]/<yx²-y> contains exactly one representative polynomial where every term has the form k·x^n or k·x^n·y^m where n<=1 and m>=1. (But the "at most one representative" could be fiddly to prove rigorously instead of just vibing it).
Since x and xy both have that canonical form, they cannot be in the same coset.

[crystal vale]

I don't get n ≤ 1 part

[tribal moss]

If you have a term with more than one power of x (and at least one power of y), you can reduce it down using yx² = y.

[crystal vale]

Yes

I got it

Because x^2 = 1 in that quotient ring, right?

[tribal moss]

No, only yx² = y.
Powers of x that haven't been multiplied by y yet are all different.

[crystal vale]

Yeah

[crystal vale]

And what does it mean by x and xy have canonical form?

[tribal moss]

By "canonical form" I meant the form I described as "polynomial where every term has the form k·x^n or k·x^n·y^m where n<=1 and m>=1."

[crystal vale]

Okay

So x and xy are not the same coset

[tribal moss]

Yeah.

[crystal vale]

and we have to show y ≠ xy, how are they related ?

[tribal moss]

Sorry, I got confused between x and y again. For y != xy it's the same: they are in different cosets because both y and xy have the canonical form I described earlier.

[tardy hedge]

Can you not just say y ≠ xy since xy-y is not a multiple of yx^2-y?

Like do u need to argue deeper WHY its not a multiple of that

[rocky cloak]

Could just argue by degree even

Or evaluation like I think Tropos mentioned earlier

[tardy hedge]

Because polynomials can only be equal if they have same degree?

[rocky cloak]

Yeah

[tardy hedge]

How do we usually think about degree of multivariable poly

[rocky cloak]

Well, either degree in a single variable (k[x, y] = k[y][x], so yx^2 - y has degree 2)

or total degree, i.e. yx^2 is a degree 3 monomial

The former being most relevant here I guess

[karmic moat]

I think total degree is more common

[tribal moss]

Yeah, the more intricate arguing I'm talking about is for when doesn't have enough intuition (or doesn't trust what one has) about what is a multiple of what.

[karmic moat]

I think another way is you can define the degree of x to be (1,0) and the degree of y to be (0,1) ie a Z^2 grading

Then you can define the degree of x^n*y^m as (n, m). I remember using something like this in the context of log canonical thresholds but maybe I remember wrong

[tribal moss]

The quotienting makes that more tricky.

[karmic moat]

Yeah I’m just rambling diff gradings u can put on multivar polys

[tribal moss]

(I tried writing down an argument based on keeping track of degrees modulo the generator, but in the end that turned out to simplify to just evaluation :-)

[crystal vale]

I see

Thanks a lot @tribal moss , you helped me a lot ❤️

[thorn jay]

You can grade the n-var polynomial ring using the free monoid on n variables

[tardy hedge]

Boi.

[weary frost]

not exactly, the problem is if det(M) = -1 for example and F = R (the real numbers) then there is no proper number a for this to work (you'd need a^2 = -1). Instead you use the subgroup diag(a,1,...1), matricies that are diagonal with a in the top left corner and 1 elsewhere on the diagonal. This also has trivial intersection with SL(n,F)

[covert cliff]

genuinely hate the semidirect product

[next obsidian]

:c

[tardy hedge]

c:

[knotty badger]

:(

[thorn jay]

:c

[serene acorn]

:<

[tardy hedge]

:p

[karmic moat]

:c

[thorn jay]

Genuinely love the semidirect product

[knotty badger]

it took me a while to understand it

but once I got the universal property i was golden

[karmic moat]

Genuinely neutral about the semidirect product

[thorn jay]

Universal property?

[knotty badger]

https://ncatlab.org/nlab/show/semidirect+product+group

[thorn jay]

Oh yeah

The shit is the arrow category lmao

Ah okay

Love that you understood the universal property but not the intuition that the group action is the conjugation and that's how the multiplication is defined

[knotty badger]

That’s how I understood the intuition

Through the universal property

[thorn jay]

Your brain works in interesting ways

Not a bad thing

Reminds me of someone..

[knotty badger]

who?

[thorn jay]

Someone called Ity/alumite, she was on this server briefly a while ago

[knotty badger]

hm, never met her :(

[thorn jay]

lucky

Rahhhh why is no one answering my question in advanced-algebra I am shitting myself

[south patrol]

I just dont think it is something many will have touchsd lol

[chilly ocean]

If K_1 is a commutative ring and f is an epimorphism from K_1 to a ring K_2, given 2 elements a, b from K_2, since f is an epimorphism, there exist c,d in K_1 such that f(c) = a and f(d) = b

So we got that:

ab = f(c)f(d) = f(cd) = f(dc) = f(d)f(c) = ba

Hence, K_2 is commutative

[south patrol]

Epimorphisms of rings are not necessarily surjective

Unless you are using epimorphism to mean surjection, in which case it is probably worth me pointing out that it is ambiguous

[chilly ocean]

I got a different definition then

[south patrol]

Ah sure

Then yes this works

[chilly ocean]

Okay

[chilly ocean]

Now I am curious about this

What would be an epimorphism of rings?

