groups-rings-fields

[wicked zephyr]

so "s" is the square root of $t^2$, which is $t$

[cloud walrus]

Potitov06

[next obsidian]

This is basically the same as C (x)_R C

[wicked zephyr]

aham

[next obsidian]

Which we know is C^2

So try to mimic how you prove that

But in this new setup

[wicked zephyr]

Im sorry, but what is R here

[wicked zephyr]

here

[dim widget]

Wait guys guys I don't think you want R[s]/(s - t^2)... that's just R since t^2 is in R

You want s^2 - t^2

Which looks weird because t is not in R.

[wicked zephyr]

yes

[dim widget]

But that's the point.

[next obsidian]

Oh wait yes

You’re totally right

[wicked zephyr]

yes, minor confusion

[next obsidian]

s squared = t^2

Thanks haha

But right so R = k[t^2] and the algebra is k[t]

Which we wrote as R[s]/(s^2-t^2)

In the case of the field R, and the field C

C = R[s]/(s^2 + 1)

s “=“ i

Because it’s a square root of -1

[wicked zephyr]

^yes

[next obsidian]

And we know (or at least I do) that C (x)_R C = C^2, and maybe you know how to do that computation as well

[wicked zephyr]

Ah so R is real numbers okkk

[next obsidian]

So if you know how to do this computation, you can try to apply it to this new setup, because they look very similar

[wicked zephyr]

ok

[next obsidian]

Does that make sense?

[wicked zephyr]

i dont really know where does the (x) come form

[next obsidian]

Both are “tensoring adjoining a square root with itself”

That’s the tensor product

[wicked zephyr]

I'm trying k[t]\otimes k[t]

[next obsidian]

(x) = \otimes

[wicked zephyr]

OK

[next obsidian]

It’s just how I write the tensor product

Haha

[wicked zephyr]

Sorry haha

freaking notation

[next obsidian]

Okay I gtg

[wicked zephyr]

NOW IT GET IT

[next obsidian]

But this is the key

[wicked zephyr]

I was like: ??

$\mathbb{C}(x)_{\otimes \mathbb{R}}\mathbb{C}$

[cloud walrus]

Potitov06

[wicked zephyr]

thinking about that

[next obsidian]

Lmfaooo

[wicked zephyr]

thank you, i will now proceed trying it

[next obsidian]

Cool

[formal ermine]

can someone please explain the characteristic polynomial of an element in a field extension in a simple and intuitive way?

no but seriously, what's an intuition behind it?

[dim widget]

It is a way of generating an algebraic relation that an element satisfies using linear algebra.

[chilly radish]

It's also just a power of the minimal polynomial

It turns out

[formal ermine]

why do we care

[dim widget]

Because you can extract the norm, trace, etc. from the char poly and because you can find the degree of the extension which that element generates by checking for repeated roots.

[chilly radish]

It's also much simpler to calculate than the min poly

Since it's just linear algebra

[formal ermine]

wdym with the last part?

[chilly radish]

As long as you know a basis for the extension you can calculate it

if a is your element in an extension E/F, then the char poly is just m_a(x)^[E:F(a)]

So by seeing how many times a root repeats you can deduce the degree of F(a)/F

[charred crescent]

do y'all have an linear algebra books that you'd recommend

for proofs-based linear algebra

[chilly ocean]

the one by friedberg, insel, and spence is good

axler's book is great as long as you supplement it with some material on systems of equations and determinants

axler's got a weird philosophy about the whole subject that results in a really strange treatment of some of the most important stuff

[long geyser]

I'd also vouch for hoffman & kunze and halmos

[charred crescent]

of these suggestions, are there any in particular that are good for self-study?

[long geyser]

halmos and axler wrote their books with functional analysis in mind, while h&k is more algebraic

[long geyser]

probably not, but as someone who self-studied linalg using more than 1 resource helped me personally, so my advice would be to not confine yourself to just one of these resources

but everyone learns differently

[charred crescent]

okay those are good points

and btw is this the channel to ask about proof-based linalg q's or is there a better one for that

[long geyser]

I think generally #linear-algebra is the better place

[charred crescent]

sweet, thanks

[solar shore]

did my test today y’all, went well

hoping for a good score but i lw demolished it feelin good rn

[warm ember]

is there any relation between M and R

if not i dont understand

btw is this wrong channel for this

[agile pine]

this is the right channel

[chilly ocean]

it's the right channel for the first question and possibly the wrong channel for the second

[warm ember]

o

#linear-algebra ?

[agile pine]

M and R are related in the sense that you have the existence of the bilinear scalar multiplication map R x M -> M

[warm ember]

M not necessarily subgroup of R?

[agile pine]

nope

[warm ember]

oh ok

[warm ember]

what about this

[agile pine]

yeah as objects M and R need not be related in such a way

[chilly ocean]

go through the axioms and find out which one fails

[agile pine]

^

[chilly ocean]

it should be straightforward

[warm ember]

ok

[chilly ocean]

well, no, it is straightforward

start with the simpler ones

[warm ember]

theres no identity right

[chilly ocean]

[warm ember]

ye theres no identity

ty

[agile pine]

the napkin is a nice text you're in good hands

[warm ember]

orz

[north parrot]

If I have a base field F, and quadratic extensions K and L, and through black magic I got that K x K is isomorphic to L x L, is K isomorphic to L? I feel like the answer to this is no (we cannot have nice things) but I am struggling to come up with a counter example, does anyone have any insight? Thanks!

[vestal snow]

Can someone help me out with 2?

[next obsidian]

I believe you want to show that there are infinitely many intermediary sub fields

And show that this contradicts being a primitive extension

[white oxide]

yo this might be too general/off-topic, but a few days ago i emailed my professor asking for research opportunities over the paper and she responded that she didn't have any projects for undergrads to work on, but i could possibly read a few papers. she asked me what my interest is - how specific should i be? i've really enjoyed algebra as a whole, and i suppose field theory is very interesting (i haven't even learned any galois theory yet lmfao) but i don't know if that's specific enough orrr

[upper pivot]

its best to give as much information as you can. I did the same to one of my prof and we just ended up doing a reading course instead of research which is fine right

you dont wanna end up in a project where you cant contribute oof

[white oxide]

yeah true but tbh like i know of no ongoing projects that are like accessible to someone who is like my level of math lol

which is pretty low

like in textbooks all i see are oh we're still working on the classification of nonabelian groups

and all of these open problems

that obviously i can't work on

[upper pivot]

right, so maybe the priority is like, building up the knowledge

[white oxide]

so i have no idea where to start

yeah true

[upper pivot]

and doing reading courses with profs are usually also good for networking+ grad apps

[white oxide]

yea that's the reason i emailed in the first place LOL

[upper pivot]

hmm right so does your school have like

grad algebra sequence?

[white oxide]

as in like classes?

