@TheGreatDuck You say "I know it is true so I'm personally comfortable saying theorem." My point is not that a theorem is a true statement, but that a theorem is a true statement together with a proof. If noone has bothered to write a proof, then it is a conjecture. To say that you are trying to prove a theorem is to say that you know (from some other source) that the statement is true and has been proved, but that you don't know how to prove it.
In Morocco people love soccer so I watched it a fair bit there (that's what was playing on TV 90% of the time you went into a cafe, the other 10% was news)
@Alex then by that reasoning people who havent seen proof of the fundamental theorem of calculus cannot call it a theorem because they haven't personally seen a proof.
Lol along those lines, I remember there was some talk regarding the recent algorithm for graph isomorphism because there was a non-zero dependence on the classification of finite simple groups
First we will define the left and right derivatives as follows:
Left derivative of $f(x)$: $f(x)^+ = \lim_{h \to 0^+} \frac {f(x+h) - \lim_{a \to x^+} f(a)}{h}$
Right derivative of $f(x)$: $f(x)^- = \lim_{h \to 0^-} \frac {f(x+h) - \lim_{a \to x^-} f(a)}{h}$
Now let us define the set of piecew...
(Schreyer's hypothesis, that the outer automorphism group of a finite simple group is solvable, is only known at the moment by checking off from the list in the classification)
@Alex the guy starting sending me messages telling me not to call it a theorem for some reason or to otherwise give a citation. I didn't even ask his opinion. He is kind of being pushy.
@Symposium my point is that it doesnt matter. I'm just a noob asking mse questions. I'm not gonna stumble upon anything that hasn't already been buried under ten feet of papyrus as ted put it
@Alex i asked to drop it 10 minutes ago but here we still are
@TheGreatDuck You are the one who keeps discussing it. But, since you seem not to want to actually drop it: yesterday, you asked why your question was being downvoted, and why people were voting to close it. You were implicitly asking for the opinions of anyone in the room. I suggest that (1) calling the result a "theorem" when you don't know where a proof can be found is confusing, and it might be better to call it a "conjecture" and (2) the titles of the questions are vague and click-baity.
@XanderHenderson no I was specifically asking a specific person why the post was being marked a duplicate simple because the first paragraph was copy/pasted.
@TedShifrin in an attempt to get the discussion back to math, would you recommend starting with something that deals with algebraic curves first in some detail (e.g. Fulton, or maybe something about the connection with Riemann surfaces, e.g. Miranda) or just do general varieties from the start?
Personally, @Mathein, I'd start with curves. They're interesting and have a rich source of results. Surfaces get very complicated. And general varieties ... one doesn't know much.
@LeakyNun the category of $k$-algebras is the category of monoid objects in the monoidal category of $k$-vector spaces with $\otimes_k$ as the monoidal structure. You only have to check that the equivalences for vector spaces behaves nicely with tensor products and the rest is automatic
@MatheinBoulomenos to be clear, the proof with universe issues is to take the points identified by all maps with hausdorff codomain; the proof with transfinite recursion takes the equivalence relation generated by x~y iff any open set of x intersects with any open set of y, and then quotient it Ord times
@Alex I wonder the utility of learning the material in Neukirch from different places (class field theory from Nancy Childress' book etc) vs learning it from Neukirch?
@Symposium I like Neukirch a lot. (One has to note that there are three books of Neukirch on ANT, I'm talking about "Algebraic Number Theory") The exposition is very clear and insightful, I think he does things in the right generality until you go deeper. I can't speak about the English translation, but the German original also is really good stylistically. He also has a chapter on "Riemann-Roch theory" that develops a notion of genus, divisors etc. for number fields closely modelled to the analogies with Riemann surfaces, which is great if you're intereted in that kind of stuff. I haven't …
@Zee: "gluing" usually suggests sheaf theory, but because of the specific differential geometry reference, he might have something more specific in mind.
@Mathein: If you're going to start learning algebraic geometry, divisors and linear systems are a key idea.
@Symposium Some things to note: Some exercises are _really_ hard. There's one in the first chapter where I still don't know how on earth you can solve that with the methods that are in the text up to that point. Also his approach to class field theory is quite elementary. That can be good thing or a bad thing, depending on your point of view. I personally think that having some machinery can clarifiy things, even if you need some dry building up to that. (Also if you want to go deeper in ANT; then the more advanced methods like Galois cohomology are worth learning, anyway)
@Alex it was asking to show that the Pell's equation has nontrivial solutions. That follows from either continuous fractions or Dirichlet's unit theorem, but I would never come up with either of that
@MatheinBoulomenos Let $F$ be a field, and fix an algebraic closure $\overline F$. Let $\Sigma$ denote the finite Galois intermediate fields $\overline F/E/F$, and consider the map $\varphi : \operatorname{Gal}(\overline F/F) \to \displaystyle \prod_{E \in \Sigma} \operatorname{Gal}(E/F)$. Is the image closed?
@LeakyNun consider the map $\prod_{E \in \Sigma} \operatorname{Ga}(E/F) \to \prod_{E \in \Sigma} \prod_{E' \in \Sigma, E' \subset E} \operatorname{Gal}(E'/F)$ where we send $\sigma \in \operatorname{Gal}(E/F)$ to the product of all restrictions of $\sigma$ to intermediate Galois extensions contained in $E$. $\operatorname{Gal}(\overline{F}/F)$ is the preimage of $\prod \prod \operatorname{id}$ under that
saying that this maps to the identity is just a complicated way to define that the elements in the product $\prod \operatorname{Gal}(E/F)$ have compatible restrictions
it's actually sufficient to know that if you have to parallel continuous maps $f,g: X \to Y$, where $Y$ is a Hausdorff space, then the equalizer $\{x \in X\mid f(x)=g(x)\}$ is closed in $Y$, because it's the preimage of the diagonal $\{(y,y) \mid y \in Y\}$ under the map $f \times g: X \to Y \times Y$
@TedShifrin You were at Berkeley in the 70s, right? I know that Berkeley was/is a huge department, but is there any chance that you overlapped with Bruce Blackadar?
Those were really good courses. I don't think that I would have made it through real analysis at UCR without having taken that material from Blackadar.
First we will define the left and right derivatives as follows:
Left derivative of $f(x)$: $f(x)^+ = \lim_{h \to 0^+} \frac {f(x+h) - \lim_{a \to x^+} f(a)}{h}$
Right derivative of $f(x)$: $f(x)^- = \lim_{h \to 0^-} \frac {f(x+h) - \lim_{a \to x^-} f(a)}{h}$
Now let us define the set of piecew...
