groups-rings-fields

[latent anvil]

The hint

κ(p) = field of fractions of A/p is finite dimensional over k

[vestal snow]

But you took the fraction field

[latent anvil]

yup

[vestal snow]

If we had that A/p is finite dimensional over k, then I can see how this would follow

[latent anvil]

How what would follow?

I'm confused by what you're saying

[vestal snow]

that p is closed

[latent anvil]

well we have that for sure

A/p is a subalgebra of A_f/m

And the latter is finite dimensional over k

But your hint is talking about the residue field

which is the field of fractions of A/p, not A/p

[vestal snow]

Wait

what is the residue ring?

[latent anvil]

There are two pieces of terminology

The residue field of a local ring (R, m) is R/m

The residue field of a ring A at a prime ideal p is the field of fractions of A/p, or equivalently A_p/pA_p. Sort of generalizing this, the residue field of a scheme X at a point x is the residue field of the local ring O_{X, x}

[vestal snow]

Ah that's what was confusing me

[latent anvil]

The hint is phrased weirdly tho, since a finite dimensional integral domain over a field is a field

Like it's just as true if you replace "residue field" with "A/p"

anyways the point is you can embed A/p into A_f/m

Which gives you finite dimensionality and then closedness of p

[vestal snow]

I still have a few questions about this

Like with that hint

[latent anvil]

sure

[vestal snow]

closed points of Spec B = max'l ideals M of B. Let B=k[x_1,...,x_n]/I. Using the correspondence, B/M = (k[x_1,...,x_n]/I)(m/I)=k[x_1,...,x_n]/m for some maximal ideal m of k[x_1,...,x_n] containing I. k[x_1,...,x_n]/m is clearly a finitely generated field

[latent anvil]

yup

[vestal snow]

Why even include the residue field?'

[latent anvil]

I don't know

I mean you also need the other direction

If the residue field is fg then the point is closed

[vestal snow]

I mean

[latent anvil]

But the proof I have in mind immediately says "since frac(A/p) is finite dimensional so is A/p"

[vestal snow]

I can just prove that if A/m is a field then m is maximal therefore closed

[latent anvil]

wait what?

[vestal snow]

Instead of proving p is closed iff Frac(A/p) is a finitely generated field, I jusst prove that p is closed iff A/p is a field that is finitely generated

[latent anvil]

I mean sure but then in the main argument there's more work to be done

[vestal snow]

Yeah I can see that

[latent anvil]

The thing I said proves A/p is finite dimensional over k

Yeah

[vestal snow]

I do not like this proof

It's really weird

[latent anvil]

What you've said here is basically Zariski's lemma

[gritty sparrow]

Well, if you remove the part about A/p being a field (which is a bit much to include) it is. Also in your proof, just to clarify there is no need to go yo the fraction field of A/p right?

[vestal snow]

So I think here's what happened: I misinterpreted the hint which made proving the hint really trivial. However, this misinterpreted result is not sufficient to prove the main problem

[latent anvil]

I agree with that interpretation

[vestal snow]

Even though the misinterpreted hint is still correct

[gritty sparrow]

Nvm i think you mentioned this

[latent anvil]

the proof as it stands is noticing something sort of weird/interesting

The local ring O_X,x doesn't change as you go closer to open subsets containing x

And the hint is saying if X is actually a variety, you can detect whether x is a closed point from that local ring

so it doesn't matter whether you think of x as a point of the whole space Spec A or the neighborhood D(f)

[vestal snow]

Shamrock I think there's an easy way to do this

Assume that my original definition of residue ring

Then we get that A/p is a subalgebra of A_f/m

In particular, A/p is a submodule of A_f/m

The latter is a finite dimensional vector space

Thus, A/p is a finite dimensional vector space

However, it is also an integral domain as p is prime

This forces A/p to be a field

a x _: A/p -> A/p will be injective for all non-zero a

Thus, an isomorphism

That means it will be invertible

Does this sound okay?

[latent anvil]

yup

That's one way to justify this claim about integralness

any particular element of an integral extension is contained in a finite extension, and then your proof works

[vestal snow]

Thanks

[chilly ocean]

Associates in R (reals) are given by two equivalence classes: R* (non-zero reals) and {0}, am I right? I'm hesitant about {0} being a class of associates because my textbook omits that. According to the definition of associates : a and b are associates in R if a = ub for some unit u in R, it makes sense. 0 = u0 for any unit u in R so it is an associate to itself only.

[nova plank]

Yes

All nonzero reals are units, and all units are associates of each other.

And in any ring, 0 is its own sole associate

[chilly ocean]

Thanks !

[chilly ocean]

In C[x], the set of all polynomials with complex coefficients, the class of associates are {0}, C* (non zero complex constants) as all units are associates, and C[x]. The last class comes from the fact that every polynomial has an associate which is the polynomial itself multiplied by a non-zero complex constant. Is this correct?

[final pasture]

it doesn't look correct

each class is supposed to be disjoint

[chilly ocean]

Hmm you're right, its an equivalence relation

[final pasture]

yes

[chilly ocean]

So ig its just the whole of C[x]

[final pasture]

I don't think so. Are X and X^2 associates ?

[chilly ocean]

Nope.

shika

So what Ive found is that for every p(x) in C[x], a*p(x) is an associate to p(x) where a is a non-zero complex constant

[hidden haven]

That is correct

[chilly ocean]

{ap(x): p(x) in C[x], a in C*} is a class

[hidden haven]

yes

[final pasture]

no

I mean

I get what you mean

and what you mean is correct

but written like that, most people would read it as "the set of all multiple of all polynomials in C[x]"

[hidden haven]

Chshika

[final pasture]

and this is just the entirety of C[x]

[chilly ocean]

yh right thats what confuses me

[final pasture]

stop that

[chilly ocean]

Nah wait

[final pasture]

ledog is already annoying enough with that

[hidden haven]

ye, you dont write p(x) inside the set builder form

[final pasture]

^

[chilly ocean]

its not the set of all multiples

[hidden haven]

p(x) is a fixed polynomial

[chilly ocean]

im multiplying polynomials by constants

[final pasture]

For each p(x) € C[x], {ap(x) | a € C*} is a class

[chilly ocean]

aha

[final pasture]

and these are all the classes

[hidden haven]

[chilly ocean]

thanks

[hidden haven]

Chnp

[final pasture]

(imo the question sucks, the answer is pretty much the def once you know that units in C[x] are elements of C* )

[hidden haven]

you can notice that each equivalence class other than {0} contains a unique monic polynomial and inflate word count by writing that

[final pasture]

[hidden haven]

ie the set of equivalence classes is {[p(x)] : p monic}

[nova plank]

nonzero equivalence classes**

[hidden haven]

[nova plank]

[final pasture]

the set of equivalence classes is {[p(x)] : p(x) € C[x]}

[hidden haven]

[final pasture]

how to answer ill-posed questions

[hidden haven]

ill posed or just easy

[final pasture]

Actually, it works with any question

Question: "Let G be a finite group and p the smallest prime factor of |G|. If H is a subgroup of index p, show H is normal."
Answer: "the set of equivalence classes is {[p(x)] : p(x) € C[x]}"

[hidden haven]

[nova plank]

The truth value of RH is [for all z in C, zeta(z) = 0 implies Re(z) = 1/2 or z is a negative even integer].

