A, B are rings

f:A--B is a surjective ring hom

so you want to show a bijection between {ideals in B} and {ideals in A that contain the kernel of f}?

hmm i want to understand that statement better

from what i understood now

@Mathei are you here?

yes

taking any ideal of I in B

the preimage of I

Want to help me with some Galois theory?

@KasmirKhaan yes

is an ideal in A, but that statement sais that the preimage contain the kernel of f

so prove it

@AlessandroCodenotti okay

hmm so i did understand it right?

I was not sure about that

kind of busy, but I'll try

let me try it again

@MatheinBoulomenos thanks for help btw :D and good luck with your study

@KasmirKhaan I don't know whether you have proved that they contain the kernel of f

did not yet ><

that is what am gonna try now

So I did an exercise asking to prove that $E/F$ is Galois if $F$ doesn't have characteristic $2$ and $|E:F|=2$

@KasmirKhaan ok

@LeakyNun You r right. Officially I am a post graduate in Mathematics. I am poor in proof techniques. I really don't know, how to correct it. :'(.

I used the fact that the characteristic of $F$ isn't $2$ while showing that the extension is separable and I don't think I needed it while showing that $E$ is normal

But aren't finite degree extensions of finite fields always separable?

it is not needed to show normality, that's correct

So I is an ideal in B, its preimage is an ideal of A that we know

I need to show that ker f in contained in the preimage of I

hmm

I dont know how to start that argument

the intersection of all the preimages of all the ideals of B

should be the kernel of f

if i would belive the statement that i want to prove

so as sets , if we draw circles , they all intersect at the kernel of f

What does it mean that ker f is contained in the preimage of I under f?

you're thinking way too complicated

@AlessandroCodenotti finite characteristic need not mean finite

well it means if I pick an element in kernel of f, it should be mapped to I

Oh, of course

OHHH

I see :D

all elements of ker f will be mapped to 0

and since I is in ideal of B

it must contain 0

Let $F=\Bbb F_2(t)$, then $F[X]/(X^2-t)$ is the counterexample, that was the next exercise

Did I get it right mathein ? :D

Because $X^2-t$ is irreducible but $\sqrt{t}=-\sqrt{t}$ so it doesn't have distinct roots

@Kasmir yes, now you know why the preimage contains ker(f)

@LeakyNun Now I really understood , why did you ask that question :). It is really a school level question. I should be careful about reading all assumptions. Thank you.

@MatheinBoulomenos Yes :D

@MatheinBoulomenos evrry ideal contains 0, and the preimage is the fiber over 0

so those elements in ker (f) are automaticlly there :D

yes

I notice that many of the things I dont get is because of lack of understanding the defintions

But Ill try to get better at that :D

thanks again mathein :D

and thanks leaky :D

@KasmirKhaan so you’re using pro terms now :p

@LeakyNun haha =p

if you mean fiber , that took me a while to get why they called it fiber not just a preimage of 1 element ><

right

If I remove the pinch point of a pinched torus I just get a cylinder right?

yup

just making sure :)

you get a twice punctured sphere, which is homeomorphic to the open cylinder

Hi @Balarka

Hey @MatheinBoulomenos

Apparently Tate cohomology does have a topological interpretation in terms of $BG$, but I know basically no homotopy theory, so I don't understand it

@BalarkaSen o people with no shapes

is it true that $|\sum_{n=1}^N \dfrac{z \ ^ n - z \ ^ {n+1}}{n+k}| \le |\sum_{n=1}^N z \ ^ n - z \ ^ {n+1}| $ ? $z\in B(0,1)$

@MatheinBoulomenos That's pretty cool

Anyone know much about Okounkov-Vershik approach to rep theory of the symmetric group?

i need to prove that $\sum z \ ^ n / (n+k)$ converge for all $k $ and $z\in B(0,1) - \{1\}$ , someone can help? the hint is to multiply the partial sums by $1-z$

@Liad en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem

that takes care of the interior

not sure about the boundary

maybe you can try to see how the hint can be used? we get $\sum_{n=1}^N \dfrac{z \ ^ n - z \ ^ {n+1}}{n+1}$ this seems like a telescoping sum

it doesn't really telescope due to the denominators

yea

hence my other question :P

if it can be bounded by the telescopic sum

I just wouldn't use a telescoping sum. Ratio test, root test, Cauchy-Hadamard all give you the convergence radius easily

yea they are, but there is a problem with $|z| = 1$

Hey @Balarka, @Liad, and @Mathein!