[upper pivot]

ye

[white oxide]

i'm taking a graduate level algebra class next sem

[upper pivot]

i see, right you could still do a reading course with em but maybe after the grad algebra sequence? idk if thats viable bc that depends on the year. You could do the reading course now too

[white oxide]

yeah i'm a second year, would prefer to do it maybe over the summer cuz imma apply to some research shit this fall

[upper pivot]

right i see, yeah REUs have like dedicated projects accessible to ugs so thats good

[white oxide]

yeah i was gonna do a REU

[upper pivot]

so I would say do the grad algebra sequence next year, maybe the 2nd semester of next year do the reading course side by side

but ofcourse your professor will be able to tell you better what to do here, so tell her the details you just told me

that you are a 2nd year right now, and that you are taking grad algebra next year, and etc

and maybe she will tell you "ok finish this first than you can read this with me" etc

[white oxide]

ah yeah i suppose that's true, but this was the first algebra class i'm taking and she's saying that i can do some reading i suppose so i'm hoping to get to work with her asap

what are like some examples of reading that profs will have undergrads do?

like i'm a little confused

as in to specialized research topics or just

"advanced" material

[upper pivot]

the latter in my experience

like yall will choose some topic and some source on the topic and read it and meet weekly etc

maybe she will involve a few other students (I usually prefer this as I am a social learner)

[white oxide]

oh yeah maybe

can i just say like

i'm interested in galois theory?

or is that too general

[upper pivot]

sure

but i suspect she will tell you to do the grad algebra sequence first which will undoubtedly teach it

but after that you can go into more depths of it

[upper pivot]

Undergrads usually cannot narrow down further than this so this is fine btw

[white oxide]

oh yeah that's totally fine

i'll also be studying some grad level algebra over the summer

[white oxide]

ok ok thx

thank you!

[upper pivot]

mhm

[frigid lark]

Lol, both questions from Lang

[novel parrot]

How would i show that Z[X]/(X^2 - 7) is a domain?

directly or by (X^2 -7) is prime

[carmine fossil]

Well X^2-7 generates a maximal ideal

So you get that is a field

[novel parrot]

well ill need to prove that

how can i prove that (X^2 - 7) is a maximal ideal?

showing that (X^2 - 7, f(X)) = Z[X] or (X^2 - 7) ?

[frigid lark]

I didn't do it this way. I think you can show that [k(x,y) : k(x^(1/p),y^(1/p)] = p^2, and that for any a in k(x^(1/p),y^(1/p)), [k(a) : k(x^(1/p),y^(1/p)] leq p.

[carmine fossil]

Ok, so you know (X^2-7) is not Z[X]

[frigid lark]

Do you have Eisenstein?

[carmine fossil]

Consider an element f not in (X^2-7), then you get remainder of f is in new ideal generated by f along with X^2-7

[novel parrot]

yes

its irreducible

[frigid lark]

yeah that's equivalent to prime here

[novel parrot]

ah

yes

mb

[novel parrot]

the remainder would be a deg 1 polynomial

so someting like (X^2 - 7, aX + b)

[slim kayak]

A geometric group theory question ig: I am given a set of non-trivial words $A = { g_{1}, ... , g_{n}) }$ of F(X) with the property that for $\epsilon_1, \epsilon_{2} \in {-1,1}$, that $| g^{\epsilon_{1}}{i} g^{\epsilon{2}}{j} | > max(| g^{\epsilon{1}}{i} |, | g^{\epsilon{2}}{j} |)$ holds whenever i is not equal to j or if i=j and eps1=eps2. The bars denote the reduced word length. I am supposed to show that these elements generate a subgroup isomorphic to $F{n}$. My idea was to use the ping-pong lemma using the set consisting out of reduced words obtained from finite sequences of elements in A such that two neighbouring words aren't inverses of each other. Then one forms $M^{\pm}{i}$ as the subset of reduced words whose equivalence class is the same as the sequences that start with $g^{\pm}{i}$ as the first word in the sequence. I don't know how to show that these sets would actually be disjoint from each other and if they aren't, what other idea to try.

[cloud walrus]

Kerr

[novel parrot]

we would want to show that this is Z[X] or (X^2 - 7)

[carmine fossil]

I don't think this would work for Z[x]

[novel parrot]

but its not (X^2 - 7)

[slim kayak]

What is a condition that given some ideal A of a ring R, that R/A is a domain?

[novel parrot]

yes needs to be prime ideal

i want to show that

showing its irreducible works

[slim kayak]

Right, and Z[X] is factorial so prime elements and irreducible elements are the same.

[novel parrot]

but i wanted to learn the other method that @carmine fossil was showing

[carmine fossil]

nvm that method won't work

I was assuming (X^2-7,aX+b) would give you (1)

Doesn't always happen

[novel parrot]

(X^2 - 7) isnt maximal right?

[carmine fossil]

Yeah it's not

[slim kayak]

Did you assume Z[X] was a PID for that argument?

[novel parrot]

Z[X] isnt euclidean right

[frigid lark]

Noether was a principle figure of algebra, therefore Noetherian is Principle

QED

[carmine fossil]

Well I thought Z was Q( that is I assumed all elements in Z are units)

Biggest mistake

[slim kayak]

Ah, just a localization away its probably fine

[novel parrot]

how would i show that (X^2 - 11) is a maximal ideal? in Z[X]

by showing that (X^2 - 11, aX + b) = (1) ?

[slim kayak]

If it was Z would be a field, so no.

[novel parrot]

oh

we would always divide monic polynomials?

[slim kayak]

whoops, meant to be a ">" instead

[void cosmos]

suppose that it is not , it is contained in some other ideal ( not necessairly principal ) and then arrive at a contradiction

[fervent gyro]

During a proof, I saw a reasoning that went: "If K (the residue field of a local ring A) is a free A-module, then A must be a field" and I'm having trouble seeing why that is the case, or finding such a result online

[novel parrot]

how exactly?

[void cosmos]

are u sure it is even maximal

[upper pivot]

if its contained in some bigger ideal, say I, then I contains an element not in (X^2-11). Try using that to show I has to be the whole ring basically.

[novel parrot]

no not really

but i assume it is lol

[wraith cargo]

ermmm it's maximal among principal ideals

I'm not sure it's principal generally

[novel parrot]

idc about principal

[void cosmos]

the ideals that would contain (x^2-11) would be ideals of Z[sqrt(11)]

[novel parrot]

so if an ideal contains I

[upper pivot]

actually yeah disregard what i said

[novel parrot]

oh

[upper pivot]

I forgot bezout doesnt work in Z[x] lol

[wraith cargo]

A better way to check if it's maximal is to see if Z[X]/(X^2-11) is a field

[novel parrot]

the question says to prove this

[void cosmos]

yes ^

[novel parrot]

but i thought showing maximal would be easier

ill try that

[wraith cargo]

Does it ask to check if it's a field or to prove it's a field?

[novel parrot]

It says "Is Z[X](X^2 - 11) a field?"

[void cosmos]

😄

[novel parrot]

lmao

[void cosmos]

it is not

[novel parrot]

oh ..

[void cosmos]

do you see why

[novel parrot]

ill need to find an element that isnt unit

right

[void cosmos]

yes

[novel parrot]

or could showing that the ideal is not maximal could be easier?

[void cosmos]

do you see how this is the same as Z[sqrt(11)]?

