groups-rings-fields

[fading field]

i use nvim which is a bit overkill, and i don't recommend that if you're worried about it being overwhelmign, but when I type preamblehomework it literally auto-completes to the whole preamble

[sonic coral]

that’s the one thing i told my undergraduate professors, is that i wish they taught me about overleaf to write proofs instead of word

it’s so much nicer and easier

[vapid vale]

learn latex lol

[steel light]

Maybe if I have more reasons than just math to use a proper TeX editor 🥴

[vapid vale]

it is very very easy

math is an enormous rreason!!

[sonic coral]

latex is very easy one you get a file from someone that has the formatting done for you

[vapid vale]

^

[fading field]

i mean that is the only reason you would ever do it, how long do you have left in your math career?

[sonic coral]

because the commands are the same shortcuts word uses

[fading field]

you won't use MS word forever

[steel light]

No but like I barely do math is what I mean, I'm an engineering student so all I need is very basic stuff that Word has been more than enough for. Math exercises with all the funny symbols are just a once-every-few-months/weeks kick when I'm on break and then I get busy with school again

[sonic coral]

i didn’t know latex existed until i got to graduate school and it was the best thing i’ve ever learned it’s so easy compared to word

[steel light]

I'll be starting a thesis soon enough hopefully though so maybe that will give me a bigger reason to learn TeX

[sonic coral]

i hate that i used word for all of undergrad lol

[cobalt heath]

Who knows though

[vapid vale]

learn it now so you'll know how to write your thesis lol

[steel light]

Blegh

[vapid vale]

it is genuinely surprisingly easy, just find someone elses template that has a bunch of useful math shortcuts

and things are very easy to google

plus most important it looks sexy

[crystal vale]

Characterize those positive integers n such that any Abelian group of order n is
cyclic.

My guess is n must be square free number

[fading field]

have you already learned the fundamental theorem of finite abelian groups?

[crystal vale]

Yes

[fading field]

the key here is that if m and n are relatively prime, then $Z_m \times Z_n = Z_{mn}$

[cloud walrus]

smay

[fading field]

if and only if actually

[crystal vale]

Yes

So if n is square free and G is Abelian group of order n then G is isomorphic to Z/nZ.

But let G is an abelian group of order n= n_1....n_k, if there is prime p such that p^2 | n.
Let n= p^2×m
Then take Group Z_p ×Z_p × Z_m then G is not cyclic because there is no element of order n.

[south patrol]

yes nice

[crystal vale]

Okay, thank you

[nocturne bone]

Can anyone explain why J/I = (5,0)?

Nevermind I figured it out right after posting haha

[crystal vale]

Let e and f are the orthogonal idempotent element in ring R. Then the intersection of left ideal generated by e and left ideal generated by f is trivial.

Let x in the intersection of left ideal of e and f.

Then x = ne + re, where n is an integer and r in R. Similarly, x = mf + sf.
Now xe = x but (mf +sf)e = 0. Hence x =0.

Is it correct?

[south patrol]

Hm im unsure why you're using n and r

You may as well just write x = re = sf

But then yes the proof is correct

x = xe = sfe = s0=0

[crystal vale]

Because R has no unity

[south patrol]

Okay sure

[crystal vale]

Finite dimensional vector spaces are Artinian.

Because, submodules are subspaces and if there are descending chain and S_(i+1) is a proper subset of S_(i) then it's dim(S_(i+1) is less than dim(S_i), right?

Then eventually it will stop at {e}.

Is it correct?

Is it necessary they will stop at {e, no ?

[south patrol]

Well it sort of depends how you word stuff

You can say like, if it decreases strictly then there can only be dim V inclusions at most

Another funny thing is that a direct sum of finitely many artinian modules is artinian. so you reduce to showing that k is artinian as a k-module ie an artinian ring. and that is obvious

lel

I guess I'm unsure what to say about "stop at {e}" because this is sort of a thing by contradiction

But sure, the maximal length chains do

[crystal vale]

I am talking about a minimal element but yes not necessarily {e} is a minimal element in chain

[long obsidian]

I found this term I haven't heard before.

Say (R,m) is a local ring and S the set of ideals of definition I of R such that there exists natural numbers n and k such that m^n \subset I \subset m^k

Does anybody have an example of such an ideal?

[crystal vale]

Let R be the set of real numbers and R be Q-module, then it is not Artinian and Noetherian module because R is infinite dimensional vector space

[crystal vale]

Proper subset?

[next obsidian]

This is just saying that I generates the m-adic topology. In a Noetherian ring this is the same as saying that sqrt(I) = m, so you can come up with examples using that

Take for example (x^2,y) in k[x,y]_(x,y)

[long obsidian]

Ohh okay thanks I'm gonna read more about the m-adic topology then but that the radical of the ideal is m kinda makes sense thank you for that example as well

[patent girder]

A bit of a silly question: I computed the size of G/H to be 3*2^k, where I have k+1 uses of third iso theorem. Ie I do like G/H1 * H1/H2 ... H_k-1/H and then I used the first iso thm to find the size on H_i/H_(i+1). I'm struggling a decent bit to find coset representatives; is there a like standard way to do it by reverse engineering the proof of those theorems.

[dim widget]

Seems like you can just take the representatives for each factor and take products

[patent girder]

Im struggling to find the representatives for $H_i/H_{i+1}$ which are like $\begin{bmatrix} \mathbb{Z}_2 & \mathbb{Z}_2 \ 2^i\mathbb{Z}_2 & \mathbb{Z}_2\end{bmatrix}$...would it just be like $\begin{bmatrix} 1 & \ 2^i & 1\end{bmatrix}$ and something else, but that something else isnt that apparent to me

[cloud walrus]

Zander

[patent girder]

Like the other thing I pick is the identity matrix right

But then I'm like confused

Actually I just can't count mb

[crystal vale]

Is it necessary that the short exact sequence only has this form ?

0-> M_1 -> M_2 -> M_3 ->0

[rocky cloak]

Not sure which short exact sequence you mean, when you say 'the short exact sequence', but that is the general shape of a short exact sequence yes.

[crystal vale]

If I have an exact sequence 0-> M_1 -> M_2 -> M_3 -> M_4 ->0
Then , is it short exact sequence? I am asking about the length of the sequence

[rocky cloak]

No, short exact sequences have by definition at most 3 nonzero terms

[crystal vale]

Okay that was my doubt, thank you

[limber sequoia]

how would you say conscisely "this fact is enough to decide ...."

example:

F is of characteristic 3, as the existence of one of the primal fields in it is enough to decide that

... to be certain that...

not sure how to say it, trying to say that being an extension of a prime field is iff to the relevant characteristic without saying this clumsy sentence

\mathbb{F}_{3} is naturally contained in F by 1\mapsto1, so F is of characteristic 3, as having a certain characteristic is iff to being an extension of the relevant prime field.

[dim widget]

It suffices to check that p = 0 in the prime field of R

this would be the normal way to say it

[limber sequoia]

I want to show the proffesor that I understand that this relation is iff

[hardy basalt]

Why do we call the axioms of groups "axioms"? What would change if we simply called them definitions?

or properties

To be able to call these properties axioms, shouldn't groups appear independently of these axioms in other contexts as well? Otherwise, why don't we just call them the definition or properties of groups?

[loud merlin]

It's an abuse of language but kinda harmless imo

[coral spindle]

The axioms of group theory are effectively the same as the axioms of set theory: we require groups to satisfy these axioms in the same way we require models of set theory to satisfy the axioms of set theory. There is no difference.

[coral spindle]

Dunno what this means.

[hardy basalt]

thankss

[tacit mauve]

Let $G$ be a group and $S$ a finite subset of $G$ that has a total order $\leq$ and such that $$\forall g\in G, \forall s\in S, gsg^{-1}\in S$$

[coral spindle]

The proposition you've written is incomplete. What about gsg^{-1}?

[tacit mauve]

Can you help me prove that for all $s_1, \dots , s_r \in S$, there exist $t_1,\dots ,t_r \in S$ such that $t_1\leq \dots \leq t_r$ and such that $s_1 \dots s_r = t_1 \dots t_r$

[cloud walrus]

TimourX

TimourX

[tacit mauve]

Yeah sry it's fixed now

I managed to prove it in the case $r=2$, but I'm stuck with higher $r$

[cloud walrus]

TimourX

[tacit mauve]

Don't hesitate to ping me if you have an idea of how to tackle it

[ornate tiger]

how to prove this is a subgroup for GL_3

i'm a newbie with abstract algebra

so first thing i'm going to multiply the 2 matrices

can i say that a = the matrix they gave us

but b = the same but instead of k i chose r

or what should i do

becuase they say k is in Z so how can i chose r

@spice gulch help ?

ok so what now ?

r+k in Z ?

yeah

ok so ig we need to check 3 things right

ab in Z

we need to check the Identity and the Inverses now right

[daring nova]

rsh

[ornate tiger]

yeah

ok wait a sec let me put what u said in my head i'm slow

why is a^-1 is in L we didn't find that we found that a matrix b is in L but finding this means that also a^-1 is in L ?

ok ok but why do we need to check if it's not empty

like what i know is to check Identity and inverse and closed under product or whatever

we didn't check if it's non empty

so we finished the proof

ok how to do that

oh ok cool i can just use symbolab

yeah cool we know that -K is also in Z

right ?

it's true

printf(" I'm smart ");

ok i have another question they asks us if it's abelian

i can see it's abelian when i do ab=ba

but the problem is we know that GL_3 is not abelian

so what can i say it's not abelian because GL_3 is not abelian even though that ab=ba satisfies

[lusty marlin]

A non-abelian group can have an abelian subgroup

[ornate tiger]

oh ok thx

[coral spindle]

A tree is not green, so how can its leaves be green?