[delicate orchid]

Omg RH solved?!?!?!

[nova plank]

Yes

I am rich

[delicate orchid]

Epic...... win.....

[cloud walrus]

Wew Lads Tbh

[upper pivot]

N is always ambigious smh, does it or does it not contain 0

[nova plank]

I was avoiding latex

Because lazy

[delicate orchid]

Fair

[nova plank]

Also what John said

[final pasture]

it contains 0 and anyone saying it doesn't is someone I'll insult, any remarks ?

[nova plank]

I thought winning the prize money without latex would be more impressive

[delicate orchid]

It doesn’t have 0 so I have a nice example of a magma when I need one

[upper pivot]

if you do any anal NT for instance, N does not contain 0 lol. it really depends on the context

which is why its ambigious

[final pasture]

currently insulting you

[nova plank]

$$0 \not \in \bN, 0 \in \bN_0 = \omega$$

[cloud walrus]

Lunasong the Supergay

[upper pivot]

i personally write

$\bN-{0}$ or $\bN\cup {0}$

[cloud walrus]

JohnDS

[upper pivot]

no ambiguity

[delicate orchid]

I use this notation

[nova plank]

This is so bad, I want to cry

This means that with either interpretation of N, you are definitely being redundant in a ridiculous way

[delicate orchid]

They’re right tho it’s not ambiguous!!

[upper pivot]

yep thats the point :)

[nova plank]

[tough raven]

And doing a very impressive job.

[final pasture]

ik ik

[tough raven]

owo Nitro

[minor badger]

i got a question

HOW DO PEOPLE STUDY MATH FOR 8HOURS A DAY 5 DAYS A WEEK GAAAH

;-;

[delicate orchid]

We don’t :trollface:

[tough raven]

It's 7 days a week
(To be clear, JK)

[minor badger]

My brain hurts so bad and i have only been at it for 4 hours

[delicate orchid]

I usually only study 3 hours at a time

Past that point it’s just diminishing returns

[minor badger]

Im using pomodoro, so study 25min and break 5 min. Otherwise i stroke

i guess ADD isnt helping lel

[delicate orchid]

I got ADHD bro I know your pain

I find listening to music really helps me focus so maybe try that?

[minor badger]

do you also have that stress just nullifies your meds?

[delicate orchid]

I’m not on meds

[minor badger]

Oh, i am heh. Yeah music works sometimes but more often than not it takes my attention.

[delicate orchid]

That can be a problem

Maybe just take a longer break? Like people don’t work in jobs for more than 4 hours without having a lunch break or something

[final pasture]

Wait I'm getting confused about something dumb .
Let F be a field, B = (b1, .., bn) a basis of F^n, and M € Mn(F) with Cj being the j-th column of M.
Assume we already defined det_B(v1, ..., vn) with each vi € F^n (like determinant of n vectors is already defined).
Is det_B(M) defined as:

• det_B(C1, .., Cn) with each Cj interpreted as vectors of F^n ?
• det_B(v1, .., vn) with vj = C1j b1 + .. + Cnj bn ?
  If I'm not bullshiting, those aren't the same, because in the second case we always have det_B(In) = 1 while in the first it's not the case

[vestal snow]

Here's what works for me:

1. Take breaks, but don't watch anime/play vidya/something you really like because you end up taking much longer breaks than you need. I usually take snack breaks where all I do is eat and focus on that for 20-30 minutes.
2. You need to make it a habit. It will be very tempting to skip days at the beginning, but you need to resist this temptation and make a habit out of it.
3. Keep your phone away away when doing math (I keep it in another room).
4. Don't do math if you're sleepy. On a similar note, avoid doing math in a position/environment that you associate with sleep.

May or may not work for others

[carmine fossil]

You define det_B(M) as det_B(Mb_1,Mb_2...Mb_n)/det_B(b_1,b_2...b_n)

[final pasture]

isn't this literally dividing by 1 ?

[carmine fossil]

You might use a different basis

[final pasture]

wut

[delicate bloom]

basis is by definition not linearly dependent

[final pasture]

det_B(b_1,b_2, .., b_n) = 1

how does that help ?

typo

meant 2

[carmine fossil]

How is det_B(I_n) not 1 according to first definition?

[minor badger]

its the first one ?

[carmine fossil]

Suppose det_B((1,0),(0,1))=1
Then det_B(I_2)=det_B((1,0),(0,1))=1

[final pasture]

det_B((1,0), (0,1)) = 1 where does that come from

B isn't the canonical basis, it's any basis

I'm not saying it's not true

but I don't see how it is true

[carmine fossil]

2 seems more accurate

[final pasture]

Y'all don't seem to agree lol, I'm even more confused now

[carmine fossil]

I think the common assumption with computing det_B(C) is that Cb_i=c_1ib1+c_2ib_2...

As in C is taken to represent a linear transform from that basis to itself

[final pasture]

ok so someone answered me on another server

the right definition is the second one

and its value doesn't depend on B, since for any B, det_B(Id) = 1

and the first one isn't equivalent

[carmine fossil]

So,this

[final pasture]

i.e det_B(In) != 1 with the first def

taking for example F = R, n = 1, B = (2) then det_B(In) = det(1) = 1/2 with the first definition

[minor badger]

yeah okey it could be the second option

if B would be an orthonomal basis then i think 1 would work too

[final pasture]

it doesn't, c.f my example just a msg before your msg

[minor badger]

but B=(2) is not a basis?

oh nvm

well orthonormal, means length one and pairwise perpendicular, your example isnt normal

[final pasture]

oh i've read orthogonal, sorry lol

but I still think that's false

but I'm too lazy to search for a counter example lol

[chilly ocean]

How do I fill this table?