Hi @Daminark

@Daminark Heyyy

(That's pretty gooood)

@Daminark hi

Wait I don't get it Balarka. Also how's it going for all of you?

It's the classic old Idubbbz meme

Oh ffs idubbbz... Kek

Can someone give me an example of how the long exact sequence of homology groups of a relative pair $(X, A)$ where $A \subset X$ helps me compute the relative homology groups?

Consider, eg, the pair $(D^n, S^{n-1})$

You could take a look at the piece $H_i(S^{n-1}) \to H_i(D^n) \to H_i(D^n, S^{n-1}) \to H_{i-1}(S^{n-1}) \to H_{i-1}(D^n)$

The homology of the disk vanishes (we're above degree 0)

So you get a short exact piece $0 \to H_i(D^n, S^{n-1}) \to H_{i-1}(S^{n-1}) \to 0$

That means $H_i(D^n, S^{n-1})$ is isomorphic to $H_{i-1}(S^{n-1})$

Run an induction argument to show these groups are zero if $i \neq n$ and $\Bbb Z$ if $i = n$.

That's useful.

@MatheinBoulomenos Wow pretty cool. Link it to me!

@BalarkaSen mathi.uni-heidelberg.de/~schmidt/papers/rims.pdf

awesome example. but what if the connecting homomorphism isn't so nice?

@Mathei Thanks!

@gian The connecting homomorphism has a pretty explicit description, actually

You know how elements of $H_n(X, A)$ are represented as relative $n$-cycles $\xi$ in $X$?

Yes

Namely, $\xi$ is a chain so that $\partial \xi$ is inside $A \subset X$

The connecting homomorphism $H_n(X, A) \to H_{n-1}(A)$ is nothing but the map $[\xi] \mapsto [\partial \xi]$

Oh I see

It's not obvious that this is what it is from the algebraic proof by Snake lemma

(algebra sucks, yes)

@BalarkaSen :O

are you T R I G G E R E D?

>8(

@Balarka A N G E R Y R E A C C

-_-

@MatheinBoulomenos can you explain to me the evaluation homomorphism ?

Snek lemma is love, Snek lemma is life

R[x] --> R

like how does it work

@BalarkaSen, unfortunately that trick doesn't work on my problem :(

@KasmirKhaan you fix one element in $a \in \Bbb R$, then you just plug that into every polynomial

i.e. $P(x) \mapsto P(a)$

hmm

I'm trying to compute the local homology at the pinch point of a pinched torus

is that important?

do I need to spend time on it for the exam or just move on?

evaluation homomorphism are very useful in dealing with polynomial rings

@gian Oh yes you need a different trick for that

oh okay :D

You want to apply excision theorem

we might have a question on that , only 3 Days left for me Before exam, want to use them as productive as I can on rings

Oh okay, I'll be back then :D

You can do it geometrically too

Geometrically?

Just draw a picture and pretend it's obvious

Yeah relative chains in $C_n(X, X - x)$ are singular simplices with image containing $x$

@MatheinBoulomenos STAHP

Geometrically? Okay bye I'm out of here

@Daminark #ragequit

cya

Really I'm staying around but I just wanted to do that real quick

You wanted to do that for a long time, didn't you?