[dull marsh]

[lusty marlin]

Whether a group is abelian or not can be determined by seeing whether ab=ba for all a and b in the group. Why does it matter whether that group is a subgroup of a non-abelian group or not?

[ornate tiger]

well uh not good enough because i can say i'm a human why my daughter is not a dog so it depends on the topic ig becuase in calculus if f and g are Continuous then f+g is also Continuous

[coral spindle]

You don't have to repeat yourself

[coral spindle]

The point here is that being Abelian or non-Abelian is determined by the scope: as KnightWatch says, it only matters what the group you're looking at is, not any overgroups

[ornate tiger]

yeah i'm talking with boy

[lusty marlin]

It depends on the definition of the property at hand, and you seem to not have understood what it means for a group to be abelian

[coral spindle]

Similarly to figure out whether or not a leaf is green you don't need to look at the entire tree :)

[lusty marlin]

And I'm talking to anybody and everybody who might have the same misconception as you

[hollow tartan]

AG wouldn't like your sentence

[coral spindle]

Fortunately he is dead so I am free to say true things.

[limber sequoia]

can you check my work for this exercise?

was trying to do it in a more abstract way and keeping it rigorous

[south patrol]

First bit is great

Second bit is correct but can be simplified very slightly

You can just say "the inclusion Z[sqrt(2)] -> Q(sqrt(2))" rather than giving a formula, and just remark that Q(sqrt(2)) is generated by sqrt(2) over Q by the earlier part to get surjectivity

[limber sequoia]

thank you !

[sly crescent]

Every group has an Abelian subgroup and a non-Abelian overgroup

[dull marsh]

Trivial group and Cayley's theorem moment (plus some small bits)

[sly crescent]

Trivial group and direct product with S₃ moment

[dull marsh]

Or that yeah

[frank cosmos]

what is the point of algebraic closure here

why cant we prove part 1 of thm by defining an isomorphism E to K to be the identity on k and send beta_i to alpha_i

and then by this logic shouldnt the second part apply for any extension of K, not just algebraic closure?

[chilly ocean]

Can anyone check my solution please? The question is as follows:

Let $G$ be a group and $H \leq G$. Let $gH$ be a left coset of $H$ in $G$. Prove that the set ${k^{-1} : k \in gH}$ is a right coset of $H$ in $G$. Deduce that, if the number $n$ of left cosets of $H$ in $G$ is finite, then there are exactly $n$ right cosets of $H$ in $G$.

My solution is: Given $k^{-1}$, this is the same as $h^{-1} g^{-1}$, by the definition of the left coset $gH = {gh : h \in H}$. So,
$${h^{-1} g^{-1} : h \in H} = {hg^{-1} : h \in H} = {k : k \in Hg^{-1}} = Hg^{-1}$$

Since $gH \mapsto Hg^{-1}$, and $Hg \mapsto g^{-1} H$, since this is invertible (and thus a bijection), the cardinalities of the sets are equal. We conclude that $|gH| = n = |Hg^{-1}|$

[cloud walrus]

45

[chilly ocean]

oh yeah, and $h^{-1}$ "changes" to $h$ since for every$ h^{-1$} there exists a corresponding $h \in H$

[cloud walrus]

45
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly ocean]

just realized this talks about the cardinalities of gH and Hg^{-1} being equal, rather than the number of cosets... oops

[dull ginkgo]

No,No

This is good

Think,

There is a bijection between cosets given by multiplication by an element… so they all have the same size

[chilly ocean]

Could you elaborate please?

[dull ginkgo]

There is a bijection between $aH$ and $bH$ given by multiplying elements of the first by $ba^{-1}$. So any two cosets have the same size, and they are distinct. And you showed cosets of the “other side” have the same size, and are in fact 1to1 with cosets on the “other side”

[cloud walrus]

Miz

[dull ginkgo]

So there must be the same number

You’ve shown a stronger statement actually lol

[chilly ocean]

I did that completely unintentionally lmao

Just thought a bijection could show that they’re the same cardinality haha

[dull ginkgo]

There’s a bijection between the set of left cosets and the set of right cosets even if there’s infinitely many

[chilly ocean]

So I did the question correctly? No mistakes?

[dull ginkgo]

If it’s finitely many then bijection implies sets have the same size

[wraith swan]

I got really confused. First everything was just functions on sets, you add an operation to get a group. There exist special functions called homomorphisms. Then you look the cat Grp and it's all homomorphisms. And now I can't wrap my head around "functions" from a set to a group any more...

When someone writes Hom_Set(A, G) where G is a group, is that actually Hom_Grp(F(A), G)? or does the free group not work like that? Nothing makes sense any more 😛

[lusty marlin]

Hom_Set(A,U(G)) for a group G is the Hom-set of functions from A to the underlying set of G (U being the forgetful functor here)

While Hom_Grp(F(A),G) is the Hom-set of group homomorphisms from F(A) to G

[wraith swan]

Are they isomorphic?

[tribal moss]

Yes, due to the free and forgetful functors being adjoint.

(Or put another way, the natural isomorphism between those two hom-sets is what make the free group "free" in the first place).

[wraith swan]

Ah ok, cool. Sorry the book doesn't go into this. This was just me thinking about these functions

[wraith swan]

Oh it's denoted F(A) as F is the free functor 😛

[crystal vale]

In the book they define local ring is a ring with only one maximal ring.

Now I want to prove that a ring is local if and only if set of all non-units is an ideal.

Let the set of all non -units is an ideal.
Let this set be S.

Now S will be maximal, because if there an element a in R-S , it is unit.

And if there is distinct maximal ideal then it contains an element a such that a not in S. Then it imply a is unit then it contradicts it's maximality.

Now let R be a local ring with maximal ideal m. And let S be the set of all non-units.

We know that for (a)≠(1), there is maximal ideal such that (a) contained in it.

But there is only one maximal ideal. Thus all non-units are in m.

And every element of m is non-unit.
Thus, S = m implies that S is an ideal.

Is it correct?

[dull ginkgo]

yeah

Units in general are the elements that don't lie in any proper ideal

Good for characterizing some types of rings

[frank cosmos]

bump

[crystal vale]

Yes

[crystal vale]

Okay, thank you

[dull ginkgo]

It’s also good for characterizing units in localizations or quotients

[crystal vale]

Let A be a ring and a_1,...,a_n ideals of A. Define a homomorphism f between A and direct product of A/a_i such that x->(x +a_1,..., x + a_n ).

Then I want to prove that if a_i and a_j are co prime for i≠j then f is surjective.

If I have f(x) = (1,0,..., 0) then for (r, 0, 0,...., 0) how can I find y such that f(y) = (r, 0,....., 0)

[rocky cloak]

Being coprime means that
a1 + a2 = A. That means that there exists an element in a2 that is 1 modulo a1. Argue similarly for the other ideals and use that 1*1 = 1

[rocky cloak]

I don't know if you're assuming your rings to be commutative, but you have to be a little bit more careful in the noncommutative case.

As an element can have a left inverse without being a unit, so it's not necessarily the case that every nonunit lives in a maximal ideal.

[crystal vale]

Yes, I used a commutative ring with unity

[crystal vale]

But my doubt is if I proved that there exists x_1 such that f(x_i)= (0,0,0,.,1 ,.,0) ( 1 in ith coordinate) then f is surjective

[rocky cloak]

Yeah, that's right.

Just try to see how you could construct a general element (r1, ..., rn) from such xi

[crystal vale]

(r_1,...., r_n) = r_1(1, 0,....., 0) + r_2(0,1,...., 0) +....+r_n(0,......,1)

But I don't have any idea how to find y such that f(y) = (r_1,..., r_n )

[tacit mauve]

Let $G$ be a group and $S$ a finite subset of $G$ that has a total order $\leq$ and such that $$\forall g\in G, \forall s\in S, gsg^{-1}\in S$$
Can you help me prove that for all $s_1, \dots , s_r \in S$, there exist $t_1,\dots ,t_r \in S$ such that $t_1\leq \dots \leq t_r$ and such that $s_1 \dots s_r = t_1 \dots t_r$

[cloud walrus]

TimourX

[tacit mauve]

I only managed to prove it in the case r=2

[rocky cloak]

But you just did....
r1x1 + r2x2 ....

[crystal vale]

Yes but f(ra ) ≠ rf(a) , right?

[rocky cloak]

f(ra) = rf(a)

[crystal vale]

How?

Got it

I thought in general it is not true, right that if f is homomorphism then g(ra)≠rg(a)

But here f is satisfy thus so it works

[rocky cloak]

I guess it depends homomorphism of what.

True for homomorphisms of R-algebras for example, like f is in this example.

[crystal vale]

Yes

Thank you

[errant wedge]

dont need help but are the first two literally just bash, or will i learn smth or am i missing something that makes it immediate?

dont tell me what it is if thats the case, just bash or no

[cobalt heath]

Is U_n unit group of Z/nZ

[errant wedge]

yes

[cobalt heath]

You can likely learn something

[errant wedge]

ok

thanks :)

[cobalt heath]

Good luck!