[wraith cargo]

finding a non unit is easier

[novel parrot]

its a root

[void cosmos]

yes and you can just use the natural map to find ur isomorphism

[novel parrot]

oh

[void cosmos]

now can u find an element that cant be inverted

in Z[sqrt(11)]

if u can then thats ur proof

that it is not a field.

[novel parrot]

ill try this

[void cosmos]

hence (x^2-11) is not maximal in Z[x]

maximal ideals in Z[x] always have the form (p,f(x)) where f(x) is irreducible mod p

with p prime

[novel parrot]

the element that maps to sqrt(11) is not unit right

[wraith cargo]

Prove it!

prove that sqrt(11) doesn't have an inverse (this isn't too difficult)

[novel parrot]

ok

otherwise Z[sqrt(11)] = Z

[wraith cargo]

?

[void cosmos]

what

how did u do that

[novel parrot]

nvm lol

one sec

[void cosmos]

just suppose it does

suppose there exists an element b in Z[sqrt(11)) such that sqrt(11)*b = 1

[wraith cargo]

Write 1=(a+b sqrt(11)) sqrt(11)

[void cosmos]

b is in Z[sqrt(11)] so how does it look like

and then simplify

[wraith cargo]

and show that a + b sqrt(11) isn't in Z[sqrt(11)]

[void cosmos]

^

[novel parrot]

kk

first i prove that Z[X]/(X^2 - 11) = Z[sqrt(11)]

[slim kayak]

Hm, why is it?

The kernel of the evaluation map isn't necessarily principal here

[novel parrot]

mapping X to sqrt(11) gives isomorphism

with kernel (X^2 - 11)

$\sqrt(11) * (a + b\sqrt(11)) = 11b +a\sqrt(11)$

[cloud walrus]

ActiveChapter

[novel parrot]

and that cant be 1

[wraith cargo]

yeah

b would have to be rational

[novel parrot]

it cant be 1 because we can never cancell that sqrt(11)

yeah

[frigid lark]

I think this is a restriction of the isomorphism b/w Q[x]/(X^2 - 11) and Q(sqrt(11))

[novel parrot]

whats an example of a principal ideal in Z[X] that is maximal?

[slim kayak]

Ah, that makes sense then.

(2)?

[frigid lark]

(57)?

[novel parrot]

sorry principal with X

degreee atleast 1

[wraith cargo]

(x-2)

tho wait

hmmm

[upper pivot]

its as momen said, the maximal ideals are (f(x),p) where f(x) is irreducible mod p

try showing that these are maximal idea.

[wraith cargo]

this would just give us Z which is a PID not a field so this doesn't work

[novel parrot]

i se

[slim kayak]

I guess you could just add x to my and parrot's example though, (2,x) and (57,x) containing (2) and (57) properly.

[novel parrot]

yeah

[void cosmos]

(2,x) is not prinicipal sadly

[novel parrot]

is there principal ideals that are maximal?

[void cosmos]

no

[novel parrot]

none?

[void cosmos]

this argument we did

with Z[sqrt(11)]

is not very dependant on the number 11

[novel parrot]

yeah

[void cosmos]

right

[novel parrot]

but

what about a polynomial with no roots?

[void cosmos]

good question

prove this

[slim kayak]

A constant?

[void cosmos]

F[x]/(p(x)) is a field iff p(x) is irreducible

given F is a field

Z is not a field

[void cosmos]

but you can do this

and now consider this counterexample

p(x) = x^2+1

and Z[x]

(x^2+1) cant be maximal otherwise Z[i] would be a field which is not

same argument

as 11

and as any element

reallyh

[novel parrot]

interesting

[void cosmos]

harder problems are like

asking if Z[sqrt(11)] is euclidean

or a pid

i dont know how to do those ";D

probably we do it the same asd we do Z[i]

by approximating +-1/2

[upper pivot]

you can actually tell when Z[sqrt(d)] is a UFD or not. But the answer is its often not

so you work with the "ring of integers" of Q[sqrt(d)] instead which are UFD

[void cosmos]

yea

ig using definitions from ANT would work ;D

[frigid lark]

I think there are some nice results for Z[sqrt(-prime)]

But I didn't really care

And forgot them

[void cosmos]

algebra is pretty

[novel parrot]

we could show that there are non principal ideals

[dim widget]

There aren't really

[void cosmos]

wdym

by definitoin any element can be a principal ideal

just (it)

u were asking if those can be maximal

[upper pivot]

yeah even finding class number for like, imaginary quadratic fields are hard

[novel parrot]

if we want to show Z[sqrt(D)] is not euclidean

[void cosmos]

too hard for me

[novel parrot]

same

[void cosmos]

it is for D = 11

[novel parrot]

yeah

[void cosmos]

and D =-1

[novel parrot]

Z[\sqrt(11)] is euclidean?

[void cosmos]

yes

[formal ermine]

what are we trying to solve right now

[void cosmos]

and Z[i] obv

idk if the same argument for Z[i] can be used to prove Z[sqrt(D)] for D squarefree

like approximating and shit but idk tbh

[upper pivot]

not really

the Z[i] arguement falls apart pretty fast

[frigid lark]

I was thinking about Z[i]/wZ[i], for w in Z[i]

[upper pivot]

its an open problem when the ring of integers of Q(sqrt(d)) is a UFD for d>0 i believe

we know the answer for sqrt(-d), these are the heegner numbers

and euclidean is a much stronger condition than UFD right

[dim widget]

The usual argument about approximating lattice points can show that for n = -11, -7, -3, -2, -1, 2, 3, 5, 6, 7, 11, ... some list ending at... 73 the ring of integers of Q(sqrt(n)) is euclidean with the norm as the norm function.

But we know more generally under GRH that rings of integers which are PIDs are euclidean wrt some euclidean function.

[frigid lark]

Sorry for spreading missinformation

[dim widget]

But the euclidean function will not be the norm.

[upper pivot]

oh huh didnt know that, thats an interesating result topos

the GRH thing

[dim widget]

Yeah it is a weird result as well.

People know some unconditional things as well. But I think beyond the cases like Z[i] where the lattice is so regular that it's really easy to prove that division works, it's useful to just give up on proving convergence of euclid's algorithm and use more fancy geometry of numbers.

[upper pivot]

right

[novel parrot]

if R is euclidean then R/I is also euclidean for any ideal right

any element of R/I can be sent back to R

[void cosmos]

this would imply Z[sqrt(D)]] is always euclidean

[novel parrot]

so we can do the division and send it back?

[novel parrot]

oof

[upper pivot]

yeah its not true in general

[slim kayak]

You want to perform euclidean division on representatives?

[novel parrot]

yeah

[upper pivot]

but quotients of PID are PID which is the important result

euclidean exists as a tool to figure out when something is PID

[slim kayak]

Someone up to speed with their geometric group theory? I'd have a question about nielsen-reduced sets or rather sets that are almost nielsen-reduced

Given some U such that N0 and N1 hold, supposedly it forms the basis of a free subgroup with U as its basis.

The arguments about that I can find is that using N2 one can write any u as l(u)m(u)r(u) where l is the longest segmenet cancelled on the left in any product and r is the longest segment cancelled on the right, that m must have non-zero length. But apparently one can still show that claim with only N0 and N1?

