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1:29 AM
What's the intuition behind closed and exact differential forms ?
What I mean is, how do you think about forms, and what's the rough mental picture in your head such that you can determine whether it's closed or exact (or, as in most of the cases, neither) without doing calculations ?
The analogous objects in homology, viz boundaries and cycles are very easy to intuit/picture in head and it's trivial to see when a chain is a boundary or a cycle or none from the image. I'm unable to think of a similar description in cohomology
[cc @BalarkaSen maybe ?]
 
It's best to treat differential forms according to the way you see them used in practice, rather than as individual geometric objects in themselves (I find, at least).
There are geometric ways of picturing some differential forms, but I don't find these visualizations all that useful in practice.
As well, they sometimes lead to some misunderstandings.
 
ok, to be more concrete: for $1$-forms in a Riemannian manifold exactness is probably not very hard to see if you imagine them as little arrows (with the canonical identification of the tangent space and cotangent space at each point) then it's roughly equivalent to having no cycles, but I have no intuition for $1$-forms closedness (ik curl should be zero. I don't see what to look in the picture to conclude curl should be zero w/o computing) or $2$-forms exactness/closedness
 
For 1-forms, try imagining it as a hyperplane field.
That is, if $\alpha$ is a 1-form, think of it as $\ker\alpha$, geometrically.
 
what does exactness translate to in that picture ?
and I guess you're referring to thinking of them as something like: en.wikipedia.org/wiki/Contact_geometry#/media/…
 
Yes, that's exactly how I am suggesting to imagine them.
 
1:45 AM
why that viewpoint is better than the arrow viewpoint (i.e, think of $\alpha$ as $(\ker \alpha)^\perp$) ? tbh that picture looks a bit weird
and how do you think about $2$-forms on a surface ?
 
It's not "better"
Neither are very effective in the long run.
 
by "better", I mean does it help to see anything ? The arrow viewpoint is good for seeing closedness
@anakhro so how are you supposed to think about them ?
 
You aren't supposed to think about them in a particular way.
That was what I mentioned above: "It's best to treat differential forms according to the way you see them used in practice, rather than as individual geometric objects in themselves (I find, at least)."
 
I mean, in practice I see them being written in a coordinate chart as $\alpha = \sum f_I dx_I$, not sure if that's what you meant.
 
Where have you seen differential forms being used?
 
1:49 AM
I'm learning about deRham cohomology
so the reason I asked this was that I was trying to prove that $H^*(\mathbb{R}^n \times \mathbb{R}) = H^*(\mathbb{R}^n)$. So what to do is that it suffices to show $\pi_* \circ i_* - 1$ gives zero map
so we construct an operator $K$ such that $ \pi_* i_* - 1 = Kd - dK$, analogous to the prism operator in homology
but I found the prism operator construction to be very pictorial, and this $K$ to be very symbolic construction
so that's why I was asking, if I'm missing some obvious pictorial interpretation
 
In de Rham cohomology, the use of differential forms is (in my opinion) rather linguistic. It is the best way of phrasing what is happening geometrically. See something like tungsteno.io/post/exp-goal_de_rham_cohomology
But note that now when you see differential forms being used, you should not default to imagining them this way. You might see them appear elsewhere in places where their use is better geometrically described in another way (e.g. symplectic forms or contact forms).
The point is that differential forms give us a convenient language to describe and use for a variety of geometric problems.
 
ok, I see, I think I get your original point (that they don't have any canonical geometric interpretation). What in your opinion is a good way to think about/build intutioin about forms in this context ?
 
In the context of de Rham cohomology?
 
Yes
 
I am hesitant to say this, and maybe it's not quite the best advice, but in my limited experience with co/homology theories, you start to not care about the basic definition and more just about the collection of results and rules you develop as you go on.
That is, the more and more you do, the less you will actually be thinking about forms and more about the structure of the spaces and how cohomology works with operations between spaces.
For me, geometry takes the backseat to algebra.
And then you just won't care about forms.
Balarka and Ted might kill me for saying these things
 
2:02 AM
Do you know the de Rham theorem?
 
@anakhro I have experienced that to be somewhat true while learning simplicial/singular homology -- with my (very, very limited) little experience with them I think of eg $H_1$ as "number of holes" instead of thinking about chains explicitly. So I guess I should just "get used" to forms so that I don't need to think about them explicitly. Thanks !
@Thorgott that deRham and singular cohomology are same for manifolds, that one ?
 