[rocky cloak]

I guess the broader theme to be learned is "when is Un cyclic"

[knotty wing]

hi guys i asked this in a help channel couple hours back and got no responses 😭 im gonna paste my qn here

what's an example of a subset of R2 such that it is not equal to the algebraic variety of the ideal generated by it in R[x,y]? i might be confused about the concept but i really cant think of the example here

[viral mountain]

In an ordered field, I'm able to prove that if $x > 0$ and $xy > 0$, then $y > 0$ by showing that $y \leq 0$ is impossible (i.e., by contradiction).

But is there a more direct way to prove that $y > 0$?

[cloud walrus]

skeptician

[dull marsh]

If you've shown x > 0 implies 1/x > 0, then you could use that

[knotty wing]

i remember the way that my textbook defined an order was to take a subset P of F and call it positive, and this subset is closed by definition, if u have some sort of similar definition already then maybe u can approach this by the defintion?

[dull marsh]

Ah that result is by contradiction though I think

[knotty wing]

and ya u will maybe use the 1/x >0 too

[viral mountain]

yeah but showing that $1/x > 0$ is the exact reason I'm first proving $x > 0$ and $xy > 0$ implies $y > 0$ 😅

[cloud walrus]

skeptician

[dull marsh]

Yeah you will have to do it by contradiction then I think

[rocky cloak]

Just any set that isn't an algebraic set. Like the set of points (n, 0), n ranging over all integers for example.

[errant wedge]

I dont feel like I learned anything

7 is a generator for U_22

U_15 has no (single) ones

[knotty wing]

what makes this not an algebraic set? my textbook doesnt really cover this

[rocky cloak]

Something that isn't the vanishing set of any set of polynomials.

Perhaps easier example would be {x : |x| < 1}. This is an open set, but the vanishing of a polynomial is always closed.

[rocky cloak]

Well, I guess you have to ask yourself why U15 isn't cyclic. What does it look like, what distinguishes 15 from 22.

It might be more insightful if you look at more examples...

[hollow tartan]

I think the takeaway is chinese remainder theorem

[errant wedge]

hrmmm

[rocky cloak]

That is certainly very relevant

[knotty wing]

this is an interesting example, and i see why here, but how come (n,0) is not a vanishing set?

[rocky cloak]

Well, try to determine which polynomials vanish on this set. Do these polynomials vanish only there?

[crystal vale]

let R be the commutative ring with unity, now let f = a_0 + a_1x +...... + a_nx^n be nilpotent in R[x]. I want to prove that a_i is nilpotent elemnet in R.
we can deduce that a_n is nilpotent in R because f is nilpotent.
then f - a_nx^n is also nilpotent element in R[x] since nilpotent set is an ideal.
Then we will get all a_i is nilpotent.

is it correct?

[knotty wing]

oh right because they will vanish for non integral values too right?

[dire siren]

yes

[noble lynx]

what is the intuition for R^n = R^m for n \neq m as modules for non commutative rings

[rocky cloak]

The basic idea is that if e is an idempotent of R, then R breaks as a direct sum
R = Re (+) R(1-e)

Now in the commutative setting Re and R(1-e) are never isomorphic, but in the noncommutative case this can happen (just look at the ring of nxn matrices for example).

So if R breaks up into an infinite product of things, that are all isomorphic, then R^n will still be that same infinite product

[wraith swan]

I read a small part of the book about the Abelian free group on A and decided to try to figure it out myself. Since every element commutes, it is enough to count the occurrence of every element in A. So I thought about the functions A -> Z and then the group Hom(A, Z) with the additive operation on the codomain, which is abelian as well, this was mentioned in a previous paragraph.

What I do not understand is that when trying to prove that this is the free ab group you get into problems constructing a homomorphism, since for example the function x ↦ 1 cannot be written as finitely many injections i : A to Hom(A, Z).

But that Hom(A, Z) is itself an abelian group

[coral spindle]

for example the function x ↦ 1 cannot be written as finitely many injections i : A to Hom(A, Z).
I don't understand what you're saying here

[wraith swan]

Like why is (x ↦ 1) an element of the group Hom(A, Z) while there doesnt not exist an element in the abelian free group with inifinitely many elements from A?

[coral spindle]

Oh, I see.

That's because Hom(A, Z) is not the Abelian free group on A, typically. It's too big.

Try a subgroup.

[wraith swan]

Right, I get that I just have to limit F^Ab(A) only to functions that map finitely many elements of A to elements other than 0

But why can the free group not consists of them while another abelian group Hom(A, Z) can contains for examplle (x ↦ 1)?

[coral spindle]

Groups do not support infinite sums.

[wraith swan]

Or is that because of the limitations set by the canonical injection?

[coral spindle]

Remember the universal property

[coral spindle]

Let's say I have an alphabet A = { a_1, a_2, ...} that I build G = Hom(A, Z) out of

And as you say let's consider the element i which is just constantly 1.

Now I choose a map (of sets) A → Z given by, again, constantly 1.

So there should be a corresponding group homomorphism. Where does it send i?

Your conclusion should be that no such thing exists. We cannot add together infinitely many elements in a group, since we are only given a finitary operation.

[wraith swan]

Sum_i g(i), but as you said that does not exist

So what makes it that an infinite sum does not exist but an infinite product does? Since Hom(A, Z) is isomorphic to Prod_{a \in A} Z, right?

[coral spindle]

That's not an infinite product inside the group.

That's just a new set that we made.

[wraith swan]

oh right uh I mean Z^A = Z x Z x ... x Z

[coral spindle]

Again, that's just a new set we made

The thing you wrote on the right hand side is an abuse of notation, but nonetheless there's no problem.

[untold basalt]

something that came up in Sylow's theorems

[wraith swan]

Can't I have a universal property indexed over some infinite countable set? Such that

[coral spindle]

Yes of course

[wraith swan]

And what about the direct sum?

[coral spindle]

That sure is a thing

I'm not sure what your question is.

[wraith swan]

Me neither 😛

[slim kayak]

If they exist they have that universal property, no matter what your family is indexed over

[wraith swan]

Really confused now, I also read that the direct product and sum are isomorphic but that the infinite direct sum and product are not.
And I think the infinite direct sum is isomorphic to the free abelian group...

[slim kayak]

free abelian group

There are coproducts in Grp but the situation is a little bit messier iirc

In Ab the difference comes down to considering the elements of each: elements of a direct sum are finite formal linear combinations and elements of the direct sum are tuples

[wraith swan]

And the direct product can be an infinite tuple?

[slim kayak]

yup!

after all the pointwise group operation gives it the right structure for the universal property

[wraith swan]

right

[slim kayak]

For the universal property of the direct sum you can only send in elements from each individual group in the diagram and add finitely of these

but there is no such thing as infinite products/sums inside a single group, so the universal property can only care about finite sums

[wraith swan]

This assymetry in the duality is weird

[slim kayak]

(forgot the negative in the statement, fixed)

[noble lynx]

for this, can we just say submodules of D^n are direct sums of ideals and since they are principal, just taking the generators gives you a basis for the submodule?

[wraith swan]

Well thanks Boytjie and Kerr! I'm going to contemplate while taking a walk, but I don't think all the pieces will fall for me today 😛

[slim kayak]

Sure, how do you know they are direct sums of ideals tho?

The proof I have seen in my courses was a lot more complex, so I am pretty sure there is some more subtlety required to conclude that submodules over a PID do take that form

[noble lynx]

yeah actually I dont think direct sum of ideals is as immediate to prove as I thought it would be

[crystal vale]

Let R be a commutative ring with unity.

Then nilradical is a subset of Jacobson radical, right?

For any x in nilradical, x is nilpotent element then for any y, xy also nilpotent element.

Then 1 + xy is a unit in R.
And I have one result that x in Jacobson radical if and only if 1 -xy is a unit for all y in R.

Hence, x in Jacobson radical.

Is it correct?

[chilly ocean]

yes

Or since the jacobson radical is the intersection of all maximal ideals, while the nilradical is the intersection of all prime ideals

[south patrol]

Yes

[chilly ocean]

a question about rigor; for small groups, such as $C_2 \times C_3$ and $C_6$, if you're asked to check if they're isomorphic, would it suffice to draw a multiplication table and indicate that these are structurally "the same"? Does it depend?

[cloud walrus]

45

[south patrol]

I would just write down an explicit isomorphism and that is bound to be faster anyway

[slim kayak]

Very much rigorous, but its still preferable to think about other ways of showing why they would be isomorphic which can generalize to problems where group tables just arent tractable at all

[south patrol]

Idk if it is really rigorous, in that you have to indicate what the isomorphism is

but then once you get to that point you have written down the isomorphism anyway

and just checked that the function you propose is a homomorphism in a very inefficient way

[wraith swan]

I think I get it 😄 Only half way my walk.. Thanks!

[crystal vale]

yes

[noble lynx]

hmm, I think I got another proof that seems a bit too easy to be correct. R^n decomposes as the direct sum of R^(n-1) and R^n/R^(n-1) and then by induction, R^(n-1) is free and use the fact that R^n/R^(n-1) is cyclic?

[slim kayak]

I am not sure what that is trying to prove?

[noble lynx]

well if you know R^(n-1) and R^n/R^(n-1) is free then every sub module of the direct sum decompose as the direct sum of sub modules of R^(n-1) and R^n/R^(n-1) and then just use the inductive hypothesis (since we know free module of rank < n has free sub modules)

[chilly ocean]

gotcha, thanks a bunch!

[south patrol]

Np

Here a special thing to look up is the Chinese remainder theorem

That generalises this

[slim kayak]

Perhaps consider if the direct sum/submodule argument works when R is a field. R^2 is the direct sum of R with itself, so the vector space generated by (1,1) ought to be some direct sum. Is it tho?