Nah I just Mathein's thing and wanted to continue the train along a different direction

conjectures are false until mathematically proven

@MatheinBoulomenos when we have two polynomials , and we want to find the ideal generated by them , how does it work?

it's generated by the gcd

same as integers?

hmm

you follow Euclid's algorithm

if we have I = x^2+x and J = x^2+2x

for example

these are easily factored

Let F be a field and I , J are elements in F[x]

I don't know how to work that out

can you show me? ( I did not spend alot of times on rings , just want to get some feel now Before exam, am a bit too late but ill give it a try ) =p

I=x(x+1) and J=x(x+2)

so gcd(I,J)=x

and if i wanted to find a generator for I+J

that would be (x) ?

yes

this is not obvious to me, what is the anology with integers?

or there any?

@AlessandroCodenotti "uscire" "esco" "esce" @_@

(cc @MatheinBoulomenos)

but you should really pay attention to the difference between an element and the ideal generated by it

@MatheinBoulomenos deja vu

What is the ideal (4)+(6) in the integers? @Kasmir

by (x) i mean , kx such that k in our field

well that would be (2)

@KasmirKhaan that isn't how you generate an ideal

@Mathein "an ideal"

you have to have x^2 as well

leaky are you sure?

@KasmirKhaan @_@

(x) is the set of all x*f, where f is any polynomial

so hmm

precisely

from I and J

we can get anything then

exept constant

am thinking right?

@KasmirKhaan an ideal is a ring (why?)

ideal is a subring also

so is a ring by defintion

why is it a subring

because of closure under addtion

and multiplication

why multiplication

well (I,+) is a subgroup

@Idle Do you actually think that the truth value of open questions is 'false' and then suddenly they become true when a proof is produced?

and a in I and r in I , ar in I

@KasmirKhaan where is this from?

but r can also be in R and still the Product in I

definition of ideal

right

leaky

@KevinDriscoll I was refering to the collatz conjecture, still not proven.

good

let me focus on my polys

:D

then x in I, x in R, x^2 in I

and computational verifications are just partially authentic.

@idle Ok sure, so in that specific case do you think that the collatz conjecture is false, and then if someone produces a proof it suddenly becomes true?

@Idle the point is that there's no "false until proven true", you have "known true", "known false", and "unknown"

I said this before wait ...

so once we have x

we have all x^n

@KasmirKhaan precisely

what I dont get is that what does F[x] mean

when F is not specified

@KevinDriscoll yes, with a dustless mathematical approach.

any field?

@KasmirKhaan F usually denotes fields

not just checking the first billion cases with some PL.

@Daminark the "unknown" isn't even a stable situation.

unlike "known false and true"

@idle Why do you think the truth value of the underlying claim has anything to d with whether a proof has been produced? I mean its a claim about the natural numbers. So the true or falsehood of the claim turns on the properties of the natural numbers. But a proof or counter-example to the collatz conjecture doesn't change anything about the properties of the natural numbers. It merely makes them apparent. So how can a proof change the truth value of the underlying claim?

@KasmirKhaan What do you mean by "I don't get what F[x] means"? :P You don't understand the notation or what?

I am really not sure

@AlessandroCodenotti hi

Hi

Like from the first view, i thought we take elements of R

like if we have x

(x) = Rx

@LeakyNun ?

@AlessandroCodenotti "hey you've changed the vowel" "hey you've changed the consonant"

I dont know exactly what are the elements generated by a polynomials

for integers i know for example (2) are multiples of 2

It's not stable, we hope to move things from unknown to known as much as possible, just that truth or falsehood of a statement is just there, and knowledge is our access to that aspect of the statement

@KasmirKhaan Surely it's best to first understand the objects your talking about before you start talking about subobjects?

@KasmirKhaan F[X] is a ring for any field F

hmm help with that?