Yes
 
It's been like 4 years since I learned about differential forms, and I have not yet gotten fully used to them. They are super versatile and appear everywhere in geometry. You slowly see more and the more you see and the more you do with them, the more comfortable you get with their existence.
 
In particular, integration of (cohomology classes of) forms along (homology classes of) chains induces a perfect, bilinear pairing between de Rham cohmology and singular homology
 
@Thorgott yeah, I know about it, but I have zero intuition about singular cohomology. It's (spaces of maps of spaces of maps of simplex to a space to a ring) too hard for me.
 
2:08 AM
I think it's pretty nice just to think about cohomology in terms of duality.
It's not uncommon to encounter duals in linear algebra.
The nice thing about dualizing homology is some things become more apparent (the ring structure of cohomology).
 
@anakhro Sorry, which duality are you talking about ? Poincare duality ?
 
I just mean the definition of cohomology that arises from taking the dual of the (homological) chain complex
 
What I'm trying to say is that boundaries/cycles make geometric sense in singular homology and the de Rham theorem explicitly translates that into de Rham cohomology as dual thereto. Forms are something you integrate, chains are something you integrate over. But I don't really know any of the singular stuff, so I can't say anything more meaningful about this.
 
3:13 AM
@anakhro: If you read any of my papers (or those of Chern, the master), you discover that there's a lot more to forms than just cohomology theories. Sometimes forms aren't closed, but they're still important. And even when you have a closed form, sometimes it's more important to have a particular representative of the cohomology class than to just have a class. But to me differential forms are extremely geometric when you're doing basic connections in differential geometry (Riemannian or not).
 
Did I say that cohomology was the only thing to differential forms?
 
I didn't read all the stuff. I just saw a lot about cohomology.
 
It was seemingly the only thing the other person had dealt with forms for.
 
3:39 AM
Hi everyone !
Can someone please tell me what is the condition for common root of two quadratic equations ? Actually I know the condition , but I want the proof of the condition
 
4:15 AM
@ronakjain sum of the roots is $-b/a =\alpha+\beta$
and $\alpha=\beta$
$$(a_1b_2-b_1a_2)(b_1c_2-c_1b_2)=(c_1a_2-a_1c_2)^2$
please ignore the above one
 
@Yuvraj question dikhao
 
konsa?
 
that which you are discussing?
 
othing is on discussion
 
can you please point out?
I want to solve it
 
5:11 AM
@Yuvraj hi
 
 
4 hours later…
8:51 AM
@Lelouch I don't have a unified picture for differential forms in my head. It depends on what I want to do.
I have explained this a thousand times to various people so I don't want to talk about it in detail once again, but it's sometimes a very good idea to imagine a $k$-form $\omega$ as eating little $k$-dimensional parallelepipeds on your manifold and spitting out numbers.
That is, $\omega_p(X_1, \cdots, X_k)$ is a value on the infinitisimal parallelepiped spanned by $X_1, \cdots, X_k$ on the tangent space $T_p M$, which you can also imagine as a a very small truncated parallelepiped on $M$ by flowing along various $X_i$'s and $[X_i, X_j]$'s.
In this interpretation $d\omega$ has a nice meaning; $d\omega$ evaluated on a $(k+1)$-dimensional "flow parallelepiped" is an alternating sum of $\omega$ evaluated on the various $k$-dimensional faces. This comes from the formula $d\omega(X_0, \cdots, X_k) = \sum_i (-1)^i X_i \omega(X_0, \cdots, \widehat{X_i}, \cdots, X_k) + \sum_{i, j} (-1)^{i+j} \omega([X_i, X_j], X_0, \cdots, \widehat{X_i}, \cdots, \widehat{X_j}, \cdots, X_k)$.
This is also the infinitisimal interpretation of Stokes' theorem. Try to understand all of this for a $1$-form, and the formula $d\alpha(X, Y) = X \alpha(Y) - Y \alpha(X) - \alpha([X, Y])$, with the $\varepsilon$-flow pentagon spanned by $X$ and $Y$.
Closedness and exactness can be interpreted in this language just like cochains in singular/simplicial cohomology
This language is somewhat useful; for example it'll immediately say that $d$-invariance of the space of differential forms on a distribution aka being a dg subalgebra is equivalent to the space of vector fields tangent to a distribution being bracket-invariant aka a Lie subalgebra, which relates to the way @anakhro wanted to think about forms
In geometry if $\omega$ is the matrix of connection $1$-forms then curvature is computed by $d\omega + \omega \wedge \omega$, and the above interpretation immediately tells you this has to do with holonomy, which relates to @Ted's point of view
 