[noble lynx]

I mean that subspace would be isomorphic to R?

(1, 1)R = R \oplus 0

[chilly ocean]

Yes that's basically how you can prove it. The case for n = 1 is clear by definition. Then you choose a basis of n elements of M and then define M' to be the module generated by the first n-1 basis elements, and then you get a short exact sequence 0->M'->M->R->0. (where the map on the right is the projection on the last coordinate, call pi_n)Then if N is a submodule of M you get a induced ses
0->M' \cap N -> N -> pi_n(N) -> 0 where the right is a submodule of R, so it's free by induction, and the module on the left is free by induction as well. The sequence splits as the right module is free

I think that works

(i hope)

[slim kayak]

Sure, but for that you have to find a basis of the original space so that the direct sum you wrote can be interpreted as an internal direct sum

[noble lynx]

hmm, what would be wrong with the external direct sum in this case?

[slim kayak]

Well, to show that it is free submodule is effectively just asking if you can find a basis which then generate the submodule internally

[noble lynx]

freeness is preserved by isomorphism, so shouldnt I be okay with just proving an isomorphic M is free? unless you want an explicit basis

[slim kayak]

I am mostly concerned with that the strength of the statement "every submodule of a free module is a direct sum of ideals" is as strong or stronger than the theorem you are trying to prove

[noble lynx]

it is stronger and yeah I dont think internal direct sum works in that statement

[noble lynx]

I havent learned exact sequences but I'll make sure to come back to this when I do thanks

[chilly ocean]

oh

[noble lynx]

*also I think my statement about ideal is false

[chilly ocean]

What exactly?

[chilly ocean]

I think you do need some kind of splitting statement, otherwise you don't know that it actually decomposes as a direct sum though

[noble lynx]

"every submodule of a free module is a direct sum of ideals"

wait nm, I think it is equivalent in a PID now

[chilly ocean]

this doesn't really make sense

Like not as internal direct sum

But I mean sure since (1) is an ideal which is just the ring

[tribal moss]

But it would make sense to say "every submodule of a free module is isomorphic to an external direct sum of ideals".
I'm not sure about when that is true, though.

[chilly ocean]

Isn't that always true?

[noble lynx]

in a PID I think

[chilly ocean]

Since R is an ideal

Like any free module is just isomorphic to a direct sum of R

Which is an ideal

[south patrol]

It's true over a PID - in general submodules of free modules needn't be free otherwise

[tribal moss]

It's obviously true for a free module itself, but that is not the same as saying it's true about its submodules too.

[chilly ocean]

Oh yeah

I mean thats what we proved before

Like in a PID

[noble lynx]

btw, when I wrote my original question I did think it was internal, but then realized it just wasnt true

[slim kayak]

Checking my notes from a previous course the proof is essentially the same as nine's one, just without the language of exact sequences

It does however require checking if the basis you construct that way is linearly independent

[crystal vale]

Let a be an ideal, a ≠(1) in a ring A. Then if a is intersection of prime ideal then a = r(a).

Since, a is a subset of r(a).
Now let x in r(a) then x^n in a for some positive integer n.

Since a is intersection of prime ideal, x^n in all prime ideals implies that x in all prime ideal.

Hence, x in a imply that r(a) is a subset of a.

Thus, a = r(a)

[south patrol]

Yes, nice

You can also deduce it by another method: you can check thaat $r(I \cap J) = r(I) \cap r(J)$ and $r(\mathfrak p) = \mathfrak p$ for any prime $\mathfrak p$. Try proving it using these two facts

[cloud walrus]

Homological Potato

[chilly ocean]

I don't think this holds for infinite intersections though (I mean that radicals commute only with finite intersections, so it'll be difficult to use that)

[south patrol]

Ah good point, I assumed we had finitely many primes but that wasn't stated

[crystal vale]

Okay but any hint how we can prove that if a = r(a) then a is intersection of prime ideals

[chilly ocean]

You can define the radical as an intersection of prime ideals

Do you know that the nilradical is the intersection of all prime ideals?

[crystal vale]

Yes

[chilly ocean]

Then look at the ring A/a

And then by the correspondence of prime ideals in A which contain a and prime ideals in A/a, you'll get a characterisation of r(a) as an intersection of prime ideals

[crystal vale]

But then r(a) is the intersection of prime ideal which contains a, right?

[chilly ocean]

Yes

[crystal vale]

But they didn't mention that the prime ideal must contain a

[chilly ocean]

I don't understand what you mean

[chilly ocean]

As you said, this will tell you that r(a) is the same as the intersection of all prime ideals which contain a, so if a=r(a) then...?

[crystal vale]

Then a is intersection of prime ideal which contains a but I want to prove that a is a intersection of prime ideals, are they want to intersection of some prime ideals or all prime ideals?

[chilly ocean]

If you know that a is an intersection of prime ideals, then it is already the intersection of all prime ideals which contain a

It's the same thing

[crystal vale]

I know a is intersection of all prime ideals which contain a, how can I show a is intersection of all prime ideals

[chilly ocean]

It's not

[crystal vale]

But in question they want a is intersection of prime ideal

[chilly ocean]

Yes, of some prime ideals

Not all

[crystal vale]

Okay

[chilly ocean]

You have p=r(p) for all prime ideals, but p isn't the intersection of all prime ideals

[crystal vale]

Yes got it

Thanks a lot

[chilly ocean]

no problem

[barren sierra]

surely this is a mistake right? Consider $p = 2$. Then $\frac{3}{5}, \frac{10}{7} \in R$ but $\frac{3 \cdot 10}{5 \cdot 7} \notin R$ as $\gcd(30, 35) = 5 \neq 1$.

[cloud walrus]

Spamakin🎷

[barren sierra]

(I just need a sanity check lol)

but what would even be the intended problem

[lusty marlin]

I think they meant p doesn't divide b when a/b is written in its reduced form

30/35 = 6/7

[coral spindle]

Indeed they do say that

But it really doesn't matter anyway, because 30/35 = 6/7 no matter what, and 6/7 is certainly in R.

[barren sierra]

ohhh they meant in reduced form

I see I see

[coral spindle]

In reduced form is the same as saying gcd(a,b) = 1

Modulo signs, I suppose.

[barren sierra]

yea true I was being dumb

[coral spindle]

No I think it's a minor abuse of set-builder notation, I think it's understandable

really what they mean is:

$\build{a/b}{a, b \in \bZ,\ \text{etc etc}}$

[cloud walrus]

Boytjie

[fading field]

wtf cursed TeX

is build a custom command

[coral spindle]

ye

[fading field]

i am totally gonna steal it though

[coral spindle]

Bc in some texts I want to use | as a separator and others I want :, so I make commands for generators, set-builder notation etc that uses it so I can just redefine it when I want

Saves a lot of time

why write \left\langle blah \mid blah \right\rangle when you could write \buildgen{blah}{blah}

[worn spindle]

Damn I need to get your sty file XD

[coral spindle]

Sure I'll put my must-haves in #latex-help

[barren sierra]

I have similar commands

Well I think some package gives me \set for this stuff

And then I have a command that give me ideals / generated stuff

And even a command that gives me probability's and expected values with the vertical bar

But in all of these the vertical bar I actually type out with | if I want it and it autoscales

https://github.com/Spamakin/templates/blob/main/hw/style.sty

lines 442-505 have the relevant stuff (and some other nice code for permutations)

[noble lynx]

when a book says "find a canonical mapping" does that just mean to find the most "obvious" one?

[next obsidian]

Basically

[crystal vale]

i want to prove that in a ring A[x], Jacobson radical is same as nilradical.
i have proved that nilradical is same as Jacobson radical.
now let f(x) in Jacobson radical, f(x) = a_0 + a_1x + a_2x^2+ .... + a_nx^n.

since if R is a ring and x in Jacobson radical then 1+xy is a unit for all y in R.
we can use this result here, then let y = x, then 1+ xf(x) is a unit implies that a_i is nilpotent element for all i = 0,1,2,..., n.
Hence, f(x) is nilpotent element in A[x].

thus, Jacobson radical is equal to nilradical

is it correct ?

[south patrol]

Yes, nice

[crystal vale]

Let R be a commutative ring with unity.
R has only one prime ideal.

How can I show that x in R then x is unity or x is nilpotent.

I observe that it implies that R has only one maximal ideal, thus R is a local ring.

Then if x in R, then x is unit or 1+x is unit.

Any hint?

[rocky cloak]

Consider the ideals that don't contain x^n, and pick a maximal one among them (use Zorn's lemma for example). Then show that this is prime.

[chilly ocean]

There exists a group $G$ of order $20$, generated by two elements $x$ and $y$ such that $x^4 = y^5 = 1$ and $xyx^{-1} = y^2$. Prove that every element of $G$ can be uniquely written as $y^b x^a$ with $a \in \mbb{Z}_4, b \in \mbb{Z}_5$.

[cloud walrus]

autodidact

[chilly ocean]

!status

[flat tree]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[chilly ocean]

1

like, I know this is true because we can use that relation, but how?

should I write out an arbitrary "word" (word meaning string of elements)? but this doesn't prove it for the general case

[crystal vale]

What if I show that G is isomorphic to Z_4 × Z_5

[rocky cloak]

Yes, you can do this. Then you might want to use induction on the length of the word.

A second approach that might be cheating: since you're given that the group is of order 20, it's enough to show that y^b x^a are all distinct. But I guess this goes against the spirit of the exercise

[chilly ocean]

that's so smart

but how do I make it arbitrary and generalize it?