[south patrol]

There is a general notion of epimorphism in category theory. f: A -> B is an epimorphism if for any two maps g, h: B -> C such that gf = hf, we have g = h

If we work with sets or groups instead of rings, then this is equivalent to f being surjective

For rings it is different (e.g. the inclusion of Z in Q is an epimorphism)

But don't worry if this is very irrelevant to what you are doing lol

[chilly ocean]

It's relevant

[south patrol]

Just i think this is a reason to prefer just saying "surjective map" rather than epimorphism if that is whst you mean

[chilly ocean]

Right now I never had any experience with category theory

But in the future I will study it

[chilly ocean]

So with this you would be saying that g(B) = h(B) but we can't suppose that g(B) = C or that h(B) = C

[thorn jay]

It appears, which is fair, and I kinda expected it

Oh well

[chilly ocean]

Well, in the book I'm using to learn abstract algebra, it says that we should understand the word "epimorphism" in the context of rings as we do in the context of groups, so I understood that it means the same as it would with groups.

However, since morphisms of rings deal with 2 operations instead of 1, I always thought it's a bit weird that the vocabulary of the types of morphisms is the same

[real knoll]

what are the prerequisites for studying group, ring and field theory?

[tardy hedge]

set theory, some introduction to proofs like properties of functions and stuff

[tardy hedge]

There technically arent really that many prereqs, like probably something like multivariable calculus require more “prereqs”

Linear algebra is good to know because it provides examples and intuitions

[thorn jay]

Also provides the best examples for groups: the general linear group!

[real knoll]

Yeah, I have a very solid background in multivariable calculus and linear algebra, I might be ready I guess.

[tardy hedge]

Yea should be fine to look into it then

[covert cliff]

Im trying to prove that if we have a short exact sequence $1 \to A \to B \to C \to 1$ then the map $A \to B$ is injective

[cloud walrus]

Khush

[covert cliff]

I'm just completely stuck

can someone give me a hint

I tried starting with calling $f: A \to B$ and $g: B \to C$ then suppose $f(x) = f(y)$ but I can't seem to use the property that $\ker g = \text{im} f$

[cloud walrus]

Khush

[karmic moat]

im (1 -> A) = ker(A -> B) by exactness. What’s im(1 -> A)?

[covert cliff]

e_A

ah I see

thats smart

[karmic moat]

Yeah so trivial kernel means what

[covert cliff]

I really thought the maps 1 -> A and C -> 1 were for shits and giggles

[dull ginkgo]

That’s why it’s like a quotient basically

A injects into B, B maps onto C

[covert cliff]

so in my mind I just ignored them

[karmic moat]

Nah

[dull ginkgo]

I mean they are basically just to make the first nontrivial map mono and the last nontrivial epi

Like “caps”

[karmic moat]

Anyway while you’re at it maybe prove B -> C is surjective

[fading acorn]

$0\to A\xrightarrow{f} B$ is exact iff f is injective and $A\xrightarrow{f}B\to 0$ is exact iff g is surjective

[fading acorn]

very important things (0's on end of the sequence)

[dull ginkgo]

Short exact sequences moment

[fading acorn]

lol

[hollow tartan]

What are some common techniques, one should be familiar with for an algebra qual in general? I.e. using sylow theorems to prove G is not simple. Any other common problem solving ideas on these exams?

[karmic moat]

I suppose you could look through these questions and see what ideas are used https://www.math.harvard.edu/media/groups.pdf

The three (or four? idr) isomorphism theorems are always good to know

https://wiki.math.wisc.edu/index.php/Algebra_Qualifying_Exam

You could also look at the notes on this page about group theory

[untold torrent]

Hello guys

Can anyone tell me where can I read about inversion in permutation group I have gallian book but he doesn't have any kind of thing

[tribal moss]

Inversions are mainly a tool for proving that the parity of a permutation is well defined. If you have proved that in different ways, you may not really need them.

But if you're just curious, you could read https://en.wikipedia.org/wiki/Inversion_(discrete_mathematics)

[velvet hull]

To follow up on what Tropo said, here’s a graphical approach to seeing why the parity of a permutation is well-defined: if P is a permutation in S_n, you can look at the induced directed graph on 2n points (see image).

Then the parity of a permutation can be defined to be the parity of the crossing number of the graph

And one can observe that this is essentially another way of counting the inversions of a permutation

[rocky cloak]

It appears to be true that an epimorphism from a commutative ring would imply the codomain is commutative

(See Prop 1.3 https://link.springer.com/article/10.1007/BF02566112 )

But I hope that's not what you're actually asking, because the proof looks difficult

[chilly ocean]

Well, I'm still not at that level, as I'm sort of learning ring theory for the first time and I have 0 knowledge about category theory.

I'm still pretty far away from a point in which I could understand that, but in finite time I'll reach that

Let me access that Springer article btw

Unfortunately I can't get access to it

[rocky cloak]

Alright, just for my own benefit let me unpack this proof.

R -> S is an epimorphism of rings, and M is an S-S-bimodule. Then the first lemma is that for any m in M if rm = mr for all r in R, then also sm = ms for all s in S.

Proof: consider the trivial extension SxM where multiplication is
(s, a)(t, b) = (st, sb + at). Then there is a canonical ring homomorphism s |-> (s, 0) and for every m in M
s |-> (s, sm - ms)
is also a homomorphism.
If rm = mr for all r then these agree when composing with R->S, hence must be equal (since it's an epimorphism), so sm=ms for all s.

PROOF OF COMMUTATIVITY: consider now S as an S-S-bimodule. Then any element of R commutes with every element of R, so R is in the center as per the lemma. But then every element of S commutes with all of R, so S is in its center, i.e. S is commutative

---

This same lemma can be used to show that R -> S is an epimorphism if and only if the multiplication map

$S \otimes_R S \to S$

is an isomorphism of bimodules

[cloud walrus]

Not jagr2808

[plucky dome]

does anyone know of somewhere i can look up the generators for Z/pZ* for say, the first 100 primes?