@KevinDriscoll en.wikipedia.org/wiki/List_of_disproved_mathematical_ideas

It also changed tense and person, I'm not sure what's your point

'unknown' is, for the most part, not a statement about mathematics but a statement about human knowledge of mathematics

@AlessandroCodenotti like they are so different lol

@KasmirKhaan You say you don't get what "F[x]" is but you're talking about the ideals generated by elements of F[x]

and, of course, what humans think they know certainly can also be wrong

Yes, but we have plenty of irregular verbs

@AlessandroCodenotti also, how can "exire" give "uscire"

well I get what F[x] is

there's like no "u" in the paradigm of "exeo" in Latin

its polynomials in the variable x

and coefficient are in the field F

So by saying something is true or false before having proven it either way, you may very well be wrong. If you wait until you have a proof and say until then that we don't know, you're all good. Stability is not what we're looking for here by any means @Idle

@Idle You're confusing the ontological question of whether the statement is true or false with the epistemological statement about whether we know that it is true or false. The existence of previous claims that people thought were true but later were proved false doesn't say anything at all about whether current open questions are true or false.

@KasmirKhaan do you understand how they form a ring

@KasmirKhaan with coefficients in $F$

It just says that in the past some people believed things that turned out to be wrong

Most occurences of "x" became "sc" in Italian (x isn't a letter in the modern Italian alphabet)

If we have ´the element 1, for example.

does it mean we have 1 ,x,x^2 ... ?

@AlessandroCodenotti I know, but how does "e" become "u"

do dont have any limit of what the Powers of x are?

@KasmirKhaan yes, <1> = R for every ring R

Dunno, I'm not a linguist

@KasmirKhaan the sky's the limit

okay 1 was stupid example

yes leaky

addition is poly addtion

and multiplication is the usual polys multiplication

then surely you can see why x^3 is in <x^2>?

the units in R[x] are units in R

@KasmirKhaan the question is ill-posed, what do you mean "if we have $1$ do we have $1, x, x^2, \dots$? Do you mean "if an ideal of $F[x]$ contains $1$ then do we have $1, x, x^2, \dots$?

@ÍgjøgnumMeg come on, we all know what he meant

@Semiclassical Did you happen to see that we were wroking on a physics problem here last night at like 2AM?

and that "we" is inclusive of him

lolno

@KevinDriscoll yes, I don't abrogate this mean of just-in-time check-ups as being a background of next applications, but just making it clear that it's not a very authentic way of tackling a proof, or asserting anything.

@LeakyNun no i dont see that

@LeakyNun I recall you saying a few days ago that one should be precise when talking about mathematics

@KasmirKhaan x^2 in I, x in R

@ÍgjøgnumMeg not when he's frustrated about not getting things right

sigh

grrrrr

how is x in R?

is x a polynomial?

R=F[X] here

ahhhhhh

:D

so we take all polynomials

and coefficient in our ring R

@idle Of ocurse. But thats a veeeeery different claim from the claim that all open problems have a truth value of false and then they become true when a proof is produced.

@KasmirKhaan yes, but you're going to confuse yourself by using R as the base ring and as the polynomial ring

Okay kasmir will keep working on this and come back later

mean while I igot a question for later

@Idle if you allow the classification of "unknown", it still fits with what you're saying here

I think we all missed one more possible classification

@Semiclassical Someone was asking suppose that I'm given an initial position and initial velocity, and also a desired final position. Assume the dynamics are discrete and that at the start of each tiem step you can apply some maximum instantaneous impulse in any direction.

Suppose R is a not necessarly commutative division algebra, prove that the only two sided ideals of R are zero ideal and R

oh, that

@KasmirKhaan ok

How do you optimally choose the direction to reach teh desired final position

We aim to figure out whether things are true or false, and until we know that, we say it's unknown. If we then use an unknown fact to prove something else, we're still not sure that it's true

@LeakyNun dont want solution yet because did not try hard on that, ill keep Reading and come back later

@kevin Take a proof for a a very practical conjecture like a roadsign before an accident-friendly hotspot or a frequent situation of hazards.

@KasmirKhaan ok

@Leaky you're gonna talk about independence of the axioms, right?

@Daminark from

Yeah yeah whatever

:)

Anyway what I have in mind is more about ontological truth

@AlessandroCodenotti does your idiolect exhibit raddoppiamento fonosintattico?

