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16:01
Well, topology depends for me
Hey guys...I heard that if we have a signal of finite width $\Delta x$ in time then this signal has a width in the frequency domain $\Delta f$ such that $\Delta x \Delta f > c = constant$ but then again many finite signals in time spread to infinity in...anyhow I miss some points there...what is this property of fourier transform called?
I took two algebraic topology courses. I did really well on the first one, and dropped the second one because it was just too much
Was the second course more abstract?
i dunno the homotopyesque stuff
the mortal categories
We just did a lot of stuff and the proofs were only sketched, if mentioned at all
Yes, we did some homotopy-stuff at the end
oh that sucks
16:07
we did poincaré duality, cech cohomology, alexander duality, some knot theory, homotopy groups, fibrations, cofibrations, stuff like Brown's representability theorem, obstruction theory, Postnikov systems, localization at primes and some stuff I forgot
coolio
It just seemed to me that the prof tried to cram as much cool stuff as possible into this course
That never plays out well
but algebraic topology is really cool
I mean, we algebraists have to thank the topologists for category theory and homological algebra
hear, hear @Daminark
16:11
@MatheiBoulomenos we do?
I have had the impression that a lot of algebra is fundamentally motivated from topology
But I am not sure if that's because I am biased
@JorgeFernández I mean, that's where they historically came from, isn't it?
that is a nice example of friendship in mathematics
friendship between algae and topols
come here, alge-bro
have a algebrofist from me
16:18
I love using topological arguments in commutative algebra as well. (Perhaps some people would call those "geometric")
I pretty much literally always think about commutative algebra in terms of affine algebraic varieties
@MatheiBoulomenos Maybe you're secretly a topologist, not an algebraist :P
I don't think so
E.g. if $R$ is a Noetherian, commutative, 0-dimensional ring, then $\operatorname{Spec}(R)$ is finite
@MatheiBoulomenos that is more algebraic geometry than topology
(not that I have not myself used topological arguments in algebra papers)
Proof: $\operatorname{Spec}(R)$ is Noetherian, as $R$ is Noetherian and it is Hausdorff as $R$ is 0-dimensional. But, by a purely topological argument, any Noetherian Hausdorff space is finite
16:22
@TobiasKildetoft I don't know, feels more topological to me. I have not used the structure sheaf on $\operatorname{Spec}(R)$ at all, just the underlying topology
hmm, are you taking maxspec here?
I seem to recall spec basically never being Hausdorff
maxspec=spec, if the ring is 0-dimensional
ahh, right
it's Hausdorff iff the ring is 0-dimensional
16:24
the zero ideal is the one which causes trouble in general right
that's a dense point
@BalarkaSen Well, when it is a point, yes
in Spec
well, for an integral domain, sure
Right, fair enough
where A = { x ∈ Z: x is an integer multiple of 3 }
is {15} ⊂ A true?
thank you
can an empty set count as a proper subset?
i think it should, but cant back it up with a good explanation
also, $R$ is Jacobson iff $\operatorname{Specm}(R)$ is "very dense" in $\operatorname{Spec}(R)$
I think that "very dense" is a really funny name
Right, every prime ideal is an intersection of maximal ideals
Define what u mean by proper set
@BalarkaSen polyisoprene?
16:39
a what now?
Many same prene clearly...
wow the chat was hidden even more deeply by stackexchange
can someone help me with the first line of this answer math.stackexchange.com/a/2465685/11260
Here's a nice algebraic problem with a topological solution: let $R$ be a commutative ring with only finitely many maximal ideals and suppose that the Jacobson radical is $0$. Show that every prime ideal of $R$ is maximal.
i got the same results as him using emilio novati's approach but i dont understand michael rozenberg's approach...
@MatheiBoulomenos This is interesting, let me think for a bit.
16:47
@Peter Something specific there, or the answer in a whole?
just the first line
i get the same results
using a different approach
the first line just kinda drops out of nowhere
dunno if that is some well know fact in some circles
@MatheiBoulomenos How do you compute the Galois group of $x^5 + 15x + 12$?
Well, you can see the first line is just a nice re-ordering of the known approach.
Hi
Good, question, I need to think a bit about that
There's probably some result as such, but you can see for yourself why it's true - reorder the elipse so it's in the form we are used to
16:51
It is Eisenstein at 3 so it is irreducible
@LeakyNun Look in Dummit's paper. It's a solvable quintic.
@MatheiBoulomenos It has only one real roots and have 1+1+1+2 factorisation in Z7 as I computed, but I hope you don’t use properties that fall from the sky
@BalarkaSen Is the computation difficult without knowing that it is solvable?
I think it's a semidirect product of Z/5 and Z/4 again
It's definitely difficult to solve the quintic by hand.
I don't remember how one computes these things.
I mean the computation of the Galois group, not the roots
That's the thing; usually you compute Galois group by explicitly writing the splitting field, which you can't, in this case, without pain.
Dummit's paper has a characterization of Galois groups of quintics by a resolvent sextic.
And indeed x^5 + 15x + 12 is provided as an example
16:58
thanks, reading
@Studentmath i don't see it... $b^2*x^2 + a^2*y^2 - a^2*b^2$ with $b^2=9$ and $a^2=36$. Now what?
@MatheiBoulomenos the response you gave me was actually false mate
from our book: "The set A has four elements: { 1, 2 }, ∅, { 1, 2, 3 }, and { 1 }. For example, { 1, 2 } ∈ A. Note that 1 is not an element of A, so 1 ∉ A, although { 1 } ∈ A. Furthermore, { 1 } ⊈ A since 1 ∉ A."
Was it? I'm quite sure that 15 is an integer multiple of 3
yes, but {15} =/= 15
so if you are given a set, {1, 2, 15}, 15 exists in it but {15} doesnt
@WillNjundong you didn't ask if {15} is an element of A. You asked if {15} is a subset of A
17:01
A is not even a set of sets, what does that even mean? It's the collection of integer multiples of 3, as you wrote.
hmmm but is it though?
Indeed, {15} is a subset of A.
even if not explicitely defined in A as {15}?
Are you sure the "A" mentioned in your textbook the same as the one you wrote down?
Here it is:

A = { x ∈ Z: x is an integer multiple of 3 }
(i) {15} ⊂ A
17:04
(i) is correct. Either your textbook is incorrect, or you are reading the textbook incorrectly.
I think I'm confusing ∈ and ⊂ lol
That's possible. The point is if x ∈ A then {x} ⊂ A
alright thank you very much guys
ugh this new UI
@Studentmath i don't see it - can u link me to some references or some clue why it is obvious?
17:09
why did they switch things to the right ?
@Peter Sure thing, let me find something solid
@Peter I think that should help:
http://www.askiitians.com/iit-jee-coordinate-geometry/tangent-and-normal.aspx
@Studentmath thanks
shame about the formatting though
How do I compute the discriminant without the roots? Using resultant?
discriminant of ?
quadratic equation?
@Peter good point :)
17:16
or its related to PDE
Honestly, you learn the formulas for degree 2 and degree 3 and for anything else you use a computer
@LeakyNun Yep.
@Studentmath i like Emilio Novati's approach since it is purely algebraic and doesnt use any calculus
@Studentmath the derivation u linked does unfortunately use a simple calculus argument
@Peter I like the fact how the two, different in essence, approaches lead us to the right place :)
$256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2cd^2e-27a^2d^4+144ab^2ce^2-6ab^2d^2e-8‌​0abc^2de+18abcd^3+16ac^4e-4ac^3d^2-27b^4e^2+18b^3cde-4b^3d^3-4b^2c^3e+b^2c^2d^2$ that's the discriminant of $ax^4+bx^3+cx^2+dx+e$
17:19
@BAYMAX quintic polynomials
@MatheiBoulomenos nope.
i see
@Studentmath i might post another answer that combines the approaches of Emilio and Michael and use your link if u dont mind
Well, you can use resultants and compute the determinant of a 9x9 matrix
@Peter sure thing, I just googled it
17:21
@Studentmath what search terms did you use?
I think "line equation tangent to elipse"
2
A: Discriminant of Quintic (Galois theory)?

Mariano Suárez-ÁlvarezJust for reference: the discriminant of your polynomial is $$3125 p^4-256 N^5 p^5.$$ I computed it using Mathematica. For comparison, the discriminant of the general monic polynomial $x^5+a x^4+b x^3+c x^2+d x+e$ is $$256 a^5 e^3-192 a^4 b d e^2-128 a^4 c^2 e^2+144 a^4 c d^2 e-27 a^4 d^4+144 a^3 ...

