It's often useful to quote which part of his work you're relating your inquiry with. It may also be alright to reach out to supervisors or students who are familiar with the topic, risks apply and results vary.
I was a graduate student a decade before email. I generally bugged fellow grad students and faculty where I was. Sending a question by mail to an author was very rare unless faculty encouraged us to do so for a good reason.
I remember reaching out to a person who did a vague research paper in the brink of the 2000s. Never got taken seriously, there was a courteous response from her academic supervisor to refer to her, but no further communication on that thread. So, I spent like a month pecking every word of that paper trying to figure out how to get a result from their concept, retyping it on overleaf hoping the lack of information will be interpolated by some wild imagination that aligns to a succesful result.
I have had one question for a long time and I have not yet found its explanation/answer anywhere :(. I don't understand how the third isomorphism theorem was applied here in Exercise 7 on page no. 3 here www-users.cse.umn.edu/~felix077/download/sec9.pdf
I can prove that $Z[x]/(2,x^3+1)\simeq Z_2[x]/(x^3+1)$ by defining a homomorphism though, i.e, by showing that the natural map from $Z[x]$ to $Z_2[x]/(x^3+1)$ has kernel $(2,x^3+1)$.
It's the usage of the third isomorphism theorem that I don't understand and it seems quicker than finding an explicit homomorphism.
If yes, then I understood that. It's easy and intuitive. Keeping in view the Euclid's division, it is to be noted in Z[x,y]/(2,x,y), higher power terms will be 'zero' and infact linear terms involving x and y will also be zero so the ring is isomorphic to $Z_2$.
But my question is on the usage of third isomorphism theorem this time.
it also makes an explicit notational distinction between J and J-bar that is maybe suppressed with "(x^3 + 1)" type notation that does not expressly distinguish ideals in different rings
i'm in favor of ted's approach
of course, i also like writing down actual maps. all of the theorems about X simeq Y simeq Z are just shortcut ways of suppressing reference to actual maps in situations that come up so often that it makes sense to do that
I understood now finally :). Here's the explanation: I know that $Z[x]/(2)\simeq Z_2[x]$ by the map $\phi(z(x)+(2))=z[x]\pmod 2$ (i.e., the coefficients of z(x) are reduced modulo 2. This same $\phi$ also acts on $(2,x^3+1)/(2)$ and gives $(x^3+1)$.
leslie, re our yesterday's discussion: If $A\simeq B, C\simeq D$ and C,D are ideals of A and B respectively and given that $C\simeq D$ by the restriction of isomorphism between A and B then we have $A/C\simeq B/D$.
lucas: after inner producting, isn't it basically the 1D case, i.e. if a differentiable solution f(t) to f'(t) = k f(t) [k nonzero] has a zero somewhere, then it's identically zero. with those hypotheses, the auxiliary function g(t) = f(t) e^{-kt} satisfies g'(t) = 0 identically and so is constant by MVT or whatever. this is solving the ODE, i guess, but not with deep theorems
the field in $f$, so pick a point $p$ such that $f(p)\ne0$
Draw a short segment through $p$ perpendicular to $f(p)$
There's an neighborhood of $p$ where the value of $f$ is still approximately $f(p)$, and so if we make the segment fit in that neighborhood, $f$ everywhere on that segment points to the same side of it
Now, let $F_t$ be this dense flow, so that $\frac d{dt}F_t=f(F_t)$
Because $F$ is dense, it passes close to this line segment
Hm, I want to be able to say it passes through the line segment
I know the geometric picture, I'm just struggling with the formalism
so let's think about the 1d version
Wait, no... I think I might see where the issue is
@LucasHenrique Is there a theorem like, if I have a point with a nonzero vector, there's a neighborhood of it such that there's a homeomorphism from it to a neighborhood of $\Bbb R^n$ where the pushforward of all the vectors are parallel?
TUBULAR FLOW THEOREM
I did not know this theorem, but I saw you mentioned it above, and I Googled it and it's exactly what I want
OK so lol
Regular point. Homeomorph to parallel horizontal flow near origin. Take a circle neighborhood there. All points there flow either forward or backward to hit the y-axis. Homeomorph back. Now all points in a neighborhood of our regular point flow either forward or backward to hit the image of that axis.