[dull marsh]

That doesn't seem to be the case though

[chilly ocean]

Let $g \in G$. Then $G = \langle y^b x^a \rangle$ is what we need to prove? Or am I getting it wrong?

[cloud walrus]

autodidact

[chilly ocean]

no... that's not it

[rocky cloak]

Like by definition, if a group is generated by x and y, then every element is a word in x, y (and x^-1 and y^-1)

Then factor out some ys on the left and xs on the right, and use the relation to simplify whats in the middle

[chilly ocean]

yeah, that makes sense, but isn't choosing the x's and y's and x^{-1}'s and y^{-1}'s to be in certain positions going to change what element it is? how do we generalize it to any g \in G?

[rocky cloak]

Well, just write it out generally.

Like any element is of the form
x^n1 y^n2 x^n3 ... y^nm

For some integers n1, ..., nm

[chilly ocean]

$x^{n_1} y^{n_2} x^{n_3} \hdots y^{n_m}$?

[cloud walrus]

autodidact

[chilly ocean]

and then induction on m

[rocky cloak]

Yeah, that would work

[crystal vale]

but if every element of G can be uniquely written as y^b x^a, a in Z/4Z and b in Z/5Z, then what if we define mapping y^b x^a -> (b,a) ?

[rocky cloak]

Have you checked that this is a homomorphism? (It isn't)

[dull marsh]

That's only a bijection

[chilly ocean]

thank you!

[crystal vale]

so we need commutative to make homomorphsim ?

[dull marsh]

For G to be isomorphic to Z4 x Z5, you would need G to be abelian, does the given information about G allow it to be commutative?

[crystal vale]

yes this one is used in proof of nilradical is intersection of prime ideals

[crystal vale]

i thought somehow we can make Abelian by given relation, but it is not possible

[dull marsh]

Right, so G is not a direct product of Z4 and Z5

But it is a semi-direct product

[crystal vale]

i don't see any conclusion from it, yes but every element of prime ideal is nilpotent

[rocky cloak]

The conclusion being that if x is not nilpotent, it's not in the ideal hence a unit

[crystal vale]

but x not in that prime ideal how it implies that x is a unit

[rocky cloak]

Every nonunit is contained in a maximal ideal

[crystal vale]

yes but is that prime ideal is same as maximal ideal?

[rocky cloak]

Maximal ideals are prime

[crystal vale]

okay got it

thank you

[crystal vale]

So it implies that if N is nilradical then A/N is a field

Now if A/N is a field then I want to prove that there is only one prime ideal.

Since A/N is a field implies that N is a maximal ideal.
But N is the intersection of all prime ideals.

So N contained in all prime ideals.
Let p is a prime ideal such that p≠N then p=(1) because of maximality of N.

Hence there is only one prime ideal, N.

Is it correct?

[crystal vale]

To show that, if a and b are ideal in a Ring A, and for any set V(E) denotes the set of all prime ideals which contains E.

Then V( a intersections b) = union of V(a) and V(b).

First we prove for if p in V( a intersection b) then p in union of V(a) and V(b).

Let p not in union of V(a) and V(b).

Then there exists a_1 in a and b_1 in b such that a_1 not in p and b_1 not in p.

But a_1b_1 in a intersection b implies that a_1b_1 in p, which contradiction.
Hence, p in union of V(a) and V(b).

Now let p in the union of V(a) and V(b).

Let p in V(a), since a intersection b is a subset of a implies that p in V( a and b).
Similarly if p in V(b).

Hence, V(a intersection b) is same as union of V(a) and V(b).

Is it correct?

[crystal vale]

what does it mean by ideals of ring A forms complete lattice with respect to inclusion?

[tribal moss]

Are you asking what "complete lattice" means?

[crystal vale]

yes, i think it is related to discrete math, set theory but what is relation between lattice and ideals of ring

[tribal moss]

A complete lattice is a partially ordered set in which every subset has a greatest lower bound and a least upper bound.
Here, we consider all ideals of the ring to be partially ordered by the "subset of" relation.

[crystal vale]

so if we take set of some ideals then there will be greatest lower bound and least upper bound

[tribal moss]

Yes.

[crystal vale]

intersection of all such ideals in a set will be greatest lower bound?

[tribal moss]

Yes.

[crystal vale]

but what will be least upper bound of set?

[tribal moss]

The set of finite sums of elements of some of the ideals will itself be an ideal, and it is clearly the smallest ideals that contains all the ones you started with.

In more abstract terms we can also construct it as "the intersection of every ideal that contains all of my chosen ideals".

(In fact, it is a general theorem of partial orders that if every subset has a glb, every subset also has a lub -- and vice versa).

[crystal vale]

yes that will be upper bound of a set

[crystal vale]

then intersection is a smallest ideal which contains all ideals which are in a set

[tribal moss]

Yes.

(Though this reasoning is subtle enough that we really ought to start giving names to things. :-)

[south patrol]

Interesting, didn't know this

Do you take the glb of all upper bounds to form a lub I guess

And vice versa

[tribal moss]

Yes.

Each of your original elements is necessarily a lower bound of the set of all upper bounds.

[crystal vale]

okay, thank you

[tribal moss]

So a greatest lower bound of the upper bounds must (by definition of "greatest") be an upper bound for the original set.

[dusty ivy]

Might be a stupid question

To be a cyclic group, $ \mathbb{Z}2 \times \mathbb{Z}{10} $ would need an element $(a, b)$ whose order is the same as the number of elements in the group. The number of elements in $\mathbb{Z}2 \times \mathbb{Z}{10} $ is $2 \times 10 = 20$
How can I know the order of an element $(a, b)$ in $\mathbb{Z}2 \times \mathbb{Z}{10}$?

[cloud walrus]

Jerome

[south patrol]

So you know the order of a and b are smaller than what

You want to find n, 0 < n < 20 with (na, nb)=0, so na = 0 and nb = 0, so what candidates are available

[crystal vale]

If R is a ring with unity then there is only one ring homomorphism between Z and R?

Here I assumed that if f is a homomorphism then f(1) = 1.

Because then n->n•1

[south patrol]

yes, exactly

[coral spindle]

There is only one unital ring homomorphism

Since you're studying nonunital rings all the time you should know that we have to specify that we are working with unital ring homomorphisms.

[south patrol]

i guess notknow said that

[crystal vale]

Okay thank you

[crystal vale]

Yes

[coral spindle]

Nono he only said that the ring R is unital, not the homomorphism. There are usually many more nonunital ring homomorphisms Z → R.

[crystal vale]

But I mentioned that I assumed that if f is homomorphism then f(1) = 1

[coral spindle]

Which is an incorrect assumption in general

One really has to specify these things when working with nonunital things, they suck!

[crystal vale]

Yes but the book defines only unitial homomorphism

[tribal moss]

"Many more" requires some cooperation from R, though.

[coral spindle]

Well R was not restricted it's not hard to produce a ring R for which there are infinitely many such things

[tribal moss]

Sure.

[dull ginkgo]

Two ring homomorphisms are the same if they map generators to the same thing. Now Z is generated by {1}, and if we have two ring morphisms from Z into R, they both must send 1 to the identity of the image (ring morphism). However {1} is a generator of Z, so the morphisms are equal. Therefore there is only one map from Z into R

You stated with unity, without unity it isn’t necessarily true

[tribal moss]

See the previous longish conversation about whether the relevant concept of "ring morphism" requires f(1)=1.

[dull ginkgo]

HOWEVER it is true if two morphisms map the identity of Z to the same element of R if we are ignoring unity

In which you are basically mapping n to nx = x + x + …

[coral spindle]

I would ordinarily have no problem with this, but notknow has asked many many questions in the past where he has referred to nonunital rings without explicitly saying so, so it was not clear whether or not he was actually referring to unital ring homomorphisms

[dull ginkgo]

Do you know how to prove that if two maps are equal on the generators that the two morphisms are equal?

[crystal vale]

Yes but here I mentioned it

Yes because it is equal on generators then every element is generated by generators so it's image will be equal

[velvet lake]

If I have the matrix [ 1 1; -1 2] : \Z^2 ---> \Z^2 with coefficient in \Z. I am trying to show the image is Z \times 3\Z Can I row reduce it so I get [1,0;0,3] over \Z ? Or no, is the best you can do [1,1;0,3] because you are not allowed?? to multiply by fractions?

[coral spindle]

You cannot use non-integer row manipulations when row reducing, no

Fortunately, it's not hard to see from [1,1;0,3] that you get the image you're looking for

You can just show it manually

Or hold on a moment

[rocky cloak]

What do you really mean by the image "being Z x 3Z"?

[coral spindle]

Yeah this just isn't right lmao?

The image of [1; 1] under this map is [2; 1] so

[rocky cloak]

I'm guessing what spikey is really interested in is the cokernel being Z/3 ....(?)

In which case you can do both row and column operations

[velvet lake]

Yes

[rocky cloak]

So you can't use a row operation to go from [1, 1; 0, 3] to [1, 0; 0, 3], but you can use a column operation.

[velvet lake]

Thanks 🙂

Absolutely Brain Dead atm

what was the problem with what I said before? Coker(A) = \Z^2/im(A) No?

But Im(A) cant be Z^2

[rocky cloak]

Yes, but the image is not Z x 3Z

The image is things of the form (x+y, 2y-x)

[velvet lake]

But would you say this isomorphic to Z \times 3\Z? Wait my confusion runs deeper. Is the Z-module nZ isomorphic to Z they are right? because you just relabelled the generator. So shouldnt that cokernel be trivial()

[rocky cloak]

The cokernel depends on more than the isomorphism type of the image.