[novel parrot]

what does free group mean?

free Z module?

[void cosmos]

free object in the category of groups

if you know the categorical definitoin for free modules

then its the same but for groups

[novel parrot]

idk category

[void cosmos]

whats ur definition

for free modules

[slim kayak]

The free group over a generating set S is defined by its universal property, a popular representative is the group of "words"

[novel parrot]

a generating set that uniquely generates

[void cosmos]

wdym unique

[novel parrot]

basically, linearly independant,

any subset

[void cosmos]

a free group ,F(S),generated by a set S satsifies the property that given any function f from S to any other group G, there exists a unique homomorphism from F(S) to G such that the diagram in mind commutes

[slim kayak]

Free modules have finite combinations of elements, so they isomorphic to the direct sum of cyclic modules generated by each basis element.

Free group can sometimes be constructed as free product of groups. Free products and direct sums differ in general

Also usually not abelian either.

[novel parrot]

hmyes

if was have homomorphism from R[X] -> R such that X -> r and a |-> a

how do i show that the kernel is (X - r)

i can see that (X -r ) subsets Ker

but how to show the reverse inclusion?

any f(X) in the kernel |-> f(r) = 0

but does that mean that (X -r) divides f(X) ?

[elder wave]

yes

x-r is a linear factor of f

[novel parrot]

even if R is not field/R[X] is not euclidean?

[dim widget]

Yes even then

[novel parrot]

how?

[dim widget]

There are many ways

If R is an integral domain this is trivial

[novel parrot]

easiest way? 🙂

[dim widget]

It depends on what generality you want to prove this in.

[novel parrot]

hm

[dim widget]

If R is noetherian it's also easy

[novel parrot]

general ring

[dim widget]

Ugh

Well so there is a map R[x] \to R sending x to r

there is also the map R[x] \to R by quotienting by x-r

Clearly the second map factors the first.

So if I is the kernel of the first map then I mod (x-r) is contained in R

But then I must have a generating set ((x-r), r_1, \dots) where the r_i \in R.

But if r_i \in R then you can verify by hand that r_i cannot be in the kernel of the map sending x to r.

Because that map is the identity on R.

[novel parrot]

i c

[novel parrot]

if $X \mapsto T^2$ and $Y \mapsto T^3$ is a homomorphism from Q[X,Y] to Q[T] how do i show that the kernel is (X^3 - Y^2)

idk how to show reverse inclusion

[cloud walrus]

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

[coral spindle]

A good method in this case is double inclusion. Look at an arbitrary element of the kernel and try to reduce it down to something more manageable via elements of (X^3 - Y^2)

[novel parrot]

ok

so we could kill Y^2 and above

[novel parrot]

but then what?

[coral spindle]

So you've got a polynomial f(X) + g(X)Y

and you know it's in the kernel

[novel parrot]

yes

[coral spindle]

so think about that now

[novel parrot]

f(X) + g(X)Y = X^nY or X^n right

[coral spindle]

No

[novel parrot]

any Y^2 terms and above, we can kill by multiplying g(X)

oh

[dim widget]

Maybe it's better to think inside of k[t]

because you know in k[t] t doesn't satisfy any polynomial relations

So if P(X, Y) = 0 and we substitute X = t^3 Y = t^2 we see that after this substitution P must be identically 0

[novel parrot]

right

[dim widget]

Since monomials of different degrees are linearly independent, we see that the relevant question is for what values of a, b, c, d is t^{2a}t^{3b} = t^{2c}t^{3d}

This is just asking when is 2a + 3b = 2c + 3d for a, b, c, d natural numbers.

Now this is just like some kind of very simple algebra about the semigroup of integers under addition.

[novel parrot]

i dont understand how that helps

because we want to show that f(X) is a multiple of (X^3 - Y^2)

so if we can write f(X) + g(X,Y)(X^3 - Y^2) = terms with only Y^1 or Y^0 right

[pastel cliff]

why is it important that K remains fixed in that iso

[delicate orchid]

it isn't, it's just that there exists an isomorphism that does fix K - which is nice for le galois theory

[dim widget]

They are trying to give a criterion for "how unique" the algebraic closure of K is.

If something is unique up to automorphisms which fix K it is also unique up to automorphisms which don't necessarily fix K

But the minimal thing you can say is that it is unique up to the action of the isomorphisms which fix K.

It's the same mathematical reason why someone might say Prop: Z is a ring and not Prop: Z is an abelian group with a binary multiplication operation.

Because the former is more specific/more precise.

[tribal moss]

Beware, though: It is not unique up to unique isomorphism.

[dim widget]

It is if K = \bar{K}

Or in general if K = K^{\text{sep}}

[tribal moss]

Sure, because then "fixing K" directly says it has to be the identity?

[delicate orchid]

tfw no universal property

[dim widget]

yeah exakt

[alpine remnant]

I have a doubt about afine spaces

AGAIN

Let there be three skew lines in afine 3D space

that is, three lines which do not intersect nor are parallel to each other

let's say that they are all parallel to a given plane

prove that any straight line which intersects all three will be parallel to a certain plane, and determine that plane

How tf am I supposed to do that???

It doesn't even feel intuitive

[formal ermine]

#geometry-and-trigonometry ?

[alpine remnant]

but it's meant to be solved analytically through afine spaces

like, a set of points associated to a vector space

that's why i thought it'd be more appropriate in algebra

[dim widget]

What have you tried?

[pastel cliff]

What is $[\Q(i, \sqrt[3]{2}) : \Q]$

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

can i use tower here

Q(i, cube root) / Q(cube root) / Q

idk what [Q(i, cube root) : Q(cube root)] would be though

[formal ermine]

fuck wrong polynomial

[pastel cliff]

lol

[alpine remnant]

i'm literally so at a loss that i don't even know how to start

i've tried research and contacting peers, but everyone is equally at a loss

[dim widget]

Have you tried writing out three skew lines just to see why its true for an example?

[alpine remnant]

i'm supposed to pick a point and a vector, right?

for each

[dim widget]

Yes and you should make sure that the vectors only lie in a given plane.

[delicate orchid]

lie in a given plane?

[formal ermine]

yeah

[delicate orchid]

didn't the question just say parallel to one?

[formal ermine]

it must obviously be <= 2

can it be 1?

[dim widget]

The lines are parallel to one, which means the vector lies in that plane.

[delicate orchid]

right yes duh

sorry

[dim widget]

if you write the line as P + xV

[delicate orchid]

then u add on the point and they move out

[pastel cliff]

no dont think so

so index is 6

[dim widget]

I think a useful thing is to reduce to the case that the plane is just z = 0 and one of the three lines lies on that plane.

[formal ermine]

yeah

[pastel cliff]

to be clear, it cant be 1 bc i is lin independent from cube root 2 right

[dim widget]

Also because the large field contains a subfield of degree 2

[pastel cliff]

but lin ind. is enough right

[dim widget]

That is enough but what I said proves that they are independent.