I've found that b * b = c, but this is taking a very long time

How am I supposed to go about this?

[carmine fossil]

a*a=a implies a=e

[small karma]

Can someone help me with ques 9. I want help with the part to show generator for each subgroup.

[chilly ocean]

Why? We don't know that a * b = b, a * d = d

[carmine fossil]

a has an inverse

So (a^-1)(a)(a)=(a^-1)(a)

a=e

[chilly ocean]

How do you know a has an inverse?

[carmine fossil]

It's a group

Every element has an inverse

[chilly ocean]

It's not given that it is a group

[carmine fossil]

nvm

[chilly ocean]

We need to determine if it is or isn't

[carmine fossil]

Isn't it just impossible to finish the table

[chilly ocean]

No, it's not

But I don't know how to do it without checking at least 64 possibilities

And moving the parentheses around

Which takes a very, very long time

[carmine fossil]

It just seems like there will be too many possibilities

[chilly ocean]

I know

That's why I am asking on what to do haha

[tough raven]

Checks out, determinant of a matrix should be basis-independent

[delicate orchid]

is c*b = d = > b*a = b valid? cause you know that b*a cannot be d as d is already b*c => a*d=d thus a is the identity and is in the "centre" of whatever object this is so you then know the entire first row and column?

[chilly ocean]

I don't understand your reasoning for why b * a cannot be d

They could both be d

Unless we find them not to be d

[lavish gale]

a generator of order k is a^{\frac{n}{k}} @small karma

[delicate orchid]

I'm pretty sure that would violate associativity

well, fairly sure

or you could use light's associativity test

[small karma]

yes i know lemme show you my solution . im confused about the generator for the subgroup

[lavish gale]

also u should review fundamental theorem of cyclic groups

[small karma]

So i did this question till this point. Now i want help in how to find all generator for the subgroups. I'm getting confused in that.

Question is asking for 1 generator for the subgroups. But i want to know how to find all the generators for each subgroup.

[lavish gale]

so do u know when an element of a subgroup has the same order as the order of the subgroup it is in?

@small karma

[small karma]

It will be a generator for that subgroup. Thats the definition for cyclic group

[lavish gale]

say G is a finite group and <a> is a subgroup of order m. then <a> {e, a, a^2,...a^{m-1}}

what else can generate this subgroup?

[small karma]

Yes

<a^n> = <a^gcd(n/k)> u mean this?

[lavish gale]

yes

and order of <a^n> is m/gcd(n,m)

when is this m?

[small karma]

Yes. Gcd =1

[minor badger]

cant use solve it like soduko? i think only element each row and column

[lavish gale]

yes

[small karma]

I use the order of the subgroup right

[chilly ocean]

@minor badger no? What?

[lavish gale]

yes because it is in the subgroup

[small karma]

Like for <2> it will be <k2> = <2^gcd(k,10)>

Bcz it is additive

We use ka instead of a^k

[lavish gale]

what is k?

o

[small karma]

K is elements of that subgroup

In this case <2>

Elements

Of <2>

[lavish gale]

if a^k is an element of <a> a^k will generate <a> when k is relatively prime to m

[minor badger]

well becase you have that ac = a, cc=a and d*c=b

so then b*a = d

basically c is its own inverse

and a identity

[chilly ocean]

How did you get b * a = d?

[minor badger]

becase if else then b= c or b = a or b = d

[lavish gale]

now do u know how to compute other generators ? @small karma

[small karma]

Can u confirm me that generators are <2> =<6> = 14 = 18. For order 10.

<14> = <18>

I used a different method

[lavish gale]

Z_24 is our original group right?

[small karma]

To find generator of the subgroup

Z20

Is

This is what i did

J are elements of Z20

[lavish gale]

yes

ur correct

[small karma]

Okay i get it now. Thanks

I was getting confused with this process

[delicate orchid]

if you want to find all of the generators you need to look at the elements in each subgroup that aren't in any other subgroup, i.e. <5> = <15> because they are the only two elements that are in <5>, <10> doesn't work because <10> is it's own subgroup.

Another example is how <6> != <8> because <8> is in <4> but <6> is only in <2>

I think this is what you're after

that's how I think about it anyway, another observation is that <8> is a subgroup of <6> because the gcd of 8 and 20 is a multiple of the gcd of 6 and 20

these things all probably have proofs I'm too lazy to figure out

[small karma]

Yes but i found that method a bit long. Bcz we can use the theorem and collary to find them easily without checking every element in each subgroup

We can use this collary

[lavish gale]

is that Gallien's book?

[small karma]

Yeah

[delicate orchid]

yeah that's kinda what I was thinking

[small karma]

Contemporary abstract algebra

Joseph gallian

[delicate orchid]

I'm now wondering if my technique works for more general groups than just cyclics

[lavish gale]

right i was doing the problem u did a couple of days ago

[small karma]

I used this example

[lavish gale]

and finding subgroups generator had confused me

[small karma]

Same my teacher is so bad

[lavish gale]

dont blame ur teacher
U can learn urself

anyway gl

[small karma]

Bruh she cancelled my mid sem bcz my answer were apparently unstandard

In college

btw the ways i used are available in book. Gotta give a mid sem again

[carmine fossil]

What was your answer?

[small karma]

it was an answer for permutation group

multiplication , inverse, decomposition, and order related

[chilly ocean]

https://tenor.com/view/cat-cute-cutecat-funny-funnycat-gif-20742736

[small karma]

smh this is the one i was using then ran out of nitro

[minor badger]

@chilly ocean did you solve it?

[chilly ocean]

@minor badger Haven't been doing math.

So no.

[daring ibex]

Okay one q

Define this to be the case

And i want to consider the set ${t^{-1} x_a t = x_{\gamma(a)} \mid a \in A}$ where $\gamma$ is an isomorphism between two subgroups of $G$

[cloud walrus]

chmirppalocks

[daring ibex]

is the following reasoning correct?