OK then lol if he says "you use the tubular flow theorem to solve this" and then you haven't covered the tubular flow theorem then it is not reasonable for him to give the problem to you
But the tubular flow theorem and JCT seem like exactly the right tools to turn the vague argument precise
I'm assuming the "dense flow" only takes values $t\in[0,\infty)$, by the way
The vague argument is, make a small segment perpendicular to the flow. Density and the tubular flow theorem guarantee that the dense flow passes through the segment twice. That bit of the flow plus that bit of the segment then makes a loop, and JCT says the loop has an inside and an outside. From then on the flow is either trapped inside or outside the loop, and so it can't be dense on the other side of the loop.
Honestly, if you have to write something on the test, write literally that paragraph. All the finicky details are in setting up the tubular flow theorem, and if you didn't learn it you don't have to write it
(If the flow lets time be negative also, I think you can essentially just do this twice)
Also, this works directly on the sphere, no need to project it to the plane
By the way: it's really instructive to think about why this fails on the torus
(basically because the loops on the torus you get don't have disconnected insides and outsides)
@TedShifrin I think I understood it now.$R= Z[x], I=(2), J= (2, x^3+1)$ so $I+J=J$. $Z[x]/J= R/(I+J)\simeq \frac {\overline R}{\overline J}=\frac{Z_2[x]}{(x^3+1)}$
Pick a circular neighborhood, pick a point on the dense flow in the neighborhood, flow long enough to leave the neighborhood, and then keep on flowing until we return to the neighborhood again
@TedShifrin ok. In that case, $ I+J\subset (I,J)$ but the reverse containment may not hold as $2f(x)+ (x^3+1)g(x)$ is not in $I+J$ unless f(x)=g(x). What am I missing?
No no. I think I’m wrong.
$2f(x)\in (2), (x^3+1)g(x)\in J$ so the sum is in I+J.
Assuming the result true for now (I'll try proving the result itself shortly), I have one confusion: I get by using the result $Z[x]/(2,x^3+1)=R/(I+J)\simeq \color{blue}{\frac{\overline R}{\overline J}}$. I know that $R/I\simeq Z_2[x]$. How is it obvious from your way from here that the blue colored expression is isomorphic to $Z_2[x]/(x^3+1)$? I ask this because **it is not true in general** that if $A, B$ are commutative rings with unity and C, D are their ideals respectively such that $A\simeq B, C\simeq D$ then $A/C\simeq B/D$.
The result in last line holds if $C\simeq D$ such that the isomorphism is restriction of the isomorphism $A\simeq B$. Was this result assumed to be known?
wolg: depends on context. sometimes if f is defined on [0,L] even extension would be defining it on [-L,L] by reflecting its graph on [0,L] across the y-axis. odd extension would be defining it on the same interval by reflecting its graph on [0,L] across the y-axis and then reflecting that across the x-axis.
sometimes paired with extending the result to be periodic with period 2L.
if you want a piecewise formula for the result, you'd have to unravel the above operations on the graph symbolically. if f is defined on [0,L] the extension defines f(x) for x in [-L, 0) to be either f(-x) [even extension] or -f(-x) [odd extension].
if f(0) isn't 0 it kinda doesn't have an odd extension. maybe you fix this by redefining f to be 0 at 0.
Let $\phi: R \to S$ be a surjective homomorphism. Prove that
$\frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi} \cong \frac{S}{\phi(J)}$ for an ideal $J$ of $R.$
Obviously, $S \cong R/ \ker \phi$ (first isomorphism theorem) and $\phi(J)$ is an ideal of $S$, since $\phi$ is surjective ($\forall s \i...
I tried to prove it like this: By the third isomorphism theorem, $\rm\frac{R/ \ker \phi}{(\ker \phi + J)/ \ker \phi} \cong \rm\frac{R}{\ker\phi+J}$ so the problem now boils down to showing that
Consider the map $\rm\tau$. $\rm\tau (r+\ker\phi+J):=\phi(r)+\phi(J)$
The map is well defined as $\rm r_1+\ker\phi+J=r_2+\ker\phi+J\iff r_1-r_2\in\ker\phi+J$ and $\rm \phi(r_1)+\phi(J)-(\phi(r_2)+\phi(J))=\phi(r_1-r_2)+\phi(J)$. Suppose that $\rm r_1-r_2=k+j$ for some $\rm k\in \ker \phi, j\in J$. It follows that $\rm\phi(r_1-r_2)+\phi(J)=\phi(…
@TedShifrin I think I understand now (but I find it more convenient to mod out by $(2)$).
@TedShifrin I definitely do not stubbornly refuse to understand argument. I made an attempt to understand spivaks proof. I did not understood it, so I decided to do a proof on my own. I failed to make any progress. Then Akiva gave a proof sketch, and so far I think there is some ideas akiva used that I should have tried.