3Z is isomorphic to Z, but Z/3Z is not 0

[barren sierra]

unsure how to proceed 💀

So R[x, y] / (xy) = all elements of the form ax^m + by^n

then looking at the fractions given by S^-1

I essentially have fractions of the form ax^m + by^n/x^s as far as I can tell

[coral spindle]

surely sums of these? e.g. x^2 + x is in there right?

[barren sierra]

oh right sums of those my bad

[coral spindle]

Or what I mean to say is that x^2 and x^2 + x aren't in the same coset

[barren sierra]

yea

that's what I meant

but I guess my issue is that it seems I have 3 generators, x, y, x^-1 but R[x, x^-1] has 2 generators

[coral spindle]

Right but under the construction, y should become 0

[tribal moss]

Something will have to die in order to make it end up isomorphic to R[x,x^-1].

[coral spindle]

y/1 ~ 0/1 since x(y-0) = 0

[south patrol]

eh yeah it's okay

just it kills y but x is ok

[barren sierra]

ahhh okay

yea I came to conclusion that something must die but then I was thinking I got non-trivial kernel

[tribal moss]

We have y = 1y = (x^-1 x)y = x^-1(xy) = x^-1·0 = 0.

[barren sierra]

ok that resolves everything

[south patrol]

one way to consider this is you have the SES $0 \to (xy)R[x,y] \to R[x,y] \to R[x,y]/(xy) \to 0 $ and localisation is exact. After inverting $x$ this becomes $0 \to (xy)R[x^{\pm 1}, y] = yR[x^{\pm}, y] \to R[x^{\pm },y] \to M \to 0$

idk if that is helpful to see or not though lol

[cloud walrus]

Algebraic Potato

[barren sierra]

interesting to see none the less

[south patrol]

Basically just localisation commutes with quotients in the way you're expect

[barren sierra]

yea

[south patrol]

Well it is useful to know that like yeah

S^-1M/S^-1N = S^-1(M/N) in a canonical fashion

[barren sierra]

yup

tried that actually as well

but idk didn't come up with the sequence

[south patrol]

I got confused initially cause I wrote (x,y) lol

[barren sierra]

I knew what you meant

[dusty ivy]

Why is C2 × C10 not cyclic?

[dull ginkgo]

In a cyclic group every element is generated by one element, but consider (1,0) and (0,1) lmao

[dusty ivy]

Tyy

[tribal moss]

Every element of C2×C10 has an order dividing 10.

In order for it to be cyclic there's need to be an element of order 20.

[lusty marlin]

C_m × C_n is cyclic if and only if gcd(m,n)=1

[long obsidian]

I think if you have two polynomials $p(x)=\sum_{k=1}^m c_k x^k$ and $q(x)=\sum_{k=1}^n d_k x^k$ then I think there is a nice formula for the product such as $p\cdot q= \sum_{k=1}^{m+n} a_k x^k$ where $a_k= \sum_{i,j \hbox{ so that } i+j=k} c_i\cdot d_j$

Is there a similar formula or some sort of description for the product of polynomials in several variables.

[cloud walrus]

HausdorffT1

[kind temple]

yes, it is best expressed via multi-indices
https://en.wikipedia.org/wiki/Multi-index_notation#:~:text=Multi-index notation is a,an ordered tuple of indices

a quick search yeilds this:
https://math.stackexchange.com/questions/1166438/product-of-polynomials-in-several-variables

the OP has the correct formula

[limpid ferry]

Hey, I want some helps on how to prove that the kernel of a group homomorphism 𝜙 is a subgroup

[rocky cloak]

What have you tried thus far?

[limpid ferry]

Let 𝜙 :G->G' be a homomorphism, if K is a subgroup of G', then 𝜙 ^-1(K) is a subgroup of G. I used this property to infer that the preimage of {e'} is Ker(K) but I realized everything I have been doing is wrong because 𝜙 may not be bijective

[crystal vale]

Why do you need phi be bijective ?

[limpid ferry]

for the inverse of it to exist

[crystal vale]

No

Phi^(-1) (K) = { x in G such that phi(x) in K }

[limpid ferry]

so for an arbitrary k \in K, how do I define 𝜙 ^-1(k) ? Because there may be more than one x \in G such that 𝜙 (x) = k

[crystal vale]

Phi^(-1) (K) = {x in G such that phi(x) = k } but it is not necessary that it will be subgroup of G

[limpid ferry]

hey wait so 𝜙 is a mapping, is 𝜙 ^-1 still a mapping then?, according to your notation

[crystal vale]

Not necessary

[limpid ferry]

thanks man, I can finally rest in peace

[chilly radish]

You might wanna review some basic properties of functions. Knowing how to work with the preimage is important

[coral spindle]

We have x is in the kernel iff phi(x) = e'. So now simply apply the definition of a subgroup with this necessary and sufficient condition.

This is very much a 'just do it' proof

No tricks are needed.

[empty kernel]

I just started with an introduction to groups and fields and rings and stuff

Ok so the definition of a ring R,+,• is that R,+ has to be a group and • has to be associative and • has to be distributive to +

This translation to english is pretty wack

Ok but so anyway R,+,• can be a ring without R,• being a group?

Or am I understanding this wrong

[coral spindle]

Yes

Eg the integers form a ring

[empty kernel]

Because this would imply that there doesnt have to be inverse elements for • and also no neutral element for 1 for •

[coral spindle]

In fact the only ring for which R, x is a group is the trivial ring.

[empty kernel]

Do the integers without 1 and -1 still form a ring?

[coral spindle]

No.

In no circumstances

[empty kernel]

Why not?

[coral spindle]

What’s 3-2

[empty kernel]

OMG

stupid me ahahha

Ok but is there a ring where no neutral element for • exist within it?

[coral spindle]

Depends on your definition.

[empty kernel]

Apart from the trivial one

[coral spindle]

Usually this is defined to exist. We call these unital rings.

If you don’t require your rings to be unital, which is unusual, then eg the even integers are an example.

[empty kernel]

Oh ye i see here

Its on the next page hahaha

Oooh

I didnt understand at first but its making sense now R,• cant be a group apart from in the trivial ring because 0 is part of the set otherwise R,+,• isnt a ring and 0 being in R breaks R,• as group

[empty kernel]

Why is this under advanced maths but i have this in my first year of uni💀

[rocky cloak]

University level math is pretty advanced I guess

[empty kernel]

My book says if R{0},• is a group then we call it a “lichaam” is lichaam the dutch word for unital?

Like is that the definition of unital

[coral spindle]

No.

[rocky cloak]

That would be a division ring

[coral spindle]

In English we call that a division ring, or if it is commutative we call it a field

It is not clear if you are considering commutative rings only rn

European languages disagree on this terminology

[empty kernel]

This is what my book says

The lowest one is commutative ring

[coral spindle]

Well there you go

[empty kernel]

Aha ok so division ring is even stronger than unital ring?

My mind feels slow rn

[empty kernel]

Bc Z,+,• is a unital ring but not a division ring

Yooooo i think im starting to understand

[south patrol]

Richtig

[hidden wind]

dutch abstract algebra

[south patrol]

The only bit I don't understand is lichaam

Hm

[hidden wind]

inverse?

[south patrol]

Oh body, so presumably field

By analogy to german

[hidden wind]

right

english is the weird one out here actually

[south patrol]

Oh but means division ring in Belgium lol

[hidden wind]

[tribal moss]

That parallels the French usage, doesn't it?

[rocky cloak]

So french not German i guess

[south patrol]

Hm aren't we closer to French for field

Lol

[empty kernel]

If i translate lichaam to english it means body

[south patrol]

Yeah

[empty kernel]

But apperantly its called unital ring

[south patrol]

Hm looking it up it should mean division ring or field

[empty kernel]

Noo a field is a lichaam that is also commutative

[south patrol]

So lichaam then means division ring

[empty kernel]

Oh mb thats what i meant

[south patrol]

Are you Belgian

[hidden wind]

what’s the french for field

[empty kernel]

A lichaam is a division ring

[empty kernel]

Yes why?

[south patrol]

Nice

[rocky cloak]

I like skew-field for division ring 🙂

[south patrol]

Nah we were just discussing this

The meaning is different in Belgium and the netherlands

[empty kernel]

Oh ye my prof said that

[hidden wind]

wack

[empty kernel]

In the netherlands the terms were different smh

[south patrol]

In the Netherlands lichaams are commutative

[rocky cloak]

Corps commutatif

[cobalt heath]

F

[south patrol]

Oh yeah so English is the weird one then huh

[cobalt heath]

That could cause some confusion

[hidden wind]

so just corps would be division ring?

[empty kernel]

But in belgium we use $N_0$ for natural numbers without 0 while in the rest of the world it means the natural numbers with 0

[cloud walrus]

you_are_me

[south patrol]

Idk I think there is some debate about this still in the rest of the world lol

[hidden wind]

seems danish uses legeme for field

[empty kernel]

Belgians are defi the weird ones

[south patrol]

Personally what I really dislike is Z_+ lol

[cobalt heath]

One may say noncomm ring does not exist

[empty kernel]

Ye like wether or not 0 is in it

[tribal moss]

I don't think there's any consisntency in whether N includes 0 within any country.

[hidden wind]

in norwegian we do as in germany and use kropp for field

[empty kernel]

But is corps a division ring or a field?

