It seems that arturo assumes 1st that $f$ can be factored into $m$ distinct irreducibles, which is what I thought he was trying to show. And also that the roots $\alpha$ and $\beta$ that he chooses are roots of distinct $q_{i}$'s
The second problem I guess is just a technicality that can be dealt with, but how does he assume that $f$ can be factored into the desired $m$ distinct polynomials?
But on its own I find it kinda boring. What I'm trying to understand these days is all sort of built on Tannakian formalism ("the category of representations recovers the group"), for example, but I can't bring myself to actually read anything on it.
But I think we get to use a lot of tools. Fun tools. And I think that numbers are good motivation.
I like manifolds, for example, but proving something about a manifold doesn't sound so exciting to me on its own. If that manifold tells me something about a Galois group, then that's different.
nope, I find approximating stuff, in general, boring. I like number theory. but to really go far in it you need to know complex analysis, general topology, commutative algebra and representation theory, blah blah blah.
@MarianoSuárezAlvarez I was thinking about the title you showed me "a Theorem is simply a very organized family of examples," and have a question about its relationship to proof?
In real mathematical practice (as opposed to what the people studying logic and proof theory study) a proof is an argument which will convince the reader
lots of times, papers do not contain complete arguments
most theorems identify classes of objects which have some property
the hypotheses describe the class of objects
"a continuous function on an interval is Riemann integrable there"
sometimes, the "class of examples" of the theorem consists of exactly one object
or 5, like the platonic solids
in fact, lots of theorems arise in precisely that way: you are interested in some property of a class of objects, you find one example which has it, and you study in detail exactly why it has that property
but interesting definitions capture a set of features of objects which are sufficiently rich that you can learn a lot from those objects knowing only that they satisfy the definition
the reason why we define the abstract notion of group
is not that we are interested in the abstract notion of group
but because we are interested in a lot of objects which happen to satisfy the definition of group
initially, people worked with groups without given them that name
the set of permutations of a set is sufficiently interestng that one can study it without abstracting it into "groups"
but then people noticed that a small set of properties of the set of permutations of a set were enough to prove a lots of other of its properties
you can always ask why, because definitions are motivated
Are not the real-number properties statements about numbers that are accepted as true? and that form the basis for computation in arithmetic and in algebra.
@MarianoSuárezAlvarez When you say that those axioms are "categorical" do you mean that the axioms only apply to those objects in the category of your construction?
Another example: it is a theorem of Steinitz that the conditions «algebraically closed field of characteristic $p$ and cardinal $\kappa$» (with $p$ a prime or zero and $\kappa$ an uncountable cardinal) is also categorical
this has the consequence that every time you have two algebraically closed fields of the same cardinal and the same characteristic, they are isomorphic
Proof by rearrangement of four identical right triangles of a a very organized family of examples known as the Pythagorean theorem.
a large square, side a + b, containing four identical right triangles. The triangles are shown in two arrangements, the first of which leaves two squares a^2 and b^2 uncovered, the second of which leaves square c^2 uncovered. The area encompassed by the outer square never changes, and the area of the four triangles is the same at the beginning and the end, so the black square areas must be equal, therefore a^2 + b^2 = c^2.
@DavidWheeler This seems like it should be so easy, but I have no idea what is going on here. I'm basically trying to find a map that permutes the roots of $x^{3}-x+1$ and $x^{3}-x-1$ ?
@tb I don't understand why I can't define my function on a closed set. For example $e^x$ on $[0,1]$. I'm not sure $e^x$ is analytic and I don't know how to check but it "looks" analytic to me.
analiticity is only defined for functions defined on open sets
it means «the function has a development as a power series around each point» and for that to make sense, the domain of the function must contain an interval centered at each of its points
On the other hand, the function $f:t\in [0,1]\mapsto e^t\in\mathbb R$ is the restriction to $[0,1]$ of an analytic function defined on an open set containing $[0,1]$
(there is no such thing as «analytic from the left» like there is «differentiable from the left»...)
I was wondering does anybody remember the name of the mathematician who derived the general rule for finding $\sum_{i=1}^{n}i^k$ for any k in \mabb{ N}
There was a discussion in M.SE about this. I am not able to find it now :(
@tb So the definition doesn't work. That was a stupid question. But even so: can you tell me why I can't talk about $supp(f) \subset O$ where $O$ is open and $f$ is defined on $O$ and $supp(f)$ is compact for $f$ analytic? If $f$ is defined on an open set but has compact support it does not break the definition, does it?
@MattN Suppose $f$ is analytic and non-zero on the connected open set $O$. Then the set of zeroes is a closed discrete subset of $O$. Thus the support of $f$ is all of $O$ and hence not an interesting notion.
You can talk about it but I don't understand why you would want to do so.
an analitic function never has compact support, unless it is empty
(if f is defined on an open set U, analytic there and with compact support, then it vanishes on an open subset of U, and therefore vanishes identically)
@MattN can you formulate what you're actually looking for?
@KannappanSampath looking
@KannappanSampath I can't help you there. I always fall immediately asleep when questions are asked seeking similarities for two manifestly unrelated concepts that happen to bear the same name! There are only finitely many words!
@tb Hmm, that's right, but we still could show some differences, no?
@MattN I wanted to point out some logical differences between the two. Some facts about topological basis won't make sense at all for basis in Linear algebra, I don't know how I would get that across!
@tb I want to see why an analytic function can't have compact support. And so far it feels like I was told that that's because it can't have compact support or to use the identity theorem. Yes, I can use the identity theorem but then I don't see why, I'm just stupidly referring to a theorem.
@KannappanSampath One reason for the utility of bases in linear algebra comes down to the equivalence of two descriptions. Top down: minimal generating set; Bottom up: maximal independent set. There's no immediate analogue for topological bases.
(and, as I understand it, this is the thrust of Jyrki's comment)
As you wish. Topological bases are supposed to be small collections on which you have a good handle; they are enough to understand the open sets in a topology. There's no minimality requirement and no independence requirement (and independence would make little sense).
@KannappanSampath No: the idea is that a generating set is "large": a basis is the smallest large set (you go down: the smaller a set, the smaller the chance it generates). Conversely, another intuition is that a linearly independent set is "small": the larger you make a set the larger the chance that it becomes linearly dependent: you go up. Bases mark the "middle" of those ideas.