Would I say that it was a waste of time for me to attempt a proof on my own? no, because from my failiure I have learned some new ideas/techniques from akiva I can try out later. It also decreased my fear of being stuck.
Would I avoid Spivaks proof now? no. After I get a proof from akivas idea, I would learn spivaks proof again. Maybe starting again would make spivaks proof click for me.
@Prithubiswasleftmse Here is another idea: Let $g(x) = f'(x)$ for $x \neq a$ and $g(a)$ be the limit. It is continuous. Pick some $\delta>0$ such that $f$ is defined on $\bar{B}(a,\delta)$. Define $h(x) = \int_{a-\delta}^x g(t) dt+f(a-\delta)$ and show that $h=f$ on this interval. Then $f'=h'=g$.
Let f : A → B be a map. It naturally induces an equivalence relation on A, in which a1 ~ a2 if and only if f(a1) = f(a2). (Put another way, the equivalence classes are exactly the sets f^{-1}({ b }) for each b∊B.) Is there a standard name for this equivalence relation, or for the corresponding quotient of A?
huh. well it seems pretty clear that the polynomials a_j don't have to be the zero polynomials. for example (y^2) y^4 + (-y^4) y^2 = 0, but neither y^2 nor -y^4 is the zero polynomial so that just won't work.
That key does work on his computer unless he has a really weird keyboard
shouldnt the two mappings be inverses of eachother i.e shouldnt we have that if $\tilde{u}=T_{\#}u$ and $u=S_{\#}\tilde{u}$ then $S$ and $T$ are inverse of eachother
I am currently trying to understand quotient spaces, but I have a problem with the following equality: dim (V) = dim (V/U) + dim(U) Over R I could choose the vector spaces V = span(e1, e2, e3) and U = span(e1), then I think dim(V) = 3, dim(V/U) = 3 and dim(U) = 1. Clearly dim(V/U) = 3 is wrong.
I assumed that V/U contains 3 equivalence classes, one for each base vector of V. Where is my thought process wrong?
Are the equivalence classes wrong? E.g. some vector (a,b,0) cannot really be assigned a proper class.
I want to show that a prime ideal in Boolean ring 💍 is maximal.
If the ring contains unity then it is easy. As Boolean rings are commutative, for a prime ideal $P$ the ring $💍/P$ is both integral domain aswell as Boolean ring. The only non trivial integral domain and Boolean ring is $\mathbb{Z}...
Does knowing $\Pr(X\ |\ A)$ and/or $\Pr(X\ |\ B)$ go any distance toward pinning down $\Pr(X\ |\ (A\ \cap\ B))$? Or does the latter help pin down the former two? I have been writing a number of equivalent statements to $\Pr(X\ |\ (A\ \cap\ B))$ using conditional probability formulae, but have been unable to not only solve it given $\Pr(X\ |\ A)$ and $\Pr(X\ |\ B)$, but to even get them to appear in the formula to reduce the degrees of freedom.
ted: refused to get out of bed because she stayed up by herself until at least 9:30pm last night. had to be carried downstairs to breakfast. had to be dressed. demanded jelly beans on pick-up.
"If $A$ has probability $0$, isn't $P(X|A)=P(X)$?" I assume you are referring to this. Is this actually true? Wikipedia says this conditional probability is undefined. I can see a motivation for defining it if your $A$ is, i.e., the event that a continuous random variable takes a single value, but not so much if your $A$ is the intersection of two mutually exclusive events.
Even in the former case I'm not sure I understand your formula.
For example, the probability that you are taller than $5$ feet given that you are exactly $6$ feet surely isn't the unconditional probability that you are taller than $5$ feet?
You still conclude though that knowing any one or two of $\Pr(X\ |\ A)$, $\Pr(X\ |\ B)$, and $\Pr(X\ |\ (A\ \cap\ B))$ does not reduce the number of degrees of freedom present in the other(s)?
I should make a network graph that shows what probabilities relate to each other algebraically for all combinations of up to three events and all common operators, to get a sense of which ones are in the same "family" in terms of degrees of freedom.
I want to show that a prime ideal in Boolean ring 💍 is maximal.
If the ring contains unity then it is easy. As Boolean rings are commutative, for a prime ideal $P$ the ring $💍/P$ is both integral domain aswell as Boolean ring. The only non trivial integral domain and Boolean ring is $\mathbb{Z}...