[south patrol]

it's cute

[pastel cliff]

[south patrol]

and yes use tower law innit

[pastel cliff]

related but - showing that $x^2 - 3$ is irreducible in $\Q(i)$

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

that's 4 isnt it

but then why is it immediate

[south patrol]

You can do it just by checking there are no roots

Which is clear

Since indeed we know what the roots are in C

[rustic crown]

or instead show x^2+1 is irred over Q(sqrt(3)) which is even easier :p

[south patrol]

lol ye

hm I was trying to show like if R[t,t^-1] is Noetherian then R[t] is

Is the argument just like, well if R[t] isn't Noetherian, then R isn't, so R[t,t^-1] isn't

[pastel cliff]

what does this imply

[south patrol]

Cause we can just take an ascending chain in R and push it through to S:= R[t,t^-1], since given an ideal I of R, IS is just the Laurent polynomials with coefficients in I

[rustic crown]

it implies [Q(i, sqrt3):Q(sqrt3)] = 2, so [Q(i, sqrt3):Q] = 4, which finally gives you [Q(i, sqrt3):Q(i)] = 2

[pastel cliff]

i see that but what's the relationship between index and irreducibility

[south patrol]

The poitn is like

$[K(\alpha):K] = \deg m_{K,\alpha}$

[cloud walrus]

potato

[south patrol]

where hopefully the notation is self explanatory lol

[pastel cliff]

min polynomial of an algebraic element alpha

ohhh

degree 2

can't be factored

[rustic crown]

why don't you just say ev_1 : R[t, t^-1] --> R is surjective so R noe, so R[t] noe

uwu

[south patrol]

Yes that is true, I think I confused it w something else lol

That is what I would do yes

I basically showed that and crossed it out, why, gr

Thank

[formal ermine]

is $R^n \otimes_R M \iso M^n$?

[formal ermine]

having a brain fart here

can't we just use basically the same proof as the one for n = 1

[tribal moss]

Looks right. I think you can prove in general that tensoring distributes over direct sums.

[formal ermine]

ah right

I remember seeing that

thanks tropos

[rustic crown]

left adjoints are ntinuous

.<

walter

[agile burrow]

det

[dim widget]

walter and det uwu

[coral spindle]

sadness

Waino

no

[white oxide]

would pZ x pZ be sufficient? (for p a prime)

[coral spindle]

Have you proved it's maximal?

[white oxide]

nah i was about to try

[coral spindle]

Maybe there's some quick test you could do that shows it's maximal :)

[white oxide]

well i know you can quotient it out right

to see if it's a field

but idk how to show it's a field lol

[coral spindle]

And what is the quotient?

[white oxide]

oh wait

it doesn't have unity

[coral spindle]

It does have unity.

[white oxide]

oh right oops

i was thinking about representatives from zp x zp

uhhh

[coral spindle]

Its unit is (1,1)

[white oxide]

right

[coral spindle]

Are you saying that the quotient is Z/pZ x Z/pZ?

[white oxide]

oh nah (Z x Z)/(pZ x pZ)

[coral spindle]

Well I'll just spoil this. That is isomorphic to Z/pZ x Z/pZ.

Maybe you should try proving that

[white oxide]

hm okay

and the direct product of fields is a field

[coral spindle]

What does that tell you about the ideal being maximal?

[white oxide]

wait no

no it isn't

[white oxide]

wdym

well if it's iso to Z/pZ x Z/pZ

and that's not a field then it's not maximal

[coral spindle]

That's right.

The ideal pZ x pZ is not a maximal ideal of ZxZ.

[white oxide]

hm

this leads me to believe that there are no proper nontrivial maximal ideals

idk i'm prob wrong tho

[coral spindle]

That's incorrect

In fact, any ring has maximal ideals, due to Zorn

[white oxide]

oh i haven't read zorn yet

[coral spindle]

Details.

I will tell you this: for any n,m the set nZ x mZ is an ideal of Z x Z, and its quotient is Z/nZ x Z/mZ. This generalises, btw.

[white oxide]

hm maybe Z x pZ

[white oxide]

ah ok

[white oxide]

oh wait

i think this wokrs

cuz thats iso to {0} x Zp by what you just said

wait i'm stupid

that's not even a field

nvm

i suppose i should try to get Z/pZ cuz that's iso to Zp

[dim widget]

Yes

[white oxide]

and there needs to be a 1 in the direct product

cuz (1, 1) must be unity

[dim widget]

What is Z x Z / Z x pZ

[white oxide]

that's {0} x Zp right

oh wait

there is unity

oops

just (0, 1)

[white oxide]

i'll try proving this yeah

[coral spindle]

The general statement is that if R, S are rings then:
(1) Every ideal of R x S is of the form I x J for an ideal I of R and J of S, and
(2) (R x S)/(I x J) is isomorphic to (R/I) x (S/J).

[white oxide]

okay i'll actually try proving that, obviously a lot stronger

thanks!

[cloud walrus]

okeyokay

[white oxide]

Is this fine? for (1)

[coral spindle]

The argument about subrings doesn't make sense to me.

Ideals are not merely sub-rngs.

Furthermore, I think your conclusion that every subring of R x S is of that form is simply not true.

[white oxide]

i just wanted to show existence of such an ideal of R x S

[pastel cliff]

ignore this, linkinn to old convo

[coral spindle]

Take, for example, the subring of R x R given by {(r,r) | r in R}

[white oxide]

wait but wouldn't that satisfy my argument

cuz R is a subring of R

[coral spindle]

You claim "it is clear that any subring of R x S must be of the form I x J" but this subring is not of that form.

So no, it does not satisfy your argument.

[white oxide]

i'm confused though, isn't {(r, r) | r in R} = R x R

[coral spindle]

No

[white oxide]

how tho

[coral spindle]

Is the element (1, 2) in {(z, z) | z in \mathbb Z}?

[white oxide]

ohh i see you're requiring them to be the same

[coral spindle]

That's right

[white oxide]

ah yeah

i see now

hm ok

[coral spindle]

Try again. I suggest you use the first isomorphism theorem to prove (2).

[pastel cliff]

bringing this back but i dont see how this relates to transitivity

[white oxide]

ah ok that makes sense thx

[pastel cliff]

repost

wait nnoo

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

that's the question

[dim widget]

Didn't this already get answered

[pastel cliff]

we talked about it yeah

[pastel cliff]

.

but we got off topiic

unless i missed somethiing

[dim widget]

So for Q(\alpha_i) Q(\alpha_j) you should build an isomorphism between them

Using some presentation of these algebras in terms of polynomials

Then use Zorn's lemma (that's the standard way) to show that there exists an automorphism of \bar{Q} which does this isomorphism on these subfields

[pastel cliff]

hadnt even thought about the lemon

[dim widget]

Zorn's lemma on the set of (L, \alpha) where L is a subfield of \bar{Q} containing Q(\alpha_i, \alpha_j) and \alpha is an automorphism inducing that isomorphism

[pastel cliff]

are you using Q interchangeably with an arbitrary field

[white oxide]

if R is a finite commutative ring with unity and N is a prime ideal in R, then N is maximal since R/N is a finite integral domain iff R/N is a field iff N is maximal correct?