9:38 AM
@Lelouch I never remember why Bott-Tu proves the Poincare lemma this way. Why not do the general thing, for any two homotopic maps $f, g : M \to N$, prove $f^* = g^*$?
It's very easy to interpret the analogous prism operator in this setup
 
9:59 AM
the general case is a corollary of the particular case
 
10:53 AM
Was messing with some neural network thingy so it can generate maths stuff. It then gave this strange result
> An example of a proper class is a prime number. Every natural number is a member of the set of natural numbers, but not every prime number is a member of the set of prime numbers. For this reason, the set of prime numbers is a proper class containing all natural numbers, but not a member of the set of natural numbers.
How can you have a prime number that is a proper class
 
 
2 hours later…
12:25 PM
Is it just me, or do other users have problems with searching in chat, too.
For example, if I search for the word test in this room, it takes ages and eventually returns 504 Gateway Time-out.
 
Yes, it’s happening with me also.
 
@MartinSleziak i think its not just you
 
0
Q: Stack Exchange is intermittently failing at random areas

DaniilStack Exchange is intermittently failing at random areas. Some issues experienced by me and others: I got logged out and couldn't log in Stack Overflow intermittently goes down saying it's unavailable Search is broken everywhere Posting was broken at some point

 
12:52 PM
I got logged out yesterday and blamed my phone, but maybe it was stackexchange issue then
 
1:28 PM
Hi guys, do you have any good tips for how to identify which solution to apply to which question in mathematical statistics?(distributions, probabilities, statistic tests, t-values, linear regressions etc.)
I've got a grip on the maths, but 90% of the time I can't figure out which approach to use to which question.
 
1:41 PM
If I could ask a naive question as a physics student, if we have a curve, say, $[0,1]\subset\Bbb R$ to some subset of a surface $U\subset M$, are the following two defined to be equivalent definitions?: $$f:[0,1]\rightarrow U$$ and $$f:\Bbb R\rightarrow M.$$ My reason for asking is that if the first is a bijection, then the second has the codomain as the entire manifold rather than specifying the subset $U$, which could imply that is not surjective.
 
Those are generic functions, not definitions. What are the definitions?
 
I am probably abusing the word defined
I just mean notationally they are two somewhat equivalent ways of writing the map that defines the curve
actually this might not even be a question with an answer
 
I mean, those are two different things
if you parametrize the curve over $[0,1]$, you have something that has endpoints, you don't have endpoints on a curve parametrized by $\mathbb{R}$
 
So it is incorrect to say that a curve $\gamma:[0,1]\rightarrow U$ is a map $f:\Bbb R\rightarrow M$
I can see why this sounds silly and trivial
I guess I've just seen people define maps from a subset of a set to a subset of another set and denote the map with the domain as the entire first set and the codomain as the entire second set
 
1:52 PM
$\gamma\colon[0,1]\rightarrow U$ already is a map
 
yeah
hmm
 
it can't be another map with different domain and codomain simultaneously
 
ok that solves my problem, thank you :)
 
 
2 hours later…
4:09 PM
@BalarkaSen Wow, nice. Thanks a lot for you reply ! I've seen the similar parallelipid interpreation in Arnold, Classical mechanics (or something like that), didn't realize that closedness/exactness can be interpreted in this language. This seems to be an useful viewpoint.
@BalarkaSen Well because the special case is easy to prove, the general case follows easily from the special case
(let $s_i : M \mapsto M \times \mathbb{R}$, by $s_i(m) = (m,i)$. Then let $F: M \times [0,1] \mapsto M$ be the homotopy, since $s_1^* \circ F^* = s_0^* \circ F^*$ on $H^*$, the general statement follows).
I don't immidiately see how do you prove the general case without proving this special case as a lemma (I tried to look up online, they seems to be referring Bott and Tu lol) or what's the "very easy" analogous interpreation of Prism operator in this case
ok maybe you can do it using cartan's formula or something
 
4:27 PM
you define it by doing it in coordinates and gluing together along a partition of unity, essentially
 
yes, that reduces it to the case of $M$ being Euclidean plane. I was talking about that.
 
yeah
there isn't really much more to it
 
I mean, I was saying to construct the operator for the case when $M$ is the euclidean plane, the cartan's formula gives some intution as on what to integrate
 
ah, I haven't thought about it that way, do you just mean this by virtue of analogy or is there a way to interpret Cartan's formula so that it serves as actual motivation for the proof?
 