${t^{-1} x_a t = x_{\gamma(a)} \mid a \in A} = {t^{-1} x_{a^{-1}} t = x_{\gamma(a)^{-1}} \mid a^{-1} \in A} = {t x_{a} t^{-1} = x_{\gamma(a)} \mid a \in A}$

[cloud walrus]

chmirppalocks

[daring ibex]

i feel a bit iffy about this, but i dont see an error anywhere

so what this is saying is imposing the relations $x_at = tx_{\gamma(a)}$ is equivalent to imposing $tx_a = x_{\gamma(a)}t$ (for all $a \in A$) and this just doesn't sit right with me

[cloud walrus]

chmirppalocks

[daring ibex]

oh bruh yeah i see it, i made a computatoin mistake

tfw literally genius

[vestal snow]

This feels like its sweeping something under the rug

We can't have a strictly decreasing infinite chain of closed subsets, but we can have a strictly decreasing chain of arbitrary finite size

[plain hill]

Maybe the keyword you missed is irreducible ?

[vestal snow]

Does that make a difference?

My problem is with the "choose Y_i, then choose Y_{i+1}, then..." argument

The logic seems shaky

[plain hill]

Ah I see your point I think

If Yi is in the set, it isn't irreducible, as a particular case

So it can be written as union of 2 closed Yi+1 and Y'i+1

now one of those as to be in our set, else you just write both as union of irreducible

Let's say it's Yi+1, then the Yi sequence is descend so it stops,

Sorry for the mistakes and sloppiness I'm getting tired

[next obsidian]

Noetherian induction can be done for general posets

[plain hill]

does it help you a bit ?

[next obsidian]

Unironically, I think seeing the technique in generality shows why it’s legit

[vestal snow]

Did you have suspicions about this proof when you saw it too?

[next obsidian]

No, because I was shown Noetherian induction before

[hidden haven]

What are the suspicions exactly?

[next obsidian]

https://encyclopediaofmath.org/wiki/Noetherian_induction

I think this does a good job of explaining it, you need a poset which has minimal elements

Then you can argue that if a set F is such that for any a in F, there’s a b < a so that b is in F, then F is empty

This is rigorous because if it was non-empty you’d get an infinite descending chain

So you can set F = the set of counterexamples and then go from there

I’m not 100% sure if that proof frames it that way, I think by taking contrapositives you can frame it a different way but this is the one that makes sense to me

[next obsidian]

Okay wait every subset needs a minimal element. This is a well-ordering

[vestal snow]

We know that there is no infinite chain Y_1 > Y_2 > Y_3 > ... of closed sets. The proof says is that if we assume that there are closed irreducible sets that are not the finite union of closed irreducibles, then there is such a set Y_1. If Y_1 does not strictly contain any such set, we're done. If it does, call it Y_2 and repeat the process. At any n^th repetition, we will have a strict chain of size n (which is allowed). I don't see how the proof concludes from this.

[plain hill]

I have to say I don't really agree because I use it geometrically and see it this way also. But on the other hand having a clean formal proof as a first approach can clean up the essential argument

[next obsidian]

It’s equivalent to the descending chain condition

[hidden haven]

Axiom of dependent choice

[vestal snow]

To add more to that, the "choosing" part of the proof is not obvious to me

[hidden haven]

If you can always extend a sequence by 1, you can get an infinite sequence

It's a corollary of the usual axiom of choice

[next obsidian]

Well the choice is just from assuming that it has no strict subset also satisfying it

[hidden haven]

The question is that induction only gives you longer and longer finite chains, but never an infinite chain

[next obsidian]

Right

[hidden haven]

You need to use choice to get the infinite chain

[next obsidian]

He was also talki about how you choose Y_n+1

From Y_n

[hidden haven]

Ah yeah

[plain hill]

Do we have the same definition of irreductible ?

[vestal snow]

No I know how to choose it

[next obsidian]

Oh okay

[vestal snow]

Chmoldilocks sums up my problem with the proof

So the argument is basically this

We order chains with the relation C_1 > C_2 iff C_1 has more elements in the chain than C_2 (and the position of the common elements is the same)

[next obsidian]

You could recursively make an infinitely descending chain

[vestal snow]

And then this relation is bounded above

so it has a maximal element

[next obsidian]

I mean if at each step you believe you can products Y_n+1

Then Y_1 > Y_2 >... is an infinitely descending chain right?

[hidden haven]

Yes but that choice at each step isn't unique

And you are sequentially making infinitely many choices

Axiom of choice makes infinitely many choices only simultaneously

[vestal snow]

Proving this seems to be a bit tricky

Let me give it a shot

[hidden haven]

But that also gives you corollary of infinitely many sequential choices

You can change you relation slightly

C_1 < C_2 iff C_1 is an initial segment of C_2

Instead of just C_1 embedding into C_2

[vestal snow]

Yeah that's what I wanted to say

Like (A < B < C) < (A<B<C<D) but not (B<C) < (A<B<C<D)

[hidden haven]

Yes

[vestal snow]

And we require the chains to be finite right?

[hidden haven]

No, include infinite

[vestal snow]

The maximal element might be an infinite chain then

[hidden haven]

I mean it will still work in this case

[hidden haven]

Yeah, and that will contradict noetherian

[vestal snow]

😮

Ah so we want the chains to be strict right?

[hidden haven]

Yes

[vestal snow]

Got it

[plain hill]

I think the little argument I gave can give a quite simple proof and explains the argument. I would be okay writing it up a bit better if that can help in any way 👐

[vestal snow]

Yeah I definitely don't think this is something that shouldn't be at least mentioned in the proof

It's not very difficult, but it is important to the argument

And since it depends on the axiom of choice, it is not very intuitive

[hidden haven]

Well most proofs skip mentions of choice completely because it feels so intuitive

And dependent choice is especially deceptive because it often feels very easily provable from induction

[vestal snow]

My brain must hate AOC then

[next obsidian]

Tfw I literally didn’t understand the issue. Dependent choice zzzzzzzz

[hidden haven]

Lol I remember struggling for a few days with why dependent choice isn't a corollary of induction

[next obsidian]

I mean I understood the part that’s the sticking point

I just didn’t get why it’s a sticking point

[vestal snow]

Your intuition is too good for this problem Chmonkey

[next obsidian]

It’s just too obvious to me that DCC is equivalent to being well-founded

And then you can frame Noetherian induction as even just being like minimal counterexample

But the DCC => well-founded requires dependent choice

[vestal snow]

I have no idea what well founded means

[hidden haven]

Every subset has a least element

[next obsidian]

It just means every nonempty subset has a minimal element

Haha you forgot nonempty OWNED

[hidden haven]

Well orderings are well founded total orderings

Oh yeah I also shouldn't have said least

Only minimal

[next obsidian]

Cringe!