There's a formula for the discriminant of a quintic
but who would want to use such a formula?
@Studentmath thanks again
@Semiclassical you around buddy?
17:30
care to give me a clue with my Lin Alg hw?
@Studentmath using the same search as you, i found a pdf with the same question cgl.ucsf.edu/home/bic/calc_class/ellipse_tangent.pdf
or too busy sorta?
What’s the question?
Let $S,T:X \to Y$ be linear mappings on finite-dimensional vector spaces
Prove: if $\ker(S) = \ker(T)$, then there is an invertible mapping $L:Y \to Y$ s.t. $S = L \circ T$
Nice @Peter
Anyone seen Prof. Ted lately?
@Balarka perhaps?
17:32
Hint: a basis of $\ker(T)$ can be extended to a basis of $X$, don't see how the hint is helpful
@Studentmath but it's from 2003 so it's not current homework
@Studentmath I have seen Ted around but wasn't able to talk to him yesterday.
@Peter I would guess it is now too, but he showed some real effort and the many different approaches are quite nice
@Balarka cheers, thanks
He's definitely around.
If the product of roots of a quadratic is less than zero does it imply that it has real roots?
17:34
Might catch him tonight/day, depends where you're around the globe I guess
@Studentmath there is a sign error in the second line solution though
Author just wrote this to solve a question without giving any explanation
@Studentmath it's not y=3-2/3(x-12) but y=3+2/3(x-12)
@Abcd No. Consider -x^2-1
@Peter in the answers on Math.SE?
17:36
@MatheiBoulomenos so?
no the pdf from 2003
Michael Rozenbergs answer is correct
ah not much we can do
Well, if anyone thinks of copy-pasting from there without understanding, will be easier to know :)
@Abcd yeah, I wasn't thinking, it is correct
@MatheiBoulomenos How? can you prove it?
You can conclude this from the intermediate value theorem.
17:38
I have no clue what's that.
If $f(x)=(x-\alpha)(x-\beta)$
then if we plug in $x=0$, we get $f(0)=\alpha \beta$
which is negative by assumption
but as the $x^2$-term dominates all other terms, $f$ must eventually attain a positive value if we plug in numbers that are big enough
(You can formulate this in terms of limits)
Since it attains both a positive and a negative value, it must attain a zero somewhere inbetween (that's the intermediate value theorem)
I am sorry. I couldn't understand. Can we prove it through $\Delta>0$ for real roots?
I'm not sure
To prove: If the product of the root of an equation is less than zero then the equation has Real Roots.
@MatheiBoulomenos Sorry, I forgot to think about your problem. I think it's a consequence of Chinese remainder theorem that $R \cong R/\mathfrak{m}_1 \oplus R/\mathfrak{m}_2 \oplus \cdots \oplus R/\mathfrak{m}_n$ if $\mathfrak{m}_i$ for $i = 1, \cdots, n$ are the maximal ideals of $R$?
17:45
Yes, you can solve it like that, I didn't think about that
The topological solution went like this: We show that if the Jacobson radical of $R$ is $0$, then $\operatorname{Specm}(R)$ is dense in $\operatorname{Spec}(R)$. Proof: A basis of $\operatorname{Spec}(R)$ is given by $D(x)$ where $x$ ranges over all non-nilpotent elements of $R$. Now if for some non-zero (hence non-nilpotent, as $R$ is reduced) $x$, we have $D(x) \cap \operatorname{Specm}(R) = \varnothing$, this means that $x$ is contained in every maximal ideal.
So I guess I need to interpret coprimality of maximal ideals topologically...
you guys have been doing this for hours
@MikeMiller Sorry, not much topology to offer today :P
@MatheiBoulomenos Ah ok that is cute
Now by assumption $\operatorname{Specm}(R)$ is dense, but as it is a finite union of closed points, it is also closed
Ya
So Spec(R) = maxSpec(R).
17:51
what on Earth...
with this $$(⋃i=25Ai)∩{x∈Z:1≤x≤20}$$

would i be considering digits in the numbers or power set elements?
Hi @Ted
Hi, @Balarka. Did you ever find a non-Chern class verification of that non-parallelizability of $\Bbb P^2-\{\text{pt}\}$?
damn i butchered that one
17:55
@TedShifrin Nope! I checked the Chern class computation, it does seem to pan out.
Hi @MatheiBoulomenos :)
Yup, easily. $c_1$ of the trivial rank 2 bundle isn't $2$ or whatever ... :)
Tangent bundle of that restricts to O(1)^3 over CP^1, which has c_1 = 3, right?
im confused as to whether I should be using the digits of integers in the range 1 to 20, or the powerset elements
Erm, maybe that plus a trivial line bundle
Hmm, no.
17:58
Hey, how do we find the URL of a question. They have a search bar there instead :(
Yeah they introduced a new top bar
Mod $\mathscr O(1)$?
Oh, duh. So, Balarka: This question took me surprisingly long. Can you give a good answer?
I'm confused. You're calculating $T\mathcal O(1)\big|_{\Bbb P^1}$'s chern classes?
Balarka wanted an easy proof that $\Bbb CP^2-\{\text{pt}\}$ isn't parallelizable.
17:59
Is $c_1$ odd?
I just used Euler on $\Bbb P^1$.

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