[delicate orchid]

The smallest primitive elements are either 2. 3, or 5 (based on modularity conditions I’ll leave for you to Google) unless p = 41 (it’s 6) or p =71 (it’s 7)

I dont know why you’d use any other generators

[twilit wraith]

https://en.m.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n

Examples section

[next obsidian]

I guess sure

I think they just don’t want to be talking about this in the situations R isn’t commutative

So they just assert it from the start @sacred wharf

I mean like

The word “Euclidean domain”

Might have only been an adjective applicable to commutative rings

Just ignore it

It doesn’t matter

[tardy hedge]

Lol if u do that send it to me

Id watch it

We need more math content like that

I fw textbook review kinda shit heavy

[south patrol]

"Principal Ideal Domain" was funny to me for some reason lol

[tardy hedge]

Like who be doing textbook reviews besides that guy

Math sorcerer dum

[south patrol]

Math sorcerer scammer

[tardy hedge]

L math sorcerer

[south patrol]

"The secret to becoming a great mathematician" with zero reason i should trust him

And thsn loads of book reviews to become experts but they are all somewhat basic textbooks iirc

[tardy hedge]

Thats harsh 😂

But u right

[south patrol]

Lol yeah

[thorn jay]

Often authors state redundant information so that your mind doesn't have to manually switch to "okay, commutative ring" mode, for example

[south patrol]

Maths is fake anyway

[tardy hedge]

Fo real

[thorn jay]

Everything is redundant

Oh well

[south patrol]

Real

[thorn jay]

Except UA of course

[tardy hedge]

So True bro

[thorn jay]

It solves ACTUAL problems, REAL WORLD problems

[stone mason]

terence tao has a blog, if you ask an honest question it seems you have some chance of getting a reply

[south patrol]

Discord exists to keep geometric algebra and UA alive

[tough raven]

UA is redundant, all the isomorphism theorems are already in Dummit & Foote

[thorn jay]

Ah of course

[tough raven]

[thorn jay]

The only thing UA is good for

[thorn jay]

I think most UAers are not on discord

Cuz they're old

[stone mason]

is apostol's calculus book any good

someone handed it to me earlier today

[south patrol]

Cause theyre dead

Oh lol

[thorn jay]

Idk I've never done calculus

[stone mason]

not even calc2/3

[thorn jay]

I know somewhat how to integrate, and I can do basic analysis reasoning if

Ig

Fair

Analysis is just topology but worse anyways

[stone mason]

yea as much as i liked calc 2, it only goes so far as balckpenredpen in yt or the mit integration bee vids

fun for learning technique though

like actually working out problems

[thorn jay]

You should totally learn algebra and eventually learn universal algebra

[stone mason]

universal algebra?

what's that

[thorn jay]

You kinda just investigate the concept of an algebraic structure, and prove theorems about them in full generality, i.e. only assuming properties which are not formulated for a specific type of algebraic structure like groups or rings or fields

[stone mason]

ahh so algebra for different structures

seems cool, someone was telling me about this earlier, like what function spaces are and their axioms, and that there are different spaces with different closures and algebreic properties

[thorn jay]

In a sense? For example, linear algebra studies vector spaces, while in universal algebra you might ask "what general properties make vector spaces so nice, and how can I extract them to tell exactly when some type of algebraic structures has the same properties as vector spaces"

[stone mason]

sounds pretty abstract

[tardy hedge]

Onggg analysis is so ahh

[thorn jay]

For example, the fact that addition of vectors commutes can be generalised to the fact something called the commutator is always trivial, and (almost) everytime the commutator is trivial, you get a module, which is like a generalised vector space

[stone mason]

what does trivial mean here? sorry, new to math phrasing

[thorn jay]

Talking abt my music taste or my music lol

[thorn jay]

Do you know subspaces in linear algebra?

[stone mason]

no i have yet to take any la or odd classes

[thorn jay]

Thank you :3 music is so fun to make, I think everyone should try it sometime

[tardy hedge]

I would love to. How do u recommend starting?

Do u make edm enpeace?

[thorn jay]

Okay, basically think of the trivial space as a single point

[tardy hedge]

Like fl studio stuff?

Im going to a rave tmr actually woohoo

Theres a big hardcore scene in toronto

Happy hardcore shit

[thorn jay]

Oh shit that's super cool!

The Netherlands has a pretty big dnb darkstep scene I'm pretty sure

At least, the pioneers were from the netherlands

[tardy hedge]

Niice

[thorn jay]

NOISIA and VISION RECORDS my love

Oh damn okay

Hm

Nah i gotta choose music

Math has kept me sane but music has given me a way to express and feel emotions

[tardy hedge]

Real

[thorn jay]

Good distraction for your brain

[thorn jay]

🤝

[thorn jay]

FL is great, and the trial is not too limiting

[tardy hedge]

Isnt there some correlation between math and music ppl

[thorn jay]

Only can't save your projects but just leave them open and ur good

Esp if you're just starting out it's fine

[tardy hedge]

Cuz besides math music is my biggest interest too

Nice

[thorn jay]

There was some correlation between mathematicians and creative hobbies

[tardy hedge]

Thats interesting

[thorn jay]

Cuz in math you gotta have some notion of creative thinking

[tardy hedge]

Yeah

[thorn jay]

Ur ass is not gonna get these abstract ass things if you're not creative

[tardy hedge]

Its interesting how the general
Public dont know that about math

[thorn jay]

Academia had a stigma of being the opposite of creativity

Cuz the math in school is engineering math

Just computation problem after computation problem

My ass took advanced math, music, and geography as electives

Most random combination

Yeah, luckily that period for me was shortlived cuz I saw group theory and was like "fuck it why not sounds funny" and taught myself group theory using Wikipedia and a long ass problem sheet

Though I had motivation for a project so ig that's the reason I could even continue

I'm starting uni next year

Mostly focused on algebra though that's why I'm a UA nutjob

[tardy hedge]

Ya enpeace is cracked

Lowkey

[thorn jay]

18

Crazy? I was crazy once

I think I do more math than the average person

I have no life 😔

I met someone my age who is currently alr doing a masters though that mf is insane

Well met online at least

Crazy

Unfortunately, rather than being cracked at number theory, or measure theory, my dumbass had to get interested in that one branch of math many people don't care about 😭

Easy papers I guess

It is, but it's also pretty cool to see

Algebraic geometry and category theory are even worse

[tardy hedge]

Yea

[tardy hedge]

Damn maybe i should stay in combinatorial algebra then

[thorn jay]

Self study...

[tardy hedge]

Everyone loves the abstract shit including me

[thorn jay]

It's just a shift in thinking, sheaves

[tardy hedge]

Yea examples can be boring but theyre kinda necessary to see

They help more than u expect

When u internalize different examples

Ya thats true

Im actually bad with examples i should spend more time on examples than i do tbh

Anytime i put some energy into understanding some examples im like wow that was helpful actually

But at first it can feel tedious to look at

[thorn jay]

I was doing a project on Latin squares and wanted to prove something but couldn't do it. Then I worked out a single fairly trivial example disproving the theorem

Failed proofs can yield very interesting new insights though

[late matrix]

gulp

im very far from knowing what id wanna do from a phd but the stuff i like seems to be pretty competitive

[quiet pelican]

Spoiler alert: in pure maths, it’s all very competitive (although definitely some areas more than others)

[karmic moat]

this kinda thing depends heavily on the courses offered/who's teaching the algebra classes

like if one were to teach rings and modules using something like Atiyah-Macdonald, it would give students at least a slight sample of some alg geo if they choose to do the related exercises

some schools offer dedicated "intro to alg geo" courses meant for undergrads, which go over like basic examples of varieties and whatnot

so like mit offers this course to undergrads

https://math.mit.edu/classes/18.721/Fall2021/problemsets.html and https://math.mit.edu/classes/18.721/notes/ag-jan26-2022.pdf

you can see what kinda stuff they do here

that being said this is also mit

i don't think this kind of course is offered to undergraduate students at most schools (other than maybe the top of the top schools)

i guess this is a big roundabout answer for "not often, with a big asterisk"

(also you should not see the cracked young people like Arti or enpeace and generalize to conclude that a large proportion of students are this way/you are behind; these make up a very small proportion of students if you look at the larger dataset of every math student in the world. just places like a math discord server tend to attract these kind of ppl)

[late matrix]

rip

[thorn jay]

I need a place to neurodivergently talk about math 💔

[karmic moat]

scram

go back to crashing cars or whatever you high schoolers do

idk about your uni so i can't comment on that

i should also say that i'm saying this all in the context of US schooling

idk what it looks like in other countries

i guess since your name has "sweden" maybe you should disregard that last message. idk what it's like in sweden

[thorn jay]

Hey I graduated >:( I am at the doing drugs part
Uh, I think

[karmic moat]

though I do believe that the amount that math undergrads know is steadily increasing every year

because "undergrad research" has been and continues to be a really really huge big point of focus for undergrads

i mean if you think they're cracked then i believe you, since "advanced" means diff things for diff schools

[rocky cloak]

You studying in söta bror?

The land of tiny meatballs and furniture

Cool cool. I know a few people at Uppsala

Yeah, they're all a bunch of swedes

[thorn jay]

Someone's from the Netherlands :D
They're from urk D:

[karmic moat]

Wtf

That’s like 70% the reason I go to ikea

And ikea = authentic swedish food

[rocky cloak]

I hate going to IKEA, but the meatballs makes it worth it

Almost as good as Norwegian meatballs

Lingonberry sauce of course

Can't have meatballs without

[karmic moat]

Wtf

You people are mental

I once took a girl on a first date to Ikea

It was awesome

Because it’s awesome

[rocky cloak]

It's horrible. It's a constant struggle trying to convince my wife we don't need a new couch/bed/whatever

[karmic moat]

Thank god I dont have a wife yet amen 🙏

[thorn jay]

Dated yourself

Dated as in datum

[rocky cloak]

Lol

[tardy hedge]

Lol

Probably my favourite message ive opened the app too

On mathcord

Thats one for the books

[karmic moat]

https://x.com/nealagarwal/status/1950923269000823079?s=46

Neal Agarwal (@nealagarwal)

ikea captcha

X · Jul 31, 2025, 2:15 PM

[tardy hedge]

BRO HAS A WIFE ‼️

LORE DROP

[karmic moat]

Just saw this one

[thorn jay]

Of course, who could result the charm of the not jagr

[tardy hedge]

Ong I love learning about Jagr lore

[tough raven]

Let k be a commutative ring. Let k[t]^c denote the set of polynomials with non-zero constant term. There is a map ^symm: k[t]^c → k[t^{-1}]^c: p(t) ↦ p^symm(t) := t^{-deg(p)} p(t) (AKA "reverse the list of coefficients and substitute t^{-1} for t"). Let S be a finite subset of k[t]^c generating an ideal I. Is it true that (I : t) = I iff S^symm is a Gröbner basis for the ideal it generates in k[t^{-1}]?

Uhh, maybe let k be a field so Gröbner basis theory works.

[crystal vale]

In alternate way, I think they are construct a mapping D[Y,Z] -> D[Z], Y = 1( any non zero constant in D) and Z = Z

Right?

To show deg Y1 in D[Z] ≥ 1 when H ≠ 0

[tardy hedge]

Bro im already confused on why xy = x there

[karmic moat]

isn't it explained in the first line of the proof?

[crystal vale]

xy ≠ x, xy ≈ x, means xy and x are associate i.e, Rxy = Rx, where R is D[X,Y,Z]/< X-XYZ>

[minor gazelle]

is the proof of this theorem correct? it doesn't seem that hard though I kinda built up all the propositions that can be used for this proof... Still, is there anything missing?

the inverse map part I made a typo which should be $a\ker(f)$

[cloud walrus]

Emmaaaaa

[languid trellis]

I really don't see the isomorphism Z_5/5Z_5 \cong F_5. How the hell do we get that?

I understand that it should be a field because we're quotienting out by a maximal ideal, but why F_5?

I also see why 5Z_5 is a maximal ideal, because it is all nonunits of Z_5, and an ideal

[rocky cloak]

Well, do you see why the quotient should only have 5 elements?

0, 1, 2, 3, 4 gives a full set of representatives for the cosets. Maybe you can try to determine which coset 1/b should belong to

[languid trellis]

1/b is in k + 5Z_5 iff 1/b - k is in 5Z_5, i.e. 5 divides 1-kb

[rocky cloak]

Yupp, so say b is 1 modulo 5 for example

[languid trellis]

so, then 1-kb \equiv 1-k \equiv 0 mod 5, so we need k to be 1 mod 5?

Well we always need bk to be 1 mod 5 then

so 1/3 is in 2 + 5Z_5, which matches what I calculated by hand earlier

[rocky cloak]

Indeed. In fact similar reasoning shows that
R_m / mR_m = R/m for any maximal ideal m

[languid trellis]

Right, so there are 5 distinct cosets because given a/b in Z_(5), we are trying to solve the congruence bk = a (5) for k, and as we are working mod 5, k is either congruent to 0,1,2,3 or 4 mod 5, which gives all our cosets. And this then shows this is isomorphic to F_5 as there is a unique field with 5 elements

[languid trellis]

Are there any hypotheses on the ring R?

[rocky cloak]

Well, commutative I guess. Otherwise I don't know what R_m means

[languid trellis]

Right, and zorn tells us we can always find a maximal ideal

hmm. interesting

I can't wait for this to actually have a geometric meaning because so far this is far too formal for me

Thanks jagr!

[limber tapir]

The proof for one is wrong (or maybe just unclear) . The way you're defining the inverse map doesn't make much sense to me, for example the equality you state can't be true as the left side is in G and the right side in G'. In general, it is probably easier to just show f\tilde is surjective onto im(f) and injective. Either way though you need to explicitly state why there is a unique preimage for every element of im(f)

[minor gazelle]

I shall be fixing it now.. that’s my thought though.. it was a typo f^{-1}(f(a))=\tilde{f} (aker f) it should be a ker f right?

Then I only need to show surjection right? Since by how \tildef is constructed

Its domain is the quotient of G on Ker f so its injection and if I restrict the range of cosets to be the imf its surjection

[limber tapir]

No the map is surjective by construction as we restric to im(f) and every map is surjective onto its image. Injectivity is the thing you need to show.

[minor gazelle]

Yes!!! 🥰🥰

I only need to show kernel of it is trivial then I shall be fine? Right?

[limber tapir]

Maybe some result you have somewhere does this but I don't see a reference in the proof to such a resylt

[limber tapir]

For example, yes

[minor gazelle]

Thank you so much and algebra is so fun

[limber tapir]

Something like this is relatively simple, but even if it's clear to you it needs to be mentioned in the proof (if even just as "clearly the map is injective", but just writing down the reason is probably better)

[minor gazelle]

Yes 🥰🥰🥰

I started to like formalization because of this group theory, it’s really Lego playing and it’s like building a house but using proposition

[minor gazelle]

Will it suffices I argue the equivalence relations partition the space such that it’s pairwise disjoint hence the kernel must be trivial

[limber tapir]

This seems much more difficult than it needs to be. I'd just argue: if f(x) = f(x'), then f(x-x') = 0 so x-x' is in ker(f) so they are equal in the quotient, so the map is injective

Sorry I used additive notation

[minor gazelle]

🥰🥰❤️

[languid trellis]

I don't understand what the point of the first paragraph is. What is it trying to show?

[velvet hull]

U(Z/12Z) is not cyclic, for instance see below

so we know that U(Z/pZ) is cyclic, but proving that U(Z/p^lZ) is cyclic for all l is a nontrivial fact

[languid trellis]

Can you explain why you mention U(Z/12Z)? It's not a prime power, why is it relevant?

[velvet hull]

oh, I misunderstood your original question; the first paragraph is just an intermediate lemma for showing that certain primitive roots mod p can give you primitive roots mod p^l

not all primitive roots mod p are guranteed to be p.m.'s mod p^l, because if g^p-1 turns out to be 1 in both mod p and mod p^l then that's not a generator

so you explicitly need to rule out the possibility that pm's mod p "stop generating" the rest of the numbers when you move to mod p^l

[rocky cloak]

Say we want to show that U(Z/25) is cyclic.

Then we might guess that 7 is a cyclic generator, because it is a primitive root mod 5.

But alas 7^4 is 1 modulo 25, so 7 only has order 4. But 7+5=12 will then be a cyclic generator.

[languid trellis]

I see. we're showing the existence of a primitive root g mod p such that g^(p-1} is not 1 mod p^2. That's the point. And that such a g will also be a primitive root mod p^l.

Because then we can apply the corollary on the order of 1+ap

Okay I think I got it

Is this the standard way of showing U(Z/p^kZ) is cyclic?

[tardy hedge]

Where are these notes from?

[languid trellis]

It's reid UCA

[rocky cloak]

Pretty much I think.

At least lifting from a primitive root mod p to one mod p^2 is probably a key step in most proofs.

Then there might be a few ways to show you get a primitive root mod p^l from that

(I guess Idk what corollary 2 is xD)

[languid trellis]

Ah okay

I was wondering if there is some bigger theory that this result falls out of, because this all seems very unnatural tbh. Like Corollary 2 which states that if p \neq 2, and p doesnt divide a, then the order of 1+ap mod p^l is p^{l-1}

Perhaps i am asking for too much and I should shut up and calculate

[rocky cloak]

Well it's a special case of Hensels lemma

So that would be the general theory

[languid trellis]

Ah cool. Thanks

[rocky cloak]

Special case is maybe the wrong word, but related anyway

[tardy hedge]

looks friendly

quadratic integer ring?

i dont remember exactly that one but its probably some ring Z adjoined with some element

[languid trellis]

Yeah I think that's its intention. Just need to spend more time with it lol

[tardy hedge]

We need more friendly grad level math texts

y does everything have to turn so serious

ive been reading from this book called cohen macaulay rings by bruns and herzog and like 😢

[languid trellis]

that and pure algebra is hard and dry

[south patrol]

Algebra

[white oxide]

I'm trying to show that for $A$-modules $N$ and $M_i$, $N \otimes \bigoplus_{i \in I} M_i \cong \bigoplus_{i \in I} N \otimes M_i$. I've constructed a map $g: N \otimes \bigoplus_{i \in I} M_i \to \bigoplus_{i \in I} N \otimes M_i$ given by $n \otimes (m_i){i \in I} \mapsto (n \otimes m_i){i \in I}$. But I'm having a bit of trouble constructing an inverse for $g$. Any hints?

[cloud walrus]

okeyokay

[thorn jay]

If your text doesn't know how to motivate it then yeah lmao

[kind temple]

isn’t this LAPC?

[tardy hedge]

Whats LAPC?

[kind temple]

Left Adjoints Preserve Colimits

[tardy hedge]

Cool, i havent studied that yet

had some introduction to adjunctions in category theory but didnt have any intuition for it

[knotty badger]

it follows fairly directly from the homset definitions of left adjoint and colimit

[kind temple]

i have found that studying monads and comonads, along with some programming, gave me a little bit better intuition for adjoints

[tardy hedge]

Hmm

Yes I have heard cat theory is somehow present in programming but i also dk why

[kind temple]

types!

[south patrol]

Or just think about what linear maps out of each look likw in tsrma of multilinear maps

[tardy hedge]

Just the general theory of types ?

like data types?

[kind temple]

i’m still a learning all of this stuff, so somebody may have a better explanation, but type theory, category theory/categorical logic, and proof theory are really interconnected

[tardy hedge]

thats neat

[next obsidian]

This book is good

[tardy hedge]

ya why

[next obsidian]

I think this stuff is mostly a meme tbh

https://blog.plover.com/prog/burritos.html

[tardy hedge]

lol bruh

[tribal moss]

A burrito is a monoid in the [redacted]

[tardy hedge]

Lol

[knotty badger]

my intuition for adjunctions is bi-representable "heteromorphisms"/profunctors

[tardy hedge]

sounds nice but uh ...

[knotty badger]

like, you can consider "heteromorphisms" from a set X to a group G

defined to be functions from X to G, viewed as a set

they're called heteromorphisms cause they go between objects of two different categories, as opposed to "homomorphisms" which go between objects of a single category

luckily, though, you can "represent" these heteromorphisms as ordinary morphisms

In $\mathbf{Set}$, you can represent heteromorphisms from $X$ to $G$ as just ordinary set-functions from $X$ to $U(G)$, the underlying set of $G$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

(essentially by definition)

[thorn jay]

They got cishet in math before GTA vi 💔

[knotty badger]

In $\mathbf{Grp}$, you can represent heteromorphisms from $X$ to $G$ as just ordinary group homomorphisms $F(X) \to G$, where $F(X)$ is the free group on $X$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

adjunctions arise whenever you have "heteromorphisms" from objects of C to objects of D, which you can "represent" by both ordinary morphisms of C and ordinary morphisms of D

in this case C was Set, and D was Grp

[kind temple]

you start off with a set of atomic types, say bool, int, void, string, etc. you may also have some type constructors, like —> which allow you to form function types. mostly, you want to be able to group data together, so there are data constructors as well. one common one is cons and this allows (i’m still not actually clear on what cons is) you to form a memory unit containing one object in the left cell and another object in the right cell.

algebraic data types are sums (coproducts) and products of data types. one example of this is the list data type. a list of elements of type ‘a is either empty or it’s a cons ‘a (List ‘a).
so it’s a recursive sum type. constructing new data types like this, taking recursive sums and products of things, that’s just algebra.

the book category theory for programmers has some good exposition on this, much better than i can give

[dull ginkgo]

I am aware of the basic idea of the tensor transformation law for dyadic tensors, is there a general functorial description of this idea

Purely algebraic here, the case for the structures over the tangent bundle seem to be just locally defined by this property using the Invertible Jacobian

Like for the case of dyadic tensors, it seems to just be conjugation passing through the isomorphism between Hom(R^n, R^m) and R^n (x) R^m due to biproduct distributivity and R (x) R ~= Hom(R,R) ~= R

for commutative rings ofc

[tough raven]

The maximal ideal of this ring is generated by y, not x, right?