[hidden haven]

[vestal snow]

Now I can finally sleep

[hidden haven]

No longer Chmoldilocks

[next obsidian]

No!

You know

[hidden haven]

[next obsidian]

You had an issue with something choicey not being obvious

That says something about you

O_O

[vestal snow]

Uh oH

[sturdy marsh]

O_O

[hidden haven]

It's not direct from choice either tho xD

[vestal snow]

What might that be exactly?

[hidden haven]

Hello fellow choice denier

[next obsidian]

It means you’re annoying in class

Jkjk

[sturdy marsh]

worrying about set-theoretic issues is cringe /s

[next obsidian]

Nah my stance is

Worry about them when they matter

[hidden haven]

Thanks, that /s encourages me to worry more

[next obsidian]

But choice issues don’t matter

For example the following version of a global Zorn’s is NOT okay

[sturdy marsh]

I use things stronger than choice

[vestal snow]

I think that mentioning something like (this is dependent on AOC) would have been nice

[sturdy marsh]

do zorn on things which arent sets

[next obsidian]

Given a class S, if every set-sized chain in S has an upper bound then S has a maximal element

This is not okay

It’s okay if you go to class-sized chains

But not for set-sized ones sad

[hidden haven]

Maximal set

[next obsidian]

Exactly

[sturdy marsh]

I believe in the maximal set

[next obsidian]

[sturdy marsh]

new religion

[hidden haven]

[next obsidian]

I used this strength of Zorn’s on a hw once

💤

[vestal snow]

Brb gonna rewrite algebra without assuming choice

[hidden haven]

Tfw no cardinality

[next obsidian]

Maximal module

[sturdy marsh]

[hidden haven]

[next obsidian]

You know it took me a bit to see why global Zorn’s doesn’t result in maximal module

Then I remember

Module is a set

[vestal snow]

🦀 🦀 MAXIMAL IDEALS ARE GONE!!! 🦀 🦀

[hidden haven]

[next obsidian]

You can get prime ideals

With something weaker

I think like the umm compactness theorem

Or some shit

[hidden haven]

Weaker version of choice right?

[next obsidian]

Maybe it’s equivalent

[hidden haven]

hm

[next obsidian]

Maybe dependent choice or some shit

Dependent choice is weaker right?

I don’t know set theory

[hidden haven]

Yes

AC → ADC → ACC

[sturdy marsh]

ACDC

[vestal snow]

Is the sequence exact tho?

[sturdy marsh]

gottem

set theorists btfo

[next obsidian]

What’s acc

[hidden haven]

Countable choice

Mathoverflow says existence of prime ideals is equivalent to the boolean ideal problem

[next obsidian]

Ahhh

ACC -> AFC too

[sturdy marsh]

but what about ABC

[hidden haven]

Which is known to be weaker the choice

But not provable in ZF still

[hidden haven]

That one's extremely weak, I think a kindergartener might be able to understand it

[sturdy marsh]

Mochizuki in shambles

[hidden haven]

whomst

[vestal snow]

Moldilocks is now an academic outcast

[hidden haven]

I thought you were making a joke on the alphabet lmao

[sturdy marsh]

[tough raven]

It wasn't?

Maybe it's the Axiom of Binary Choice ...

[hidden haven]

ABC conjecture?

[young coyote]

I'm working through Abstract Algebra - An Introduction, 3E [Hungerford] this summer. Anyone's welcome to join me 🙂

[green stone]

has anyone here ever studied the Mathieu groups or any sporadic simple groups for that matter?

[chilly ocean]

Anyone know how to prove $a\cdot{b}=1$ in a field with 5 elements? I'm struggling because $a\cdot{b}=c$ or $a\cdot{b}=1$ since all other cases cause contradictions but how come $a\cdot{b}=1$ and not the former?

[cloud walrus]

Reaper

[thorn delta]

i mean, u can just check manually

[stone fulcrum]

It's not true that multiplying any two elements in F5 will give 1. Am I misunderstanding the problem?

[chilly ocean]

?

Its one of the two

How are you certain that those two do not produce 1?

[stone fulcrum]

Like 1•3 = 3

[chilly ocean]

There has to exist a multiplicative inverse in the field

[stone fulcrum]

Using 0,1,2,3,4 as F5

1•1 = 1
3•2 = 1
4•4 = 1

[chilly ocean]

Yeah

where 3 is b and a is 2

But im struggling on how to prove that

[hidden haven]

a and b are fixed elements?

[stone fulcrum]

Okay so a and b are multiplicative inverses then?

[chilly ocean]

My field is F={0, 1, a, b, c}

yes

a,b,c are fixed elements

just placeholders

[stone fulcrum]

I mean I could make a and c inverses instead

That would be the same field

I guess I'm trying to say that it feels like you're missing info haha

[chilly ocean]

What info am I missing?

[hidden haven]

So how exactly are you defining F_5

It is the set {0,1,a,b,c}

[chilly ocean]

Yes

[hidden haven]

But what else do you have in the definition

[chilly ocean]

with the usual field axioms

[hidden haven]

Because right now it's not a field

You have to define addition and multiplication

[stone fulcrum]

I can take that set, and turn it into a field, such that ab ≠ 1.

[chilly ocean]

What

then what would ab=?

in that field

[chilly ocean]

It just follows the regular field axioms

[hidden haven]

It could be c

No

[chilly ocean]

So does it matter whether ab=c or ab=1?

[hidden haven]

Field axioms do not define unique operations on this set

[chilly ocean]

it will still be a field regardless?

[hidden haven]

A lot of other things need to also multiply correctly for it to be a field

But yes both are possible

[chilly ocean]

Sorry

I should give you the full picture

I am trying to construct an addition and multiplication table of a field with 5 elements

[hidden haven]

ok I see

So you already know any fields with 5 elements?

[chilly ocean]

Well

in the course they made a field with four elements

which was F={0, 1, x, y}

so I decided to make mine F={0, 1, a, b, c}

[hidden haven]

Yeah but we need 5

Do you know when Z/nZ is a field?

[chilly ocean]

when n is prime then it is a field?

or something along those lines

I haven't taken abstract algebra yet

[hidden haven]

Ah damn

[chilly ocean]

It's for a different course, no clue why we're doing fields

[hidden haven]

This is correct yes

[chilly ocean]

oh, it should be iff

[stone fulcrum]

Not iff since there's a field of 4 elements

[hidden haven]

But without that knowledge this seems like you just have to smartly brute force it? Idk

[hidden haven]

That isn't Z/4Z

[stone fulcrum]

Oh misread mb kek

[chilly ocean]

Yeah I do

[stone fulcrum]

I will slink away again

[chilly ocean]

I arrived at the possibility of ab=1 or ab=c

but, does it matter which one i choose?

[hidden haven]

Ye both work lol

[chilly ocean]

Ok ok

[hidden haven]

Choose one and go with it

[chilly ocean]

Thank you

😊

[hidden haven]

[chilly ocean]

Wait

if i decide

to go with ab=1

will that affect future computations?

nvm ill figure that out

[hidden haven]

Yes it will

You'll get different tables for the 2 choices

(but they'll become the same after appropriately renaming a,b,c)

[chilly ocean]

another i saw was that a^2=b or a^2=c

oh wait

[hidden haven]

I think you should start with addition

And figure out what 1+...+1 is for a certain number of 1s

And then multiplication uses distributivity

[chilly ocean]

Alright

[stone fulcrum]

Something something isomorphism

We don't really talk much about field isomorphism do we?

[hidden haven]

We do

All of galois theory is about studying field automorphisms lol

[stone fulcrum]

Oh ya

slink

[hidden haven]

[chilly ocean]

Even in the addition table a+1=b or a+1=c do we also just choose one?

[hidden haven]

first just handle sums of the form 1+...+1

[chilly ocean]

?

I'm confused

[hidden haven]

like what should 1+1 be

[chilly ocean]

in this field?

[hidden haven]

yes

[chilly ocean]

it could be a, b or c

[hidden haven]

right

so pick one

[chilly ocean]

Ohhhhh

why did you start from there??

[hidden haven]

then decide that + 1

[chilly ocean]

and not like a+1 or a+a

[hidden haven]

do that and you will see

multiplication becomes really easy to define

[chilly ocean]

^

[hidden haven]

what will be 1*1

[chilly ocean]

1?

[hidden haven]

why?

[chilly ocean]

I dont know

[hidden haven]

what do the field axioms say about 1?

[chilly ocean]

Oh

im dumb

any element in a field times 1 is itself

[hidden haven]

exactly

[chilly ocean]

so 1*1 = 1

[hidden haven]

yes

so now if you can figure out what adding 1 does

you will know how to multiply things

because you know how to multiply 2 1s

and then you can apply distributivity

[chilly ocean]

Ohhh

because 1=a-1?

[hidden haven]

which you have as a field axiom

if you set a = 1+1 yes

[chilly ocean]

OHHHHH

[hidden haven]

so for example with a = 1+1

what is a^2?

[chilly ocean]

(1+1)^2?

mm

wait

It can't be 4

4 is not in the field

[hidden haven]

yes

apply distributivity on that

to compute the square

[chilly ocean]

(1+1)(1+1) => 11 + 11 + 11 + 11

[hidden haven]

yes

[chilly ocean]

(1 + 1)+(1+1)

a+a?

[hidden haven]

so once you know what 1+1+1+1 is, you will know what a^2 is

[hidden haven]

yes

thats why you start with figuring out how adding 1 works in the field

so figure out what 1+1, 1+1+1, ... are

[chilly ocean]

Ohhh

i see

oh and from the addition table if say, a+a=c then a^2=c?

[hidden haven]

that will only work if 1+1 = a

[chilly ocean]

but whos to say a+a can't be b?

then wouldn't a^2=b?

Is that another one of those things you just choose?

[hidden haven]

so what all have you filled in the addition table so far

[chilly ocean]

not much 🙁

oh

[hidden haven]

did you draw a table?

[chilly ocean]

Yes

I just have the trivial stuff down

like +0

and 1+1=a

[hidden haven]

right

[chilly ocean]

oh

we need to know 1+1+1

[hidden haven]

that gives you 1 row and 1 column

and then you have put 1+1 = a

notice that b and c so far have the exact same properties

they are not distinguishable, other than in name

right?

[chilly ocean]

they are distinct

[hidden haven]

how?

I mean in terms of behaviour

[chilly ocean]

Oh

No

[hidden haven]

right

[chilly ocean]

just unique

[hidden haven]

so now you need to define 1+1+1

[chilly ocean]

for 1+a?

[hidden haven]

thats what it will become, but the reason is not that

like in the beginning I said that you should start by figuring out what 1+1, 1+1+1, 1+1+1+1,... are

[chilly ocean]

I don't fully understand why

How does that correlate to the addition table

[hidden haven]

because you can multiply any 2 such elements

for example what is (1+1+1)(1+1+1+1)?

[chilly ocean]

well

11 + 11 + 11 + 11 + 11 + 11 + 1*1...

it would be like

12 of those

[hidden haven]

yes

exactly

so you see that if you can figure out these elements, then you already have multiplication of these elements

and heres the great thing about these elements

since you are asked to construct a field with a prime number of elements

each element of the field will be one of those

so as soon as you know what those elements are

you know what their multiplication table is

so you know what the multiplication table of the whole field is

[chilly ocean]

Okay

I think i get it

[hidden haven]

nice

[chilly ocean]

but this would just turn into 6a or something

and how does that help me

[hidden haven]

you dont have a 6 in your field

[chilly ocean]

oh

right

it would be

a+a+a+a+a+a

[hidden haven]

it will just be 1+1+1+1+1+1+1+1+1+1+1+1

dont write it in terms of a

leave it in 1s

[chilly ocean]

ok

[hidden haven]

but this is why I asked you to figure out what sums of 1s look like

once you know that

you can figure out what that big sum of 1s is

(it will turn out to be a)

[chilly ocean]

but all i know is that 1+1=a and a+1=a+a

[hidden haven]

how did you get the second thing?

and yes so far you dont know what I said

[chilly ocean]

oh sorry

a^2=a+a

[hidden haven]

so far you already know how to add 0 (on either side) and 1+1=a

right, but forget about multiplication for now

[chilly ocean]

ok ok

[hidden haven]

so you know 1+1

next step is to figure out 1+1+1

notice im not bracketing this

can you tell me why?

[chilly ocean]

its one element

[hidden haven]

no, i mean how does 1+1+1 make sense, when addition is only defined between 2 elements

im writing 3 elements here

[chilly ocean]

Oh

[hidden haven]

so im asking you to figure out what this could mean'

[chilly ocean]

hmmm

it doesn't make sense?