[formal ermine]

yes

[dim widget]

Yes :chad:

[pastel cliff]

[pastel cliff]

theyre the same as F[x]/(m(x))

where that m is the minimal polynomial of a given root

but that's the same f in the question since it's irreducible

ok i see that much but you lose me at the lemon

(yes i know what the lemon does)

[dim widget]

The lemon is just a way of taking this isomorphism and extending it to an automorphism of the alg closure one step at a time

Like we can see a way (I claim) to extend from automorphism of L to automorphism of L(x) for some x

So we use the lemon

[pastel cliff]

oh when you said "does this isomorphism"

that literally is transitivity right

[dim widget]

Uh I just mean that we have an identification of these subfields

but we want to show it comes from some automorphism

if we show that since i, j are arbitrary we get transitivity yes

[pastel cliff]

wdym alpha is an automorphism

isn't it a root

[dim widget]

\alpha: L \to L

I'm using bad notation

\alpha_i are roots

[pastel cliff]

ah

[dim widget]

\alpha: L \to L

And \alpha sends \alpha_i to \alpha_j

[pastel cliff]

so zorn gives us an automorphism of a subfield of Fbar which can be extended to a transitive automorphism of Fbar

[dim widget]

No just one that takes \alpha_i to \alpha_j

But since i, j are arbitrary we can send \alpha_i to any \alpha_j with some automorphism

so the action of all automorphisms is transitive

[pastel cliff]

ok rephrased: zorn gives us a specific automorphism of a subfield of Fbar sending a_i to a_j which can be extended to an Fbar automorphism and this can be done for all i,j

[dim widget]

Okay let's do this a different way

You start with a field L = F(\alpha_i)

You have \alpha: L \to Fbar with \alpha(F) = F and \alpha(\alpha_i) = \alpha_j

You then use Zorn to extend this from F(\alpha_i) to Fbar.

At the end you get \bar{\alpha}: Fbar \to Fbar

Sending \alpha_i to \alpha_j

[hollow tartan]

What's the idea of a grupoid? Why does the fundamental group qualifies as one?

[dim widget]

@prisma ibex

[prisma ibex]

the fundamental group is a group, it depends on a basepoint. It's natural to get rid of the dependence on basepoint by considering the whole fundamental groupoid

if you think of groups as categories with one object where all morphisms are invertible, then groupoids are like groups with more than one object

[white oxide]

Ok, I think I got (2): Define $\varphi: R \times S \to (R/I) \times (S/J)$ by $\varphi\bigl((r, s)\bigl) = \bigl((r + I), (s + J)\bigl)$. We check that $\varphi$ is well-defined. We know that the canonical homomorphism $\gamma: R \to R/I$ is well-defined; that is, if $r = r_1$, then $r + I = r_1 + I$. Similarly, since $\psi: S \to S/J$ is well-defined, $s = s_1 \implies s + J = s_1 + J$. Hence, if $(r, s) = (r_1, s_1)$, then $\varphi\bigl((r, s)\bigl) = \bigl((r + I), (s + J)\bigl) = \bigl((r_1 + I), (s_1 + J)\bigl) = \varphi\bigl((r_1, s_1)\bigl)$. It is clear that $\varphi$ is onto $(R/I) \times (S/J)$. It follows that Ker($\varphi$) = $I \times J$, for if $(i, j) \in I \times J$, $\varphi\bigl((i, j)\bigl) = \bigl((i + J), (j + J)\bigl) = (I, J)$ which is the identity in $(R/I) \times (S/J)$. Hence, it follows that $(R \times S)/(I \times J) \simeq (R/I) \times (S/J)$ by the first isomorphism theorem.

[cloud walrus]

okeyokay

[hollow tartan]

I only have some idea of categories but what would it be a counterexample where a group can't be a groupoid, and by objects you mean sub classification characterized by whatever parameter of these groups?

[prisma ibex]

every group is a groupoid in a natural way

roughly speaking a category consists of a collection of objects and a collection of maps between them, which you can compose

Like the category of all sets and all functions between them

if you have a category with only one object then the only additional data is the maps from the object to itself, which you can compose with each other

this naturally forms a monoid. If you demand these maps are invertible then this naturally forms a group

[hollow tartan]

I mentioned the above comment because it seems that the fundamental group characterizes a space to find its genus. I'm not sure although I believe this conclusion is a bit too ambicious, maybe this is a misunterstanding from what i believe the professor was going for.

[prisma ibex]

I mean yes there are several things wrong with this

[hollow tartan]

That is a very neat way of forming a monoid

[prisma ibex]

if you're talking about topological surfaces then yeah you can recover the genus from the fundamental group

in general there are many spaces which are not homotopy equivalent but which have the same fundamental group

genus isn't really a thing that makes sense in general

the other subtlety is even in the case of surfaces, you should add the word "connected" otherwise the fundamental group can't tell the difference between different connected components

like if you had a space X which is the disjoint union of a genus 1 surface and a genus 2 surface let's say

each connected component will have a different fundamental group

if you talk about the fundamental groupoid then this is okay, you just have a fundamental groupoid with two connected components so to speak

[hollow tartan]

I see thanks for clarifying my foggy introduction for this subject

[prisma ibex]

no problem!

[formal ermine]

what does "let p_0, ..., p_n, q in P^n such that no n + 1 points lie within a hyper plane" mean

does it just mean

"let blabla in P^n such that they don't all lie on a hyper plane"

[dim widget]

It means let p_0, ..., p_n, q in P^n are points such that no n + 1 points lie within a hyper plane

[formal ermine]

who hurt you

[dim widget]

A hyperplane is the zeroes of a single linear homogenous polynomial

[formal ermine]

the definition I know is that a hyperplane is a subset of the form pi(V \ { 0 }) with dim V = n

where pi is the quotient mapping

[dim widget]

Yeah that's the same

[formal ermine]

I don't think you can buy those anymore

[formal ermine]

I still don't understand the original sentence

[dim widget]

A linear homogenous equation is \sum_i a_i X_i = 0

So they're asking that for every hyperplane and every n+1 tuple of those points at least one point is not on the hyperplane.

[dim widget]

It's the same as when people say 5 points with no 4 coplanar or something.

It just means that the points are generic in some sense.

[formal ermine]

wtf does coplanar mean

[dim widget]

Lying on a plane

So 2 points determine a line

If you have a set of 3 points they're "generic" if not all 3 lie on a line

Similarly if you have 4 points they're generic if not all 4 of them lie on a plane.

Etc.

[formal ermine]

"tuple of those points"?

[dim widget]

No selection of n+1 of them

[formal ermine]

what points

what tuples of those points

[dim widget]

Ahhhh my dude

you are killing me

You took n+2 points in projective space

of dimension n

[formal ermine]

Yes

[dim widget]

You are now asking that if you take any cardinality n+1 subset of those n+2 points

[formal ermine]

ahhhhhhhhh

ok I get it now lmao

[dim widget]

That they don't lie on an n-1 dimensional linear subvariety of the original thing.

Alternatively they don't live inside of a P^{n-1} which lives inside of P^n

[lapis trail]

A group G is solvable if you keep recursively finding commutator subgroups and eventually get the trivial subgroup. I know a polynomial is solvable in radicals if the automorphism group of its splitting field is solvable, but I don't really see how the two concepts are connected.