[south patrol]

I know people who use N with 0 and Z_+ to include 0

[hidden wind]

same etymology as körper

[south patrol]

To me that seems the complete wrong way round

Lmao

[cloud walrus]

Absta

[empty kernel]

Yes

[south patrol]

But then I took a quantum mech course and they did that absta aha

[empty kernel]

Ok i gotta go study again byebye

[south patrol]

Z_>=0 and Z_>0

Goodbye

[chilly ocean]

N

[south patrol]

I need to study tooo...

[hidden wind]

i like N without 0 and N_0 with 0

[coral spindle]

I thought in German that Körper refers to commutative things too?

[hidden wind]

i don’t like to refer to N via Z

[south patrol]

It does

[rocky cloak]

I always like to think of a field as a body where addition, subtraction, multiplication and division are the arms and legs

[south patrol]

I mean merely in terms of etymology.

[cobalt heath]

I like to think of N as an infinite monoid of one generator

[hidden wind]

amazing, thank you

[chilly ocean]

🤤

[south patrol]

Like idk why English has field as opposed to body or smth lol

[rocky cloak]

It does, but english is odd in that the word 'field' doesn't mean 'body' like it does everywhere else

[chilly ocean]

Then what about division ring

[south patrol]

I think I conflated Boytje and Jagr in this cause I only looked at the colour

[cobalt heath]

I mean, a field does not feel like a body

[rocky cloak]

That's like a twisted body

[cobalt heath]

It does not feel like field as well, but eh

[chilly ocean]

😭

[south patrol]

Oh yeah rødbet lol

I heard in København they had a first course in lin alg with division rings and matrices asking on the right

[chilly ocean]

What about algebra (over field)

[cobalt heath]

Why is a ring called "ring"?

[tribal moss]

Køvenhabn.

[south patrol]

Ask Emmy noether

[cobalt heath]

Hmm, does ring look like 💍

[chilly ocean]

Cause it's a horror movie duh

[rocky cloak]

I think the etymology of both group and ring are just "synonyms for collections"

[south patrol]

Køvenhabn.

[chilly ocean]

It should be called sexier sets instead

[rocky cloak]

But my headcannon etymology is that 'ring' comes from the composition symbol. Since obviously the canonical ring would be an endomorphism ring

[south patrol]

Idk why I wrote a v

[cobalt heath]

Oh, that's good

[south patrol]

Yeah I swapped the v and b for no reason lmao

[rocky cloak]

You mean a quiver 😉

[south patrol]

Even though it is clearly rhe other way round

Mb

[chilly ocean]

Quiver?

[hidden wind]

i don’t get it

[south patrol]

Lol like kinda cursed

[chilly ocean]

I thought the sipping emoji was sth else 💀

[rocky cloak]

The only way to do algebra my friend

[chilly ocean]

Interesting...where can I know more

[hidden wind]

god the only thing i remember from my linalg course is my danish prof

[south patrol]

In any case I'd say acting on the right uses more paper

[cobalt heath]

Is this about algebra over a ring or general algebra

[south patrol]

Lol

[hidden wind]

he was incredibly cute

[rocky cloak]

Introduction to representation theory of algebras by Assem Simson Skowronski is good

[south patrol]

In an old man way or like

[hidden wind]

i will not elaborate

sorry

bye

[south patrol]

My lawyer has advised me to stop

[rocky cloak]

This is about representation theory of finite dimensional algebras (the only true algebra 😉 )

[south patrol]

Imagine being over a field

[chilly ocean]

What is truth anyways

[cobalt heath]

Meanwhile (infinite-dimensional) quantum algebras:

[chilly ocean]

Is there a only true truth

[chilly ocean]

I found elements of representation theory but not quite the one you named

[rocky cloak]

Yeah, sorry I just wrote down the title from memory. It's called elements yeah

[chilly ocean]

Ty

Is it possible for a set other than a normal subgroup of G to have normalizer as G

Like the set dosent have to be a group

Or does N(H)=G automatically imply that H is a subgroup and that too normal

[south patrol]

Well you could do silly things like G minus {1}

[chilly ocean]

Ah

[south patrol]

And presumably you can modify this to provide slightly less trivial examples

[chilly ocean]

Right cool

Thank you

Clever

[tribal niche]

how can I construct a group of order 3^4 * 5 * 7 = 2835 with exactly 7 subgroups of size 81? (i.e. 3-sylow subgroups)

is there some semidirect product trick i can use? i haven't really needed to construct things before

[rocky cloak]

Construct such a group of order 3^4 * 7 and then take the direct product with C5.

To construct the first group you just need any group were the 3-sylow subgroup isn't normal, so a semidirect product will do.

[empty kernel]

Ok im here again

Ring morfism

Let f be a morfism from R to S both unital rings

Does f(r)=s if r is the neutral element for • in R and s is the neutral element for • in S

Wait not very well explained

[rocky cloak]

Usually that's part of the definition of being a ring homomorphism, but it doesn't follow if you don't assume it

[mighty kiln]

ring ← unital ring
rng ← non-unital ring

[rocky cloak]

Consider the map f(x) = 0 for example

[empty kernel]

All it says here f(x+y)=f(x)+f(y) and f(x•y)=f(x)•f(y)

[mighty kiln]

That's a rng homomorphism

[empty kernel]

Ok so but it doesnt have to be what i said earlier

[mighty kiln]

See for instance the trivial map Z → Z

[coral spindle]

Not necessarily unital tho

[rocky cloak]

From this it doesn't follow that f preserves the neutral element of multiplication

[empty kernel]

How not

[mighty kiln]

1 is sent to 0 here

[rocky cloak]

Consider for example the map f(x) = 0

[empty kernel]

Idk somethibg goes wrong in my reasoning but idk what

[empty kernel]

Oooh ye thats it

I didnt think abt 0

Stupid zero

[rocky cloak]

Even if you ignore that things can go wrong

[empty kernel]

Wait how?

[rocky cloak]

Like you just need an idempotent different from the neutral element

For example let R be a field and S the ring of 2x2 matrices over R, then the map
r |-> [r, 0; 0, 0]

[empty kernel]

Oh

Ok wait

This prob dumb asf but how does this not work?

Cuz thats what i tought at first

You get f(x)=f(x)•f(1)

Doesnt that mean that f(1) is the neutral element

[rocky cloak]

Sure, but there can be things in the ring not of the form f(x)

But if f is surjective, this works

[empty kernel]

Ah

Ok I understand now thanks

I will prob be back in 30 min💀

[empty kernel]

Wait i forgot to ask why i asked this whole thing in the first place so it says here if f is a ringisomorfism so f must be a bijection and then it says if f(1)=1 then blablabla

[empty kernel]

But for an isomorfism f f must be surjective so this would work

So f(1)=1 for all isomorfisms doesnt it?

So isnt it enough to just say if f is an isomorfism then blablabla?

[coral spindle]

isomorphism*

Yes, if you have an isomorphism then indeed you must have that f(1) = 1. In fact this holds for any surjective homomorphism.

[winged void]

I have small question why the conjugation class of S_4 is 5

number

[next obsidian]

Is that the number of conjugation classes or something

If so, you show that conjugation classes are exactly the set of elements with the same cycle type

Do this by showing conjugation stays inside the same cycle type, and that it’s transitive within those sets

Then the number of conjugation classes = number of cycle types = number of partitions of 4

So you have
1,1,1,1
1,1,2
2,2
1,3
4

[empty kernel]

Omg bidual spaces😭

Oh lets do this oh we get a new vector space lets do it again i swear if they gonna pull out tridual spaces im gonna die

[rotund aurora]

Is there a smart way of doing (c) and (d)?

here Gamma_0(p) are matrices ((a,b), (c,d)) in SL(2, Z) with c=0 mod p

[fresh gate]

I have a question?
Is it possible to have a field have 1 + 1..... + 1 n times be equal to 0 while being infinite?

[delicate bloom]

yeah as long as n is prime, as an example F_p(x) is the field of rational polynomials with coefficients in F_p

if n is not prime you will have 0 divisors and necessarily not be a field

[south patrol]

Well for n non-prime, and not 1, you can just pick a prime dividing n

just yeah n cant be the actual order of 1 in that case

[novel spruce]

Suppose we have an ordered ring A such that every non-empty subset of A that is bounded above has a least upper bound. If A is a field, then a well-known theorem states that A is isomorphic to R. A clear example of such a ring that is not a field is Z. Do there exist other examples?

[next obsidian]

I feel like you ought to get an example by modifying Z. Like maybe some order on Z[x] where you say that polynomials of degree n > degree n-1, and then in case of a tie you order by the leading coefficient

Now if S is bounded above that means the polynomials have bounded degree, so say the maximum degree is degree n

Now let S_k be the set of (nonzero) coefficients of the x^k term over all polynomials in S, I think because

Actually I think this doesn’t have to be bounded, but this ought to still work, I think S_n has to be bounded

So we can find a LUB a_n

And then you look at like a_nx^n, is that a LUB? Not necessarily because you can have like a_nx^n + kx^n-1 for all k in Z

But in that case I think (a_n + 1)x^n is a LUB

Maybe try seeing if this example works?

I think there’s a decent chance it can, but don’t wanna work out all the details ¯_(ツ)_/¯

[novel spruce]

The subset Z is bounded above by any polynomial of positive degree (or maybe you want the leading coefficient to be positive?). Doesn't seem to have a least upper bound: one can always subtract 1 from a given upper bound to get another upper bound.