You can't define truth, anyway

@Daminark like $\vDash$?

@MatheinBoulomenos $\varphi$ is true in model $M$ if $M \vDash \varphi$

@LeakyNun what's that?

Like, from a more philosophical sense, so to speak. It's not formal so much as, we find that our axioms of a set do not give us power to prove that $\aleph_1 = 2^{\aleph_0}$, but in principle that statement still has a truth value

@AlessandroCodenotti let's take it as a no

it's where you say "accasa" instead of "a casa"

I actually dont think you can define truth value independent of a particular model

Nah, that's more common in Tuscany and south Italy

@AlessandroCodenotti so you're from north Italy?

Yep

I see

That's why I say it isn't formal, like I'm thinking more... I hesitate to say it in this way but the way Plato talks about "the form of the good" in the Republic is sort of what I have in mind.

A month old with no answers :(

Like there's some idea we're trying to capture with the axioms, and I'm wondering how much of truth is intrinsic to that idea.

AH okay. In that case I agree there is some state-of-affairs about the universe and statements about such states-of-affairs are either true or false. But I'm not into philosophy o mathematics to have an intelligent position on whether we should include mathematical proposition in our states-of-affairs of the universe, like Plato would

I mean I'm not deep either, it's just that I tend to view model theory in two lights. One as a subject of study in itself because it seems really cool, and the other as being a technical framework of truth

handsome folks

the defnition of prime ideal

is that if ab in P, means either a in P or b in P

But as far as the discussion we were having above went, whether "morally speaking" we ought to classify statements as true/false/whatever before they're being proven, I find that we're thinking about the latter more, which is why I'd rather push it more in the direction of, is this true or false in a state-of-affairs. On the whole I'm not sure if I buy completely into trying to transplant Plato's theory of the forms into math

doesnt this hold for all ideals?

Nope

hmm

@KasmirKhaan <4> in Z

and generate the witness of the counter-example yourself

Think about $\mathbb{Z}$, prime ideals are meant to represent prime numbers

so the analogy of prime ideals are the prime numbers?

(4) = 4n

but this is also 2*2

but 2 is not in (4)

hmm

great

Precisely

:D

Sometimes kasmir is smart

Okay thanks guys :D

ill keep Reading :)

@Daminark what's a prime in the category of a ring with morphisms being divisibility?

Well in rings you think about prime and irreducible as being different things, I believe

So if $R$ is a ring, you say $p$ is prime if $p\mid ab \implies p\mid a \lor p\mid b$

@Daminark that's right

You say $n$ is irreducible if $n = ab \implies a\in R^* \lor b\in R^*$

yes

"Let $S$ be a set. the law of composition defined on $S$ by $ab=a$ for any $a,b\in S$ is associative. For which sets does this law have identity? " It seems to me that if $S$ is singleton, then this law has identity, otherwise not; because, if $e$ is identity, then $ea=e$ and $ae=e$ for all $a\in S$. Am I right, or missing something? Also, what if $S$ is empty?

Hey guys, can i ask a magnetostatic question here? The people at the physics forum dont wanna help me.

can i?

hola hola

"Just ask; don't ask to ask." is the rule here.

If anyone can help, they will do so. If they can't, you get ignored.

@KevinDriscoll I think the main reason why the reverse tree is disregarded that there exists someway a group of integers that are not blackhole'd to 1. So starting from 1 as a reference just suggests the existence of an arbitrary group of integers forming an arbitrary galaxy.

Plz take a look

Hey @Alessandro!

Also @Silent let's see

So $ab = a$ for all $a,b$ is associative, which is true since $(ab)c = ac = a$ and $a(bc) = ab = a$

hi @anakhronizein

And yeah that's right, since if $a\ne b$, then $ab = a$ and $ba = b$

Usually you exclude $S$ being empty when you talk about these things

@Daminark thank you!

Ya saying that the set has identity is saying there exists an element such that..... But if $S$ is empty there do not exist any elements. So there cannot be any such element

hi @MikeMiller How are you?