[prisma ibex]

let's say the polynomial is irreducible and F/Q is the corresponding Galois extension, then Gal(F/Q) is solvable precisely if the polynomial is solvable in radicals, each time you take an n-th root this corresponds to having a cyclic group Z/nZ as a composition factor

[white oxide]

the sentence starting with similar arguments refers to these groups as subgroups of H, right?

oh

well the two ones starting with K* would be subgroups of K right

K*(H intersect K) and K*(H* intersect K)

oops

you know what i mean

[vagrant zinc]

name of book?

[white oxide]

oh yeah my bad forgot to get back to you

fraleigh a first course in abstract algebra

this is section 35, series of groups

[toxic zephyr]

so the quadratic residues mod 17 form a multiplicative group, right? is it possible for it to be cyclic? can the group of quadratic residues ever be cyclic?

i think 2 might be a generator but I'm not sure

or... i suppose any element of order (p-1)/2 would be a generator... isn't there a theorem about what orders are guaranteed in a group?

[delicate bloom]

what group do you get for multiplication mod p?

how might that restrict what possible subgroups you might get

[toxic zephyr]

oh right. U(p) is cyclic so its subgroups are cyclic

[tired horizon]

Knowing that j is a cubic root of 1 in complex numbers C

I want to find the irreducible minimal polynomial of j over Q

Knowing also that j^2+j+1=0

I did this

[frigid lark]

x-1 1/3 of the time

Nah jk

[frigid lark]

You observe that j is a root of f(x) = x^2 + x + 1, and that f only has complex roots, specifically j and j^2 so it must be irreducible over Q

[tired horizon]

Yes so I just say that theyre not in Q?

[frigid lark]

Yeah, that's enough for degree 2/3 polynomials

[tired horizon]

Okay

[frigid lark]

Reason being if they are reducible, you can factor out a linear factor which will just be x-a where a is a root of f(X), not in Q, a contradiction.

[tired horizon]

Okay

Thanks

[lapis plume]

Can anyone tell me how they get the statement circled in blue?
I just know that for any $\bar{r(x)}\in\mathbb{Q}[x]/(x^2)$, it is related to the form of $f(x)=q(x)x^2+r(x)$, but why there exists such $\bar{x}$?

[cloud walrus]

Trenton

[delicate orchid]

x^2=0 rather trivially

[coral spindle]

Remember that the elements of Q[x]/(x^2) are of the form a + (x^2) for some element a of Q[x]. You should very easily be able to see that since x^2 is in (x^2), we have x^2 + (x^2) = 0 + (x^2).

[plucky flicker]

What is the rank of a group of Lie type?

[lapis plume]

Got it! Thanks!

[dim widget]

What do you mean by a group of lie type?

Do you mean a Lie group? or are you talking about "finite groups of lie type?"

[plucky flicker]

finite groups of lie type

i'm thinking about those appearing in the classification of finite simple groups

[dim widget]

Finite groups of lie type come from reducing some algebraic groups modulo primes, so they are the F_p points of some algebraic groups. The rank of these groups is the dimension of a maximal torus in that group either as an algebraic group over that field, or the dimension of a maximal torus of the complex points of an associated group in char 0.

So for instance Gl_n(F_p) has rank n

The better way to define this is to use root systems to classify the finite groups of lie type over an algebraically closed field, then the rank of a group of lie type comes from the rank of the associated root system

[plucky flicker]

thank u

[novel parrot]

how is this done?

If R was free, then all ideals are free?

[dim widget]

R is always free

But ideals in R may or may not be free

Can you think of some free submodules of R?

[novel parrot]

principal ideals?

[dim widget]

Yes great

Since R is integral those are free modules of rank 1

What do non-principal ideals look like?

[novel parrot]

two or more generators

[dim widget]

So there is a map R^n \to I which is surjective

Coming from the generators.

[novel parrot]

but I can have infinite

generators

?

[dim widget]

That's possible too if you don't assume R is noetherian

But suppose $I \cong R^\alpha$ for some $\alpha$, where $R^\alpha$ is some direct sum of $R$'s

What happens when you pass to the fraction field?

[cloud walrus]

Topos_Theory_E-Girl

[novel parrot]

showing this is injective would show its free, right

[dim widget]

Yes but showing that it's not injective doesn't necessarily show that it's not free

[novel parrot]

if it was injective

[dim widget]

You have to show that any choice of such a surjection is not injective.

[novel parrot]

oh

[dim widget]

So okay if S is any multiplicative set then if M \to N is injective so is S^{-1}M \to S^{-1}N

[novel parrot]

are we sending I to R's fraction field?

[dim widget]

We are taking M = I, N = R and S = R - {0} in the above proposition

See if you can figure out why I cannot be free of rank > 1

[novel parrot]

what does rank mean again?

[dim widget]

Rank is the number of R's in the free module

so R^n is rank n

Every free module is R^I for I some set

where this means \oplus_{i \in I} R

which is somewhat bad notation because it looks like a product.

[novel parrot]

mapping from R to I, r -> ra

thats a surjection

when I in principal

and injection?

[dim widget]

It's an injection because R is a domain

[chilly radish]

Which is why some people choose to notate it $R^{\oplus I}$ which is slightly better

[cloud walrus]

ShiN

[novel parrot]

yea, its also surjection???

[dim widget]

Obviously yes

[novel parrot]

so R is isomorphic to principal ideals?

[dim widget]

By the definition of (r) that is a surjection

[dim widget]

Yes as a module

[novel parrot]

as Rings too??

[dim widget]

Obviously nontrivial principal ideals don't have a unit so not as a ring or anything.

[novel parrot]

but i thought R is a ring iff R is an R module ??

[dim widget]

💀

[elder wave]

[novel parrot]

[dim widget]

I don't even know what that is supposed to mean.

[tribal moss]

"Is an R-module" doesn't even make sense unless R is a ring.

[dim widget]

The really interesting exercise (from Atiyah and Macdonald) is that even if R is not a domain there is no injection R^n \to R^k for k < n.

Which is surprisingly hard.

When you can localize to the fraction field though, this reduces to linear algebra.

You should think about why!

I tried to give you a hint above.

[novel parrot]

so ra = 1 for some r ?

Since R is isomorphic to (a) as R modules

[dim widget]

No

There is an injection R \to R with image (a)

But this doesn't mean that (a) = R as submodules of R

necessarily

[novel parrot]

okay

[elder wave]

comm alg is so evil in this aspect

[dim widget]

Think about the integers: even integers are an ideal, there is an injection Z \to Z by n \to 2n. However not every integer is even.

[elder wave]

easy and intuitive looking statements can get so hard to prove

[dim widget]

I agree, I think the surprising thing is that that exercise is doable. It's easy to imagine a universe in which it's a completely intractable problem for arbitrary rings.

[novel parrot]

okay so all principal ideals are free of rank 1

[chilly radish]

I resent your assumption that rings need be unital

[novel parrot]

and for any other ideal, there cannot be any isomorphism from R^n to I ?