5:01 PM
If we are given $$(n!)^2 \gt n^n$$ then how to obtain $$\left( \frac{n+1}{n}\right)^{n-1}\lt n$$
???
 
@Thorgott There's an way to interpret Cartan's formula which gives the proof
6
A: Homotopy invariance of de Rham cohomology

user99914Now I read the book and realize that Lie derivative is introduced after the chapter on cohomology, if the order is reversed there is a very direct interpretation. You want to prove: If $f_0,f_1 : M\to N$ are smooth mapping which are homotopic, then $$ f_0^* = f_1^* : H^k_{dR}(N)\to H^k_...

 
nice, that's really cool
 
5:18 PM
What's going on in this chain homotopy is that integration over the fiber (in this case $\int_0^1\,dt$) is the adjoint of $\times I$.
That is, $\int_{\sigma\times I} \omega = \int_{\sigma}\pi_*\omega$, where $\pi_*$ is integration over the fiber.
 
5:33 PM
@Knight The given there is irrelevant. Just prove the (second) inequality by induction (starting with $n=2$).
 
5:44 PM
Ah, Ted, that is also a nice way of thinking about it
 
@TedShifrin someone gave a very nice proof: $$(n!)^2= (1\cdot n ) (2\cdot (n-1)) \cdots ((n-1) \cdot 2 ) (n\cdot 1\\ \text{we know} ~ 2(n-1) \gt n \\ 3(n-2) \gt n \\ 4(n-3) \gt n \\ \cdots $$
I failed to grasp the thing from “we know” part. Can you please help me in convincing myself that for any $n$ all those inequalities hold?
 
I don't see the point, though, since it's immediate by simple algebra and induction. That proof is wrong, I think.
Yeah, it's wrong. $3(n-2)>n \iff 2n>6 \iff n>3$. $(4(n-3)>n \iff 3n>12 \iff n>4$. This is no good.
 
I said the same thing
 
Well, you are right.
 
But he said if $n<3$ then $3(n-2)$ will not appear at all
 
5:56 PM
I just wrote an answer on main because someone was puzzled by an accepted answer to an earlier question that was total garbage.
Oh, I see. Oh, OK, yeah, he's right. Fix $n$. Then of course $n$ is greater than all of $1$, $2$, $\dots$, $n-1$.
 
Is there some convention for what "naturally" means in terms of group/ring actions?
 
I cannot find your answer on this post
 
No, no. I'm talking about something completely different, @Knight. I was talking about accepted answers that are wrong. But this person is right, and you and I were being silly.
But you haven't answered my question. Why would I want to prove the second inequality using the first?
 
@TedShifrin I relized lately that it’s the induction step.
 
It's a clever idea, using a multiplicative version of Gauss's additive trick. It's cute.
What's the induction step?
 
6:00 PM
$$(k+1)^2 (k!)^2 \gt (k+1) (k+1)^k$$
 
I thought you were trying to prove $\left(\dfrac{n+1}n\right)^{n-1}<n$?
That is easily proved directly by induction (for $n\ge 2$).
 
Okay! Now, we are more than clear :-)
@TedShifrin Can you please make it even more lucid for me?
 
@Lozansky Informally, it just means that it is natural in the everyday sense of the word. Formally, it usually means that whatever is going on can be interpreted as a natural transformation between categories, but how exactly that looks like will depend a bit more specifically on the scenario.
 
You're fixing $n$, and the numbers that appear in that product are all $<n$. For all $1\le k<n$, we have $n>k$ and we conclude that $k(n-(k-1))>n \iff n>k$, which is true.
It sounds like you messed up your logic. They're using the second inequality to prove the first. You asked how to prove the second from the first.
 
Okay! Thank you.
 
6:10 PM
@Thorgott Are all actions of commutative groups natural?
But the converse isn't necessarily true
 
most definitely not, unless you are using "natural" in a very weird sense
 
We haven't touched upon categories so I suspect a superficial understanding of it is enough:P
 
most likely, yes
@Ted btw, do you know a good reference where I can read up on that unit tangent bundle stuff?
 