[next obsidian]

:(

So true

If only this was N, then I think this works

Cuz you can’t keep going down

[rotund aurora]

Maybe you can prove that Z (it must be char 0 obviously) cannot be bounded from above, if it's an integral domain you can take the field of fractions and then it's a field with an archimedean ordering which should embed into R

and I think it must be an integral domain because if a,b>0 then ab>0

[next obsidian]

But that means it’s a subring wedged between R and Z

So it doesn’t rule out stuff like Z[sqrt(2)]

I think

I dunno

[rotund aurora]

this is ruled out by Dirichlet's theorem on irrationals, or idk what theorem

[next obsidian]

Oh yeah

Cool my chmintuition said that it bad for some reason

Wait what about Z[1/2] or something idk, I think this still is bad

I guess you can argue like this

Say you have x in R\Z

Then you can get x-n until this is between 0 and 1

Then (x-n)^k is arbitrarily small

So {0} has no LUB

[rotund aurora]

Z[1/2] does work, because it's discrete in R

I think

[next obsidian]

Nah

It’s a ring

Not a module

That’s an algebra adjoin

[rotund aurora]

ah oops yeah forgot you can multiply 1/2 all the time

[next obsidian]

I think I proved what I wanted

Oh wait

{0} does have a LUB

0

Ummm

But you get arbitrarily small stuff so

Hmmm

Idk if this immediately contradicts the thingy

I can’t off the top of my head come up with a contradiction

[rotund aurora]

I think it might be easier to just compute the closure (as in topology) of Z[1/n]

I am sure you always get some irrational

[novel spruce]

Okay {0} works indeed.

[rotund aurora]

actually this looks dense to me

[novel spruce]

The closure of Z[1/n] in R is all of R.

[rotund aurora]

so that rules it out

[dense raven]

can you construct an uncountable additive proper subgroup of R without using axiom of choice

[south patrol]

hm

[dense raven]

there might be a way to reduce this to a measure theory problem

if you can prove that such a group isnt lebesgue measurable

but not sure if this is the right approach. the only examples ive thought of arent lebesgue measurable, but a counterexample might exist

[rocky cloak]

Yes you can
https://mathoverflow.net/a/23206/157483

More examples
https://mathoverflow.net/q/35970/157483

[lilac mango]

How do I calculate in the finite field of order q=p^n in general? I know this field exists and is unique up to isomorphism, but if I wanted to do explicit calculations in GF(q), how would I do that? One way would be to find an irreducible degree n polynomial f(x) over GF(p) and quotient GF(p) over <f(x)>

However in general it seems pretty difficult to do the task of given a prime p and an integer n, calculate an irreducible polynomial mod p of degree n

So is there some better way of being able to calculate in GF(q) or some sort of fast(ish) algorithm by which one could calculate such an irreducible polynomial?

[coral spindle]

Disclaimer: I do not know if there are better ways than what I'm about to describe. Hopefully a computational algebraist can comment.

I think the polynomial construction is the most common method, afaict.

It seems that algorithms simply find a single irreducible factor of x^{p^n}-1 in F_p[x]

Of course, there is a naive algorithm for this: simply exhaustive search for factors

But presumably there is a cleverer way to do this, alas I simply don't know

[lilac mango]

What do you mean with this?

Would it be x^(p^n)-x I guess?

[coral spindle]

Ah yes my mistake

indeed

I am remembering now that there are algorithms for finding the factorisation of a polynomial in F_p. I'm going to look at the MAGMA/GAP documentation and hope that there are details of the implementation in there somewhere.

OK. MAMGA says that it uses something called the Conway polynomial

And according to wikipedia at least, there are good algorithms to compute these polynomials

From the MAGMA documentation

The primitive polynomial used to construct GF(q) when n > 1 will be a Conway polynomial, if it is available. If the parameter Optimize is false, then no optimized representation (i.e., by using Zech logarithm tables or internal multi-step extensions) will be constructed for the new field which means that the time to create the field will be trivial but arithmetic operations in the field may be slower -- this is useful if say one wishes to just compute a few trivial operations on a few elements of the field alone.
So it seems essentially that it cheats a bit using a lookup table most of the time.

GAP also uses Conway polynomials and seems to describe its method a bit more.

From the GAP documentation:

he computation of Conway polynomials can be time consuming. Therefore, GAP comes with a list of precomputed polynomials. If a requested polynomial is not stored then GAP prints a warning and computes it by checking all polynomials in the order defined above for the defining conditions. If n is not a prime this is probably a very long computation. (Some previously known polynomials with prime n are not stored in GAP because they are quickly recomputed.) Use the function IsCheapConwayPolynomial (59.5-2) to check in advance if ConwayPolynomial will give a result after a short time.

So okay. TL;DR: yes it's hard.

[rocky cloak]

According to Sages documentation, they have a lookup-table of polynomials to use for constructing finite fields. Those being the Conway polynomials

https://en.m.wikipedia.org/wiki/Conway_polynomial_(finite_fields)

[lilac mango]

Wow ok thanks

[coral spindle]

Haha all the major CASes use the same method huh

[rocky cloak]

Makes sense really. Then you can compare the outputs and stuff

[coral spindle]

But also demonstrates that this really is the state of the art

[lilac mango]

Thank you both

[lilac mango]

wdym by this?

[coral spindle]

If they all use the same method, then the way they present the results are also the same.

So you can compare the outputs directly between CASes

[rocky cloak]

Like if you're using both sage and GAP for example

[lilac mango]

Oh ok yeah you represent the field in the same way makes sense

[rocky cloak]

This reminds me that it's annoying to work with finite fields in GAP. I have to write like Z^0 instead of 1, it's horrible

[long geyser]

I need help with this
I guess if our ring is of the form a + b * alpha, where alpha satisfies the equation alpha^2 + c* alpha + d = 0, will the degree function always look like a^2 - cab + db^2? this seems to hold with Z[i], Z[sqrt(2)], Z[omega]

but true or not, I don't understand why

[rotund aurora]

this comes from calculating (a+b\alpha)(a+b \overline\alpha)

but in general, if you have a ring of the form Z[alpha] with alpha satisfying some quadratic polynomial over Z, it needn't be a euclidean ring

but even if it's an Euclidean ring, it could have other Euclidean functions, while this norm function may not work

[long geyser]

if we want to look at things algebraically then, is the complex conjugate of alpha related to the quadratic equation it satisfies?

[rotund aurora]

yes, because both alpha and \overline alpha satisfy the same equation

when alpha is real, you just change the sign of the square root

[south patrol]

basically there is a natural thing where you send smth to the product of its Galois conjugates, or equivalently the product of the roots of the minimal polynomial. this doesn't necessarily define a euclidean function thoug

but it is a useful construction known as the norm of the thing

[long geyser]

oh, hmm

idgi tho, how is this helping me with the question

[south patrol]

well it is the product of a+bw and its conjugate

so this is the norm in the complex numbers (squared)

that is useful

[long geyser]

I get that, I mean the fact that this comes from w^2 + w + 1 = 0

[south patrol]

yeah idk i think that is just so you can work back to this

once you have this it is more a computation

[long geyser]

or was that fact just so I could deduce the degree is a + bw times the conjugate, they just don't seem very related

[south patrol]

but you can use some geometric intuition

[long geyser]

I already showed that it is just the complex norm squared by writing omega in terms of complex numbers lol

I want to show that division algorithm works

[rotund aurora]

Since omega is complex, in this case the norm is the distance squared

[long geyser]

yes, that's what I wrote

so, is the division algorithm like the gaussian integer case

[rotund aurora]

I think a similar argument works

[long geyser]

I see, I'll try

[south patrol]

how did you do it for gaussian integers

[long geyser]

uh

[south patrol]

tbf there are a couple of ways to do this

[long geyser]

write alpha/beta as an element of Q[i]

[south patrol]

a geometric-brained way and a more "compute" way lol

okay great yes

[long geyser]

separate out the integer part

[south patrol]

yes that works perfectly

[long geyser]

I guess the geometric way is viewing this thing like a lattice?

[south patrol]

one way to phrase it is that like

yes

Z[w] and Z[i] are lattices

and you need to show that each element is within distance 1 of a lattice point

which you can do by a picture lol

[long geyser]

right

[south patrol]

liek for Z[i] you get a bunch of "squares" and the point furthest away is the centre of the square, which is distance 1/2 away

try smth similar for Z[w]

in fact similar arguments work more generally, but unfortunately not always, sad

like it doesn't work for Z[sqrt(-5)] (as you can check from the fact 6 = (1 + sqrt(-5))(1 - sqrt(-5)) = 3*2

[long geyser]

hmm, so this Z[omega] guy is like a diamond shaped lattice right?

[south patrol]

hm it is a bit simpler

[long geyser]

not this?

[south patrol]

sure, but that is just made up of little triangles

[long geyser]

oh

right

so horizontally the lattice points are 1 unit apart, connected above and below like equilateral triangles

[south patrol]

yes exactly

of side length 1 of course

[long geyser]

and any point is gonna be enclosed in one of these triangles

[south patrol]

yes

[long geyser]

and the distance to the endpoints has to be less than the side length

[south patrol]

oh yeah you can just do that lol

[long geyser]

what's the alternative

[south patrol]

well i don't quite see why that is the case

it isn't true in general

say you had some like septagon

[long geyser]

triangle inequality

[south patrol]

but that doesn't really work

like

[long geyser]

hm

idk I suck at geometry

[south patrol]

you can argue that the furthest point is the centre and then just work out that

[long geyser]

makes sense

I guess there is an easier way to argue

pick a point inside the triangle