[devout crow]

average non-commuter

[pastel cliff]

is there any intuition/motivation behind splitting fields

[upper pivot]

Preach shin

[delicate orchid]

When do they polynomials have all they roots

[pastel cliff]

erm

what the deuce

[ebon pine]

I and nearing my goal to finish Abstract Algebra using Judson but I don't feel very confident in my abilities to do abstract algebra. What should I do to improve it? Should I just I solve more problems using a tougher book say Herstein or something else?

[pastel cliff]

problems

find exams online if you want too

[upper pivot]

Yeah I would second that, try doing some random schools alg quals or whatever

[pastel cliff]

wew fr can you elaborate a bit pls

[upper pivot]

Idk if that’s the level you are aiming for

[novel parrot]

can you explain this part again?

[upper pivot]

But being able to do them is a good indicator. I

[novel parrot]

Couldnt I be free with infinitely many elements

[delicate orchid]

No. That’s my intuition - if a polynomial splits over a field then all of its roots are in that field

[pastel cliff]

oh okay that one is at least comprehensible

[ebon pine]

I am not sure if I wanna do analysus or algebra later plus a lot of higher maths is intertwined so I wanna sure I am good with basics

plus there's also the target of competitive exams

[upper pivot]

Yeah for sure you want to have enough knowledge to pass quals of e.g. your school maybe

[lethal dune]

Galois theory deals with automorphism of the field, so wouldn't it be nice if all the embeddings of K in it's algebraic closure actually give an automorphism? In that sense, you will have "enough" automorphism to separate the roots.

[upper pivot]

So I’d suggest looking into that probs

[pastel cliff]

this is what im reading

[ebon pine]

thanks!

[pastel cliff]

feels like a bit much lol

[delicate orchid]

Yeah it’s the smallest field required for the polynomial to split - so examining that polynomial in larger fields doesn’t reveal anything else

[elder wave]

This is basically just saying it's the smallest field s.t. all the roots of your polynomial are contained

sniped

[devout crow]

it's ok you sniped me

[devout crow]

the food chain is long

[novel parrot]

any gigabrain can explain me this question?

i didnt understand

[wraith cargo]

can you write I as a direct sum of R's?

[pastel cliff]

polynomial in the larger field right?

[wraith cargo]

that is essentially the question

[novel parrot]

but wouldnt this work for any finitely generated ideals?

[wraith cargo]

no

[upper pivot]

We should probably make some threads as there are multiple convos happening here

[novel parrot]

I = (a, b) I is not free here?

[pastel cliff]

enjoy the chaos john

[novel parrot]

i have finals this week

[pastel cliff]

same

bleaknervoussweat

[devout crow]

the splitting field of a polynomial P(x) in F[x] is just F(a_1,...a_n) where the a_i are exactly the roots of P

but like

[wraith cargo]

OK so this ideal can be written as aR + bR
Now if there's no intersection between these two then it's free (and we replace the + with a \oplus)
If the intersection is non-zero it's not free

[novel parrot]

oh ...

[devout crow]

without referencing an algebraic closure or something, it doesn't make sense to talk about the roots of P a priori

which is why you want to define it as the "smallest field such that bla bla bla"

[wraith cargo]

it's the smallest field containing all the roots of a polynomial
So like idk that feels like motivation enough lol

[pastel cliff]

thanks yall

i'll be back with better questions soon

[wraith cargo]

yeah it takes a while to get used to free modules

[novel parrot]

overall, how would you answer this question?

we know principal ideals are free

and those finitely generated ideals such that intersection are 0, are also free

what about non finitely generated ideals?

[tribal moss]

"How would you answer this question?"
"I would say: "I would say: "I would say: "I wou[NO CARRIER]

[agile burrow]

There's the rub; is the intersection ever just 0?

[novel parrot]

?

[novel parrot]

not sure haha

[pastel cliff]

r/woosh

[agile burrow]

For the sake of simplicity (and to make my point valid), consider the case of an integral domain

[dim widget]

El momento de reddit

[pastel cliff]

topos theory egirl

español?

[novel parrot]

for this question, R is an integral domain

[dim widget]

Non sono capace ma è quasi uguale a italiano

[pastel cliff]

[agile burrow]

Oh perfect, sorry I haven't read any of the context and just jumped in at your last message lol

[pastel cliff]

je suis dans ta mere

[novel parrot]

so (a) cap (b) = 0

then ra = yb = 0

for some r and y

[nimble moat]

فعلن

[novel parrot]

since domain, r and y are forced be 0

[agile burrow]

Well you're operating under the assumption that the principal ideals (a) and (b) have trivial intersection. Why should that be the case?

[novel parrot]

im not

heres the question again, i dont fully understand

[agile burrow]

Then what is this message about?

[formal ermine]

me gusta tu madre

[novel parrot]

if we had (a) cap (b) = (0) then (a,b) would be free of rank 2, yes?

[formal ermine]

https://tenor.com/view/momento-del-mogus-el-momento-del-mogus-mongus-gif-21646360

[agile burrow]

Yes, if they had zero intersection then that would be true

[dim widget]

As long as ra = 0 implies r = 0.

[novel parrot]

but you are saying that they can never have 0 intersection?

[dim widget]

and similarly for b.

[agile burrow]

Oh, good point topos

[dim widget]

because ab is in the intersection. In your case R is an integral domain so ab is nonzero.

[agile burrow]

Yes, I'm claiming that if a and b are non zero then (a) cap (b) is nonzero

For the reason topos pointed out

[novel parrot]

right

Okay, so ideals with 2 or more generators are not free, yes?

[dim widget]

In general: R^2 \to (a, b) is injective when restricted to (r, 0) and (0, r) iff ra = 0 = r'b implies that both r = 0 and r' = 0. In this case ab is nonzero, which implies that (b, -a) is in the kernel.

So ideals which require at least two generators in any ring are never free.

[agile burrow]

GOOD point

I was mildly concerned about domain assumptions but I see now

[dim widget]

The domain assumptions just make the argument simpler.

[novel parrot]

ok

that makes sense i think

[agile burrow]

Oh, I suppose if your ideal were free then your generators would necessarily be non zero divisors to preserve linear independence

[novel parrot]

this also covers when I has infinite generators aswell?

[dim widget]

If R^{\oplus \infty} \to I is an isomorphism then there is a J \subset I which is generated by 2 elements and also free.

[agile burrow]

I certainly hope infinity is at least two

[novel parrot]

If $R^{\oplus \infty} \to I$ is an isomorphism then there is a $J \subset I$ which is generated by 2 elements and also free.

[nimble moat]

ReL

[cloud walrus]

ActiveChapter

[nimble moat]

i cN aGrE

[novel parrot]

is that the whole answer to this?

we didnt need to embed R into fraction field?

[dim widget]

Yes, they wanted you to prove it a different way

I tried to explain it to you but you didn't focus on the explanation I tried to give.

[agile burrow]

I like our solution better anyway

[novel parrot]

i didnt understand 🙁

[dim widget]

But our solution is better anyway because the result is more general

Their solution uses that R is an integral domain which is unnecessary

But for the record: