Well, you put in Nullstellansatz and removed it. So, he was like, is my Nullstellansatz not woth entering this chat room, and cursed this world and in particular you.
I am working on a problem involving the Frobenius map, denoted $\varphi$. The problem says prove that $\varphi^n$ is the identity map. Does this mean to iterate the map?
It's about time!! I finally have more votes in [elementary-set-theory] than I have in [set-theory]. (I have like 50 more answers in the elementary tag... :-))
@robjohn Perhaps I remember Asaf telling us that he sketched a proof without going to its details. Students lost 15 or 20 points in there (for 100 total points) which might have been dearer for a few!
Well, I am off to walk my dog. She likes Saturdays because she goes with my wife to the place where my wife works on jewelry on the weekends. I go there to walk her in the countryside.
So suppose we have a holomorphic around the origin in $C^n$. We can uniquely write it as the product of a Weierstrass polynomial in a specified coordinate and $h$, where $h(\bar(0))\neq 0$.
Can someone explain the statement, "The zero locus of $f$ projects locally onto the hyperplane $w=0$ (where $w$ is the specified coordinate) as a finite-sheet cover branched over the zero locus of an analytic function (the discriminant of the weierstrass polynomial)
@KannappanSampath You see the thing you want is motivation. If you know algebraic topology and the like than there is more motivation for category theory.
The trouble comes in showing the surjectivity of the map
So I tried to look at what maximal ideals that contain $I(X)$ look like in $k[t_1, \ldots , t_n]$.
By the weak Nullstellensatz because $k$ is algebraically closed we know that any maximal ideal in the polynomial ring is of the form $(t - a_1, \ldots, t - a_n)$ where the $a_i \in k$.
well it mentioned Nullstellensatz so that's why I looked it up :D
@DylanMoreland Do you think it would be a good idea to try and realise every maximal ideal $m$ in the form $m_x$? Surjectivity would then be proven no?
I guess I don't see where the hitch is. It seems like you have everything you need to do the problem: for $\mathfrak m = (t_1 - a_1, \ldots, t_n - a_n)$ a maximal ideal set $x = (a_1, \ldots, a_n)$. Then each $t_i - a_i$ vanishes on $x$, hence $\mathfrak m \subset \mathfrak m_x$ and these are both maximal ideals.
@BenjaminLim They might have suggested that for the purpose of showing that the maximal ideals of $P(X)$ correspond to the maximal ideals of $k[t_1, \ldots, t_n]$ containing $I(X)$.
It's applicable because the weak Nullstellensatz tells you about ideals of the polynomial ring. You are doing a problem involving the ideals of a quotient of that ring.
@Benjamin I got to prepare for exams. I've procrastinated enough. I'll be back to check the site once after a few hours. Let's hope I'll catch you then!
@BenjaminLim It goes to its image in the quotient. But it's also true that the restriction of the quotient map $k[t_i] \to P(X)$ to $k$ is injective. Because $P(X)$ is not zero and $k$ is a field.
You see if I know that the image of $(t_1 - a_1 , \ldots, t_n - a_n), a_i \in k$ in the quotient is $(\xi_1 - b_1, \ldots \xi_n - b_n)$ for $b_i \in X$ then automatically I can say that there is a point $x \in X$, namely $x = (b_1, \ldots, b_n)$ such that each polynomial in the ideal vanishes at $x$.
I look at the preimage of this maximal ideal in the polynomial ring. I know it has this form $(t_1 - a_1, \ldots, t_n - a_n)$. Why is $(a_1, \ldots, a_n) \in X$?
If I know why the image of a maximal ideal $(t_1 - a_1 , \ldots t_n - a_n)$ that contains $I(X)$ in the quotient is $(t_1 - b_1, \ldots t_n - b_n)$ where each $b_i \in X$ I am done. That is it.
@BenjaminLim Now, you're in the situation $I(X) \subset \mathfrak m_x$. Or maybe we should use some different notation for the maximal ideal because we're considering it in $k[t_i]$. It doesn't matter.
Kannappan: Regarding your text in the proof of 2.5. (b): at the end you write "So, $A$ is closed. Since, $B\neq \varnothing$, $A \neq X$ ." I think you should write "Hence $A$ is also closed since it is the complement of an open set and it is non-empty and proper since $B \neq \varnothing$ and $A \neq X$."
To finish (e) you do something similar to what you did for (d): You assume that $X$ is not connected so it can be written as a union of two separated sets $A$ and $B$. Then we show that they are both open (in the same way as we did in one of the other proofs). Then you assume you have a non-constant continuous map from $X$ to $\{0,1\}$ with the discrete topology. Then you show that that's a contradiction.
@Robjohn: I wonder, if you have ever seen the construction of the Riemann integral as an $L^\infty$-extension of linear functional determined only on piecewise-constant functions? I've found it interesting that such extension only need a little knowledge of metric spaces and that's it
@robjohn yeah, but then I remember us having some technical proofs of existence and so on. Ok, I'll write it down and show you later, would you take a look?
@JonasTeuwen I love these Spanish-Mexican beers to be honest: Corona, Desperados and Sol (quite worse that first two) - these you can by in AH and wherever. On Gran Canaria I've also tried the local beer 'Tropical' - that was perfect
I'll check back later, I was actually going to post some solutions to the exercises in the set theory book in the other room. It would be nice if we could talk about the Fourier transform later (or whenever you have time and are in the mood), @robjohn