[next obsidian]

Sounds right

[minor gazelle]

Guys I refurbished my proof for First theorem of isomorphism, is it better now? It’s quite lengthy so I might have many slips

I didn’t want it to look messy so I made another lemma independently to show that tilde f is injective, because if I write it too messy it might obscure the main logic also risk myself being lost in symbols and this proof is already lengthy..

Oops I made a typo for 1.8 it should be $g’\ne g$, I fixed it on my note but it’s a routine proof using uniqueness to force equality

Another small mistake, should be equivalence classes maps to f(g) 🫣

[cloud walrus]

Emmaaaaa

[minor gazelle]

Rewrote

[limber tapir]

This doesn't work. If it did, you would have shown that f is injective, which isn't true in general. Instead, you need to show that if f(g) = f(g'), then f(g * g'^{-1}) = e' (which you did show), but then use this to conclude that g*g'^{-1} is in ker(f) and thus gker(f) = g'ker(f)

The conceptual idea is that modding out the kernel contracts things that map to the same element

[minor gazelle]

Got it I shall be fixing it🥰🥰

[regal mango]

I have a question in mind

Why is it like that if T:G->G' and that T^-1(G')=Gker{T} and not simply G ?

[thorn jay]

What

G • ker T = G

Assuming that T is a homomorphism

[limber tapir]

Did you do no exercises at all? If yes, then you should probably change that habit. It's extremely easy to feel like you understood everything without having understood it without doing exercises to test your knowledge

This might come back to bite you. if you have the time you should probably do some more. But this obviously depends on the person

[misty holly]

#1398993464338808872 message

[boreal condor]

yeah ur cooked

[regal mango]

If not atleast do the proof based problems or practice examples

[regal mango]

Bro is not cooked he is fried

[boreal condor]

think about it this way, out of all the questions the author could've asked, they chose a dozen that they thought would help you understand the material and broaden your mathematical skills. the exercises are core to learning material

[tardy hedge]

Preach it Brochacho 🙏

[karmic moat]

would u go to the gym and watch tutorials on how to do a bench press and say that's enough without touching the bench?

i can 95% guarantee this will come back and bite you in the ass

i did the same thing when i started doing math and it did come back to bite me

and sometimes it still nibbles

[tardy hedge]

How many exercises did u go back to do?

Did u go through everything in atiyah macdonald?

[karmic moat]

so basically i skimmed atiyah-macdonald and then went straight to hartshorne

and then my shit understanding of ring/modules kicked me in the ass

so i went back through atiyah-macdonald and did most of it

[thorn jay]

Haha L

What did you think was gonna happen

[karmic moat]

u are an L

at least my questions get answered B)

because people know what im talking about B)

anyhoo i was ambitious and young then

i.e. last year

[thorn jay]

Yurr

[thorn jay]

Oh noo commutative monoids so hard

[thorn jay]

I feel that

[karmic moat]

yeah now i am older and wiser

[thorn jay]

Until you do that stuff again 💔

[karmic moat]

who says i will ❤️‍🩹

[thorn jay]

I am a seer

[karmic moat]

go get a new crystal ball, your current one is broken

[tardy hedge]

Dramaaaaa!!

[thorn jay]

crash sound
Oh no, my balls, they're broken

[karmic moat]

saddest shit ive read all day. very emotional and moving

[thorn jay]

hoping someone somewhere ever gets that reference

[thorn jay]

I am a poet

I am endpiece mewsick

[tardy hedge]

Agreed

[glad osprey]

I'm struggling to understand how the domain of f can be defined in terms of f itself it feels circular, or is it well defined?

[kind temple]

lmao

[minor gazelle]

I’m rewriting it tomorrow all at once

[kind temple]

i think they meant ~f

[minor gazelle]

Indeed tilde f

[glad osprey]

I see

[minor gazelle]

It was a bit rushed

[kind temple]

recursively defined data types tho

[minor gazelle]

What’s this 🫣

[kind temple]

like how a list is either empty, or its an element (the head of the list) along with another list (the tail)

this was more of a meme

since the typo you made was kind of funny and had a vague resemblance to this

[minor gazelle]

That’s kinda how equivalence relations and how’s it partitioning everything right? Like you partitioned stuff inside a block one next to another

[minor gazelle]

True I was so happy with that I almost manage to get the result

It took me a while to get used to the portioning using equivalence classes, I used to do direct set manipulation🫣

[minor gazelle]

I agree! And there was a more funny type, Becasue I wasn’t used to lambda notation and here

I mapped it to f(gkerf) and felt it’s kinda coool😭😭😭

Though the partition using equivalence class is really powerful

Automatic disjointness

Gives you the whole space

Isomorphism

[tardy hedge]

You will go far if you keep up that enthusiasm

[minor gazelle]

Hopefully so🥰 and thanks.. 🎊🐈 with all the helps i received from people here ❤️

[tardy hedge]

@languid trellis that text is so friendly dude lmao

first section: "Hello!"

The author is just a bundle of joy 😍

[tough raven]

If your goal is to do well on an exam, that's even more reason to do the exercises. They are fre material to practise for the exam.