[hidden haven]

(think of numbers. if you can add 2 numbers, how do you then move on to adding 3 numbers?)

[chilly ocean]

you would first have to add the first 2

then add the third

[hidden haven]

exactly

[chilly ocean]

(1+1)+1

[hidden haven]

yes

so when I write 1+1+1 it is really abuse of notation. I am omitting brackets

[chilly ocean]

then ((1+1)+1)+1?

[hidden haven]

yes

but now I want you to justify my omission of brackets

1+1+1 could also have meant 1+(1+1)

[chilly ocean]

Its the same either way no?

[hidden haven]

yes, why?

[chilly ocean]

because of commutativity?

[hidden haven]

this is associativity

[chilly ocean]

Oh

yeah

[hidden haven]

commutativity is x+y = y+x

[chilly ocean]

because elements are unique?

[hidden haven]

wdym?

[chilly ocean]

nvm

[hidden haven]

so when I say define 1+1+1

I am actually asking you to define 2 things

(1+1)+1 and 1+(1+1)

just that since you want associativity, you will have to define them to be the same

right?

[chilly ocean]

Yeah

they should be the same thing

[hidden haven]

and notice that the 2 things are actually a+1 and 1+a

[chilly ocean]

yes they are

[hidden haven]

now what values can you give a+1

[stone fulcrum]

For example / as division isn't associative. If you write 3/2/1 mathematicians will cry

[hidden haven]

and what values can you not give to a+1

[chilly ocean]

hmmm

ok

im gonna come back to this later

its too late for me now to keep going

its 5am where i am

[hidden haven]

weirdly enough, in that case it doesnt matter

[hidden haven]

oof alright

[chilly ocean]

ill be back tomorrow 😈

[hidden haven]

[final pasture]

"weirdly" "/1" not that weird tbh, doing nothing is associative

[hidden haven]

Chshut up

[final pasture]

[zinc cloud]

im trying to think of a commutative ring with an ideal that is not principle

but i just cannot think of one

anyone got any examples

[hidden haven]

Z[x] has (x,2) which is not principle

[zinc cloud]

whats (x,2) mean i havent seeen that notation before

[hidden haven]

ideal generated by (ie smallest ideal containing) x and 2. It will turn out to be the set of all polynomials which have even constant term

[zinc cloud]

ooh is this the same as saying 2 Z[x] + x Z[x]?

[hidden haven]

yes

[zinc cloud]

gucci

[hidden haven]

(S) is notation for ideal generated by S

[zinc cloud]

oh thats cool

i havent seen it like that

[hidden haven]

kinda annoying sometimes

given that () are also used in other ways

I lost so many marks to one such ambiguity once

[zinc cloud]

yeah maybe thats why we a taught like this instead so its easier to understand, i reckon the 3rd year courses they wil use that notation

but im not planning

on venturing into that

lol

[hidden haven]

yeah depends on the course text

[fallow depot]

How would you prove that it’s an ideal and not a principal ?

0 € 2Z[x] + xZ[x]

a > -a in 2Z[x] + xZ[x] as -a in Z[x]?

[zinc cloud]

thats what im trying to work out now

lol

[hidden haven]

||suppose f(x) generates it. Then f(x) must have degree 0, otherwise the ideal it generates wont contain 2. But then f(x) is some even integer and x cant be in the ideal generated by f(x) because x is not a multiple of any even integer||

[zinc cloud]

for proving that its an ideal || could you set up a ring homomorphism from z[x] --> z/2z and then show its also a ring isomporhism with kernel of 2z[x] +x[x] ||

[hidden haven]

yes

||every polynomial maps to its constant term mod 2||

is ideal defined as kernel of a homeomorphism for you?

[zinc cloud]

nono, its not, we have it has the kernel of a homomorphism in an ideal

the ideal has its own definiton but i wouldnt know how to apply that directly to this question so im using this isntead

[hidden haven]

it has 2 equivalent definitions. A subset is an ideal iff it is the kernel of some ring homomorphism iff it is an additive subgroup of the ring which absorbs multiplicationfrom any element

both can be done here

[fallow depot]

what function should your homomorphism map be defined by and how can you show that 2Z[x] + xZ[x] is the kernel ?

[hidden haven]

this, to Z/2Z

[fallow depot]

okay thanks

does anyone know how to express a number x as a product of irreducible elements of the ring Z[i], Z[i] is the Gaussian integers

[delicate bloom]

pretty much the same way you'd represent a number x as a product of irreducible elements in Z

I suppose you're really asking, what are the irreducible elements in Z[i]

[scarlet estuary]

(which is a surprisingly tricky question)

https://math.stackexchange.com/questions/1169065/all-irreducible-elements-in-gaussian-integers

obviously that link spoils the answer

might want to try it yourself firsst

but its not obvious at a glance

[fallow depot]

yh what are the irreducible elements

and fro example how i would i represent 21 as a product of irreducible elements in Z[i]

ill give it a read

[delicate bloom]

funnily enough just as 3*7

but you just picked an unlucky example lol

do you know about the norm map yet?

[fallow depot]

okay 37?

N(a +ib) = a^2 +b^2 ?

[carmine fossil]

Yes, that

[delicate bloom]

yeah, we can use this to determine what the primes are

it turns out the primes that split are the ones that are 1 mod 4, ones that are 3 mod 4 don't

and 2 is special too

[fallow depot]

so first find the primes

[delicate bloom]

I'd say first determine what the units are

since that's important too

[fallow depot]

the units of Z[i] are just 1 and -1 right ?

[delicate bloom]

no

[fallow depot]

oh what

[carmine fossil]

See i

Do you think we have any more units?

[delicate bloom]

and how do we prove that these really are all of them?

[fallow depot]

if the N(z) = 1?

[chilly ocean]

Yup z is a unit iff N(z) = 1

[fallow depot]

so 1, -1, -i

[chilly ocean]

and i I'd say

[fallow depot]

isnt that equal to -1

[chilly ocean]

N(a+bi) = |a^2+b^2|

with z=i we have a =0 and b =1

[fallow depot]

ahhhh

righ right

okay so i have all the units 1, -1, i and -i

now what

[chilly ocean]

Yeah

Are you seeking a full characterisation of irreducibles in Z[i]?

It's quite tricky ngl

[fallow depot]

no just for a single number to be expressed as a product of irreducibles in Z[i]

[chilly ocean]

What's the number ?