6:29 PM
If $P$ is a prime ideal and $x$ a nilpotent element, then I can be sure $x^n = 0 \in P$ for some $n \geq 0$. How can I be sure there exists $k$ such that $x^k \in P$ but $x^{k-1} \notin P$?
 
pick the smallest $k$ such that $x^k\in P$
 
You might try Spivak Volume 5, @Thor. I have done the Grassmannian proof in my classes a few times, but not the sphere bundle proof.
 
@TedShifrin How to make something appear in spoiler?
 
Oh, I once figured that out, but I don't remember. Google.
@Knight. Here you go.
 
6:48 PM
I don't know what Spivak discusses when it comes to the sphere bundles/Grassmannians, but whatever it is, it's too advanced for me :P
 
@Thorgott Ah I was thinking about the counter-case when every power of $x$ is in $P$ but then the statement (all nilpotent elements are containted in every prime ideal of a commutative ring) is trivially true
 
7:03 PM
that can't happen if you permit $k=0$
cause $x^0=1$ is not contained in any prime ideal
but I don't see why you need such a $k$ in the first place; it is immediate from $x^n=0\in P$ that $x\in P$, by definition of a prime ideal
 
Ah, a prime ideal needs to be a proper ideal?
Oh it does
@Thorgott Yes, I see what you mean (I think)
One can "track back" the factors of $x^n$
 
7:19 PM
Yes, prime ideals are proper ideals such that $xy\in P$ implies $x\in P$ or $y\in P$. A straightforward induction establishes from the latter property that if $\prod_{i=1}^nx_i\in P$, there is an $i\in\{1,...,n\}$ such that $x_i\in P$.
 
Yeah
Induction is probably a more proper term than "track back" :P
 
8:08 PM
Hi everyone
 
Hi, demonic.
 
 
2 hours later…
10:04 PM
hey chat
 
hi Lucas
 
Why does Lucas come in and say hi and then say nothing more?
 
dumb question: if $\zeta$ is a n-root of 1 (i.e., $F(\zeta)$ is the splitting field of $t^n -1$), does it follow that the set of roots is $\{1, \zeta, \ldots, \zeta^{n-1}\}$?
like in the complex realm
 
Yes, of course.
Oh, wait. No.
You didn't say primitive $n$th root.
So your i.e. is wrong.
 
@TedShifrin I'm sorry. I'm quite busy since last week - had my finals, a lot of real analysis lists and the Galois theory report
 
10:17 PM
LOL, well, you don't have to come in and say hello :P
 
I always come to chit chat but... :p
@TedShifrin oh, yeah, you're right. if I add that to my hypotheses then it's obvious
thanks Ted!
 
10:33 PM
If $E$ is a finite extension of $F$ and $K$ is a finite extension of $E$ and we have bases $\beta_E, \beta_K$, then can we be sure a basis of $K$ over $F$ is given by $\beta_E \times \beta_K$?
The product is elementwise
 
yup, even works if they're not finite
 
Does this result have a name?
 
ABC
I was reading that the line is not defined because it's taken as an obvious concept, like the point.
In general I know that in a cartesian coordinates the line equation is $ ax + by + c = 0 $.
How can I say that If I don't know the definition of line?
I need to define line directly as $ ax + by + c = 0 $?

Sorry for the stupid question
 
I think it follows from the degree theorem or whatever it's called
 
@Lozansky try to write down an arbitrary element of $E$ in terms of $F$, and then an arbitrary element of $K$ in terms of $E$ ( -> in terms of $F$). so you get this result
 
10:36 PM
it's the other way round
 
yeah, that's pretty much it. degree is multiplicative in towers
 
this is usually how you prove the "tower law", which says that degree is multiplicative
 
@ABC ... they mean in classic Euclidean geometry it's an undefined term. Certainly with vectors or analytic geometry we have a definition.
 
but I don't think the fact about the bases has its own name
 
ABC
@TedShifrin which is it?
Where can I find this def?
 
10:38 PM
You just gave one. Or I can write it parametrically in vector form.
 
ABC
Ok so I can take the equation written in the precedent post and take it as a definition of line?
 
@TedShifrin but even if I use a primitive root (every $n$-root $\alpha$ is s.t. $\alpha = \zeta^r$), I do not have it guaranteed to have $n$ distinct roots, right?
 
I don't know whom you're doing this for. Of course, lines might be in higher dimensions, and then your definition is wrong. You have a hyperplane.
Of course you have $n$ distinct roots, @Lucas. That's what it means for it to be primitive. You generate the group.
 
you will always have $n$ distinct roots if $n$ is coprime to the characteristic of the field you're working in, it can fail otherwise
 
ABC
@TedShifrin ok so I need only to generalize my equation
 
10:43 PM
Your "definition" is more problematic in higher dimensions, because then you need dimensions and the theory of systems of linear equations. My definition writing it down vectorially, parametrically, works in arbitrary dimension (even infinite).
As often happens with you, I have no idea of the context of your question.
 
@TedShifrin I know it generates the group - but why does the group of $n$-roots have exactly $n$ elements? Why not a number that divides $n$?
In the case $n$ prime, it looks straightforward.
 
We're working in $\Bbb C$, right?
 
nope, arbitrary fields
 
a primitive $n$-th root is by definition an $n$-th root that has order $n$
 
Oh, I thought you said $\Bbb C$ earlier.
 
10:44 PM
so it generates a group of order $n$
 
So, of course you're right. If the field isn't algebraically closed, you might not have $n$ roots.
Consider $t^4=1$ in $\Bbb R$.
 
yeah, primitive roots don't necessarily exist
 
maybe Lang's definition is a bad one
 
ABC
@TedShifrin
I was proving that the equation of the parabola represented exactly the geometric definition of a parabola.

having done this I wanted to do the same thing for the straight line, and I discovered that there was no definition as a geometric place.
That's all
 
my problem is: if $\zeta$ is a primitive $n$-root, does it always follow that $t^n-1$ splits in distinct linear factors over $F(\zeta)$?
 
10:47 PM
If you're doing parabolas, you're doing analytic geometry (with cartesian equations), so your definition is fine.
 
yes, it does
 
Lang also assumes $\mathrm{char} F = 0$, but you can state pretty much the same results about radical extensions as long as you have $F(\alpha)/F$ separable
 
Yes, it does.
How can $\zeta^k=\zeta^\ell$ ?
 
the existence of a primitive $n$-th root implies that there are $n$ distinct $n$-th roots
 
ABC
@TedShifrin Thanks
 
10:49 PM
Sure, @ABC.
 
a primitive $n$-th root exists iff the characteristic is coprime to $n$
 
@TedShifrin $\zeta^{k-\ell} = 1$. ok.
hmm...
if $d$ is the least positive integer s.t. $\zeta^d = 1$, then $d | n$
which is pretty much the Lagrange theorem on subgroups
 
$d=n$, that's the definition of a primitive root of unity
 
my brain is not working. I really can't see why that definition implies $n$ different roots (even if I know, obviously, that anyone who call something a "primitive $n$-root of unity" expects $n$ different roots)
 
what's your definition of a primitive root of unity?
oh wait, it's in the image
hmm, this looks plain wrong to me, actually
 
11:05 PM
So, in my example, $\zeta=-1$ would be a primitive fourth root of $1$ in $\Bbb R$.
Are we sure Lang means to allow this?
 
oh wait, you are assuming characteristic $0$
 
I just gave a characteristic $0$ issue.
 
@Ted when he says "root of unity", I'm sure he means in an algebraic closure
 
Yeah, I think so, too.
 
you are abstractly adjoining one after all, we didn't fix an ambient field or anything
so this works, because, in characteristic $0$, the roots of $x^n-1$ are all distinct
 
11:07 PM
@Thorgott yeah
 
so we don't have any issues after all
 
@Thorgott why?
 
Because the polynomial has distinct roots in the algebraic closure.
 
(I promise I'm not doing this to bother you guys...)
@TedShifrin ...why?
 
Do $f$ and $f'$ have any common roots?
 
11:09 PM
because repeated roots are the same thing as common roots between the polynomial and its derivative
 
this is where the "$n$ coprime to characteristic" condition comes from
 
yeah, otherwise it would be purely inseparable, right?
in fact, $x^{pm} - 1 = (x^m - 1)^p$
 
not necessarily
yeah, that is the crucial observation
but $x^m-1$ is a separable factor
so splitting that gives an intermediate separable extension
if $m>1$, that is
 
got it. thanks @Thorgott, @Ted!
 
11:16 PM
np
 

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