[languid trellis]

some might say that

[languid trellis]

the feeling I get is that algebra rarely develops 'on its own' as a subject. Rather other subjects, like number theory or geometry or topology lead the charge and we use algebra to formalise afterwards, which is why I feel like its unmotivated. Been doing lots of number theory recently though so this problem is sorta solving itself

Like we defined what a group was way after galois' death, for instance

I think noether was the first to define a module too, and LA was only formalised in the early 1900s from what I understand

but of course we had been solving systems of equations for eons

I might do some bedtime reading on the history of algebra actually

[minor gazelle]

@limber tapir I finished the refurnishing of my proof 🥰🥰🥰

And I am trying to do the second theorem (this lemma 1.9 is wrong but I’m fixing it)

It’s pretty addicting in a way it’s muscle building and you see you have more and more propositions and how you can use them to solve difficult problems with relative ease by citing what one has built. Though rather than second theorem of isomorphism it’s more a corollary I would say

And it’s also a quite journey to be able to shrink the length of each individual claims and only get detailed and focused when big one comes

[languid trellis]

don't worry you will start hating this god forsaken subject soon enough

[minor gazelle]

Please no I don’t want to hate it 😭😭😭

But it’s fair I mean I never really studied math formally so it might turn bad🫣

[delicate orchid]

GET ME OUT 🫨 🫨 🫨 🫨 🫨 🫨 🫨 🫨 🫨 🫨 🫨

[languid trellis]

don't worry. unemployment will come soon enough

[vital jackal]

how do you guys take notes like this?
I mean what kind of programs are you using in order to get something like this

[kind temple]

going to advocate for typst as a cool modern alternative

let go of your old fashioned ways

the future is here

lol

[minor gazelle]

I kinda finished the second though

A bit detour and m is really a typo there

I barely use pen for math despite I write heavily for my literature studies

[kind temple]

precisely, a subgroup N of G is normal in G if and only if there is a group H and a homomorphism G —> H having N as its kernel

i think there are similar theorems in abelian categories and lattices

@thorn jay may have something insightful to say about this

https://en.wikipedia.org/wiki/Isomorphism_theorems?wprov=sfti1#Universal_algebra

yea, universal algebra

i’ve never actually scrolled down that far on that wiki page

[minor gazelle]

Dp labor has statistics for salary of those with background in mathematics, median is 120k though

Ai boom make it flourish

[thorn jay]

Yes, an equivalence relation on a set A equipped with an algebraic structure is a congruence relation (analogue of normal subgroup) if and only if it is the kernel of a homomorphism

Oh right the isomorphism theorems yeah

Even H/(N ∩ H) ≈ HN/N has an analogue

[kind temple]

yea, that’s pretty neat

[thorn jay]

You can consider an algebraic structure A[θ] for any congruence θ such that the subuniverses of it are precisely the congruences contained in θ

If you consider a general version of exact sequences, then, when the congruences in A 3-permute you have the generalised short exact sequence
A[θ ∩ ψ] → A[ψ] → A/θ[(ψ ∨ θ)/θ]

[languid trellis]

you must be new here

mathematics being unemployable is just a meme people say lol, i didn't mean it seriosuly

[thorn jay]

Then, if we have some element 0 we can take the language of all operations f with f(0, ..., 0) = 0, and get the actual exact sequence
0/θ ∩ ψ → 0/ψ → 0/[(ψ ∨ θ)/θ]
Or, in the case that A is 0-regular and joins of algebras are the same as joins of congruences, we get the more familiar
I ∩ J → J → (I ∨ J)/I

[elfin wraith]

So real, I’m fighting off recruiters looking for people who are mediocre at ring theory

[minor gazelle]

The result you gave me islovely

I proved the second theorem again but much neater (if I didn’t make a ton of mistakes which is likely)

Though it requires more proposition to implement proposition 1.46

[minor gazelle]

Evil and I like it

[minor gazelle]

Is it like this?

Even it’s wrong I’m not gonna fix it more for today I will do a bit of la and hanging out with friends.. please criticize my work as detailed as possible, so I can fix them and improve 🥰🥰

I was initially offered to be at medical school

[thorn jay]

What I said has little to nothing directly to do with what youre

Very different contexts

[minor gazelle]

Since i attended gymnasium and good at Latin but a while later, realized that doctor doesn’t do that much and are overpaid for what they should got. I mean doctors never develop any novel therapeutic agent they do by memorizing more.. and you deal with tons of people too. And personally I don’t think people at medical school more moral as well

[minor gazelle]

My bad thought it got related 😭

[thorn jay]

Like, it's a high level analogue of what you're doing but not much mor3

[minor gazelle]

Math is too hard this little proof I did consumed me for a day.. can’t imagine higher level 😭

[thorn jay]

Wait until you learn how learning works

[minor gazelle]

Hopefully, I don’t have much confidence in myself but thanks. . I kinda feel it’s more interesting than analysis in many sense.

[thorn jay]

Welcome to the club

[minor gazelle]

I would rather be jobless than becoming Dr

I want it! 🥰🥰 though don’t think I have that brain power but yes.. not medical Dr

Medical Dr are pretty nobodies they never invented anything

Literally some very moral pretending people without having a lot of actual contribution to earth and be fair their jobs are likely replaced by AI and I’m gonna contribute to replacement of human medical doctors

[tardy hedge]

damn u popped off emma

[minor gazelle]

I hate medical doctors I feel they are quite often some of my old classmates went to med school.. hopefully they never become my doctor

[tardy hedge]

ya i mean i think it probably attracts a lot of ppl just for the status/prestige sort of thing

[minor gazelle]

Besides medical doctors barely do a lot though I visit medical school often since the med lib is closer.. they seem to read mainly Wikipedia styled textbooks for characteristics of diseases

I mean ai does that much better

[tardy hedge]

Lol yea

Yea i dunno i mean every once in a while you see a real smart doctor but yea a lot of them idkkk

[minor gazelle]

Yes they only do memorization and have tons of time to read certainly seem clever.. but whether you want them to do an operation on you that’s pretty crucial