[fallow depot]

47

[chilly ocean]

Z[i] is a UFD so no doubt any element of Z[i] is a product of irreducibles

47 is prime

Is it a sum of two squares?

No

because 47 = 3 mod 4

[fallow depot]

it says express the number 47 as a product of irreducible elements of the ring Z[i], show that each factor of your product is irreducible

[chilly ocean]

Actually (maybe surprisingly too), 47 is an irreducible in Z[i]

I know it because I just looked at a theorem

But to check that, you have to use the norm function

[fallow depot]

It’s irreducible only if it can be expressed as a product of units ?

[chilly ocean]

No (almost) it's irreducible if whenever you write 47 as a product of two elements then one of the elements is a unit

[chilly ocean]

a product of units is a unit and an irreducible isn't one precisely

[fallow depot]

Ahhh okay

[chilly ocean]

So it can't be that both of the factors are units

[fallow depot]

It’s confusing can’t I simply write 47 = 47*1

[chilly ocean]

It seems complicated to show that 47 is irreducible using the norm because N(47) = 2209 which isn't prime and is a quite complicated number to work with

[fallow depot]

Is that any easier

[chilly ocean]

Maybe use the fact that Z[i] is a UFD and every irreducible there is a prime

Showing that 47 is prime amounts to showing that 47 is irreducible

Actually, there's a thing one can notice here

Suppose p is a prime. If p is a sum of square p = a^2+b^2 then p = (a-ib)(a+ib) so it is not irreducible since neither of the factors is a unit

That's how i knew that 37 and 47 are irreducibles themselves

[fallow depot]

As they’re primes

[chilly ocean]

Primes AND they're not sum of two squares

[chilly ocean]

if they were sum of squares then ^

[fallow depot]

To represent it as a product of irreducibles

[chilly ocean]

Its a product of one irreducible (itself only)

[fallow depot]

Because it’s prime and not sum of two squares

[chilly ocean]

Yup

[fallow depot]

Okay thanks for the help !

[chilly ocean]

Np

[fallow depot]

How do I find the units of the the ring R/I

[sturdy mirage]

That's all the polynomials in Z_2[x] with max degree 2

because f is an irreducible poly of degree 3

[fallow depot]

Nothing to do with f given

is it just [1], x and [1] + x

[sturdy mirage]

the f given matters: its irreducible in Z/2Z, and it is of degree 3

so you get a new field of 8 elements

7 of which are units (all the non-zero elements)

[fallow depot]

because 2^3, in mod 2 and deg 3

[sturdy mirage]

yes

[fallow depot]

how do i find the elements though

[sturdy mirage]

its all the 2 degree (and lower) polys with coefficients in Z/2Z

0, 1, x, x+1, x², x²+1, x²+x, x²+x+1

[fallow depot]

ahhhh okay

does that have anything to do with the coset of each R relative to I

there'll be 8 cosets

[sturdy mirage]

technically the 8 elements i wrote can be written
0 + <f(x)>
1 + <(f(x)>
x + <(f(x)>
x + 1 + <f(x)>
... and so on

[fallow depot]

okayy right

[sturdy mirage]

these are cosets in R

where <f(x)> means the set of all polynomials of the form
h(x)*f(x) , where h(x) is any polynomial

so any multiple of f(x)

[fallow depot]

all the elements in the set added w f

oh ok

the representatives for these cosets will just be the elements though

ie 0,1,x,x+1 etc

[sturdy mirage]

we're sort of splitting up the infinitely large set of polynomials with coefficients in Z/2Z into 8 disjoint sets

[fallow depot]

i see

[sturdy mirage]

yes

you can just as easily just write 1, x, x+1

or you can write [1], [x+1]

just remember that it actually means 1 + <f(x)>

[fallow depot]

if they're disjoint does that mean the group of all units are cyclic

[fallow depot]

gotcha thanks

[sturdy mirage]

this is because of the polynomial division formula which says ANY poly g(x) in our ring R can be divided by f(x):
g(x) = f(x)q(x) + r(x), where r(x) is sort of the remainder

and this remainder will always have degree 2 or less

these 8 elements are represented by this remainder

[sturdy mirage]

umm im not sure being disjoint has anything to do with that.. theyre cyclic maybe because units in any finite field is always cyclic

NOTE i am taking a course on this right now soo im no expert

[fallow depot]

oh yh the follow up is find all elements h of ring R st polynomial x*h + [1] belongs to the ideal

[fallow depot]

you seem a lot better than me lool

[sturdy mirage]

i should understand it cause i have an exam coming up soon.

[fallow depot]

good luck, im sure you'll do well !

[sturdy mirage]

i think this means that you have to find h(x) such that x*h(x) + [1] = [0], since [0] = 0 + <f(x)> = <f(x)> which are exactly the elements of the ideal

[fallow depot]

okay thank you ill give it a go

[sturdy mirage]

since theres just 8 elements you can kind of calculate each one i think 😅 thats the only way i know

for one, if h(x) = x² + 1 then
x*h + 1 = x³ + x + 1
is in <f(x)>

[zinc cloud]

why [0]?

[sinful mirage]

can someone help me fix this proof? I feel I messed up

[zinc cloud]

up here i mean

[cloud walrus]

ProphetX

[zinc cloud]

surely u would just plug each of them in and see if it is within the ideal

[sturdy mirage]

[0] is the set of polynomials of the form 0 + f(x)*g(x) , for any g(x), ie. it is a multiple of f(x), ie. it is in the ideal

right?

[zinc cloud]

I still don't get why 0 tho?

wouldn't you just see if it's in the ideal

I find anywhere that says that

[sturdy mirage]

well you have a quotient field R/I , which splits R up into 8 disjoint sets (cosets), where [0] represents all elements in I, [1] represents all elements of the form 1 + I, and so on

[zinc cloud]

...

I still don't get how u got that

[limpid edge]

I think that’s a general fact about cosets

I should learn how to prove that

[sturdy mirage]

i dont know how to.. prove it.. i thought it was just like a definition

[limpid edge]

it’s probably equivalent to the definition

[zinc cloud]

but why 0

why not x+q

x+1

surely we just have to show that gx is an element of I

[sturdy mirage]

x+1 is not a multiple of f(x) = x³ + x + 1, so that wouldn't work. I is all the multiples of f(x)

[zinc cloud]

ok but why 0 then

[sturdy mirage]

0 is a multiple of f(x)

[zinc cloud]

0 isnt a multiple

[sturdy mirage]

it is! take 0*f(x) , then you get 0