Hardy was asked what his greatest contribution to mathematics was, Hardy unhesitatingly replied that it was the discovery of Ramanujan. He called their collaboration "the one romantic incident in my life."
doens't look too demanding. You need to know what singular homology is; that a short exact sequence of chain complex induces a long exact sequence of homology; and be comfy with the double cover $p: S^n \to \Bbb{RP}^n$
Ok. In order to have a Gaussian Prime our result must be a 3 mod 4 and we can't represent it as a sum of squares. If our result is a 1 mod 4 and can be represented in the sum of 2 squares, then we don't have a Gaussian Prime. Ex. 3 is a Gaussian Prime because we have a remainder 3 in mod 4 and we can't write 3 as a sum of squares, so there's no (a+bi)(a-bi) combination.
@MikeMiller Trying to inspect the least genus (not sensible, up until now) manifold $M$ in which the Cayley complex $X_G$ of $G$ embeds. $M$ is probably a 3-manifold, I dunno.
You think? I trust it makes sense to generalize now that we tried to define genus of groups by genus of the least 2-manifold inside which $Cay(G)$ embeds.
first, no, there's no nice notion of 'genus' or even anything that behaves like it for 3-manifolds (in any sense that I know - you probably should clarify what properties you want genus to have), whose story is much more complicated than surfaces
second I'm not convinced it's a natural question to talk about embeddings of 2-dimensional complexes, unlike embeddings of graphs
Does there exist a polynomial in one variable $f(x) \in R[x]$ such that $f$ has infinitely many roots? Clearly, any polynomial over a domain in one variable has finitely many roots, so $R$ must have zero divisors for this to be a question at all. I'm also aware that multivariate polynomials over algebraically closed fields can have infinitely (in fact, uncountably) many zeros. But I couldn't quite answer this one.
I'm reminded of, say, how $x^{2}-1$ has four roots over $\mathbb{Z}/8\mathbb{Z}$. I couldn't concoct a more dramatic example though. I'm wondering if there's an obvious solution one way or another, or if there's something more exotic at play.
Given a commutative Ring $R$ of ordered pairs $(x,y)$ of reals $x,y$ with addition and multiplication defined in the following way.
$$(x,y) + (u,v) = (x+u,y+v)$$
$$(x,y).(u,v) = (xu-yv,xv + yu)$$
I already showed that $R$ is an integral domain , now i need to show to prove $R$ is a field or no...
@MikeMiller Why isn't my question research-related? There might be people continuing A.-E. studies in that direction. Indeed, that is why I ask "what is known"
@VincenzoOliva Look at questions with the tag "research" The tag is mostly used to talk about mathematical research itself, not to ask about questions that you're researching.
Frankly, I don't think the tag should exist at all.
It's important to note a possible connection with other series families by using the simple fact that $$-1-\sum_{n=1}^{\infty} (-1)^n x^n (1-x)^n =\frac{1}{x^2-x-1}$$
I have feeling that this was discussed on meta before: Solution to my question exceeds my knowledge. Does anybody remember a post there with a similar topic? (I did not find it.)
@MartinSleziak I remember that being discussed before, but abstractly, iirc. This is the only one I found so far, but I'm almost sure that it has been discussed elsewhere too.
I think I even remember something along the lines: It is ok not to accept answer. If you find out that the question is above your level of mathematical maturity, you can get to it later, after you learn the stuff.
However, it's no big deal, I just thought that adding pointers to the earlier discussion would be useful there. I tried to find such discussion, but I did not succeed.
I am glad that I am not the only person who thinks that there was something like that. (So that I do not have to doubt my own memory.)
@BalarkaSen well, more precisely, this one $$\displaystyle \int_{0}^1\frac{\log(x)}{x^2-x-1}\text{d}x$$ and as far as I can see the second one does the same thing but I didn't check the details yet.
Darn, nothing in the recorded presentation will help, either.. Can anyone direct me where to look at? I am trying to show that every sylow-2 subgroup of $D_60$ is $D_4$. Obviously it has 8 elements, yet I'm unsure how to go about showing it is non-abelian, etc. Should I try to count the number of sylow-2 subgroups first? It seems like a certain failure
Does $I_a$ have some special meaning? (Is it principal ideal generated by $a$?) I thought that they are arbitrary ideals. I do not understand why you wrote $x=ak$, if $I_a=\langle a \rangle$, that would be an explanation.
But the claim you wrote there holds for arbitrary ideals, not only principal ideals.
Why would you know there are 15 of them? (you're probably right, I just don't follow (yet))
@UserX huge indeed
@Balarka Okay, we know that sylow-2 subgroup is of order 8, right? We also know there is a sylow-3 subgroup of order 3, and sylow-5 subgroup of order 5.
@MartinSleziak I want to show that if $(I_a)_{a \in A}$ is a family of ideals in $K[x_1, \dots , x_2]$, then $V \left ( \sum_{a \in A} I_a \right )=\cap_{a \in A} V(I_a)$
I tried the following:
$x \in I_a$
$\sum_{a \in A} I_a$ consists of all finite sums, where summands belong to the ideals $I_{a} $.
@Balarka if it is Abelian then the statement is false - which I doubt it is. If it's not, all we need to do is find an element of order 4 (denote by a), and then find an element of order 2 in H\<a> and we will know it is $D_4$
@Balarka well first we need to show it can't be abelian. I can't figure out how, besides the $N(H)$ thingy, which I have no idea how to use as it will require showing a specific number of sylow-2 subgroups, and then again I have to prove it is non-abelain to find them.. and the circle goes on and on.
@MartinSleziak If $S=\{f_a \in K[x_1, \dots , x_n] | a \in A\}$ then the set of roots is $V(S)=\{(a_1, \dots , a_n) \in K^n | f_a(a_1, \dots , a_n )=0, \forall a \in A\}$
@evinda Yes, it does. If you have $V(J)\subseteq V(I_a)$ for each $a\in A$, then also $V(J)\subseteq \bigcap_{a\in A} V(I_a)$.
Btw a TeX-nical note: You can use \bigcap instead \cap when writing about intersection of a system of sets. I.e. $A\cap B$, but $\bigcap_{i\in I} A_i$ (or even $\bigcap\limits_{i\in I} A_i$).
@MikeMiller I am pretty sure there is some sensible way to define hyperbolicity in uncountable groups. An interesting question would be if there is a notion of hyperbolicity in uncoutable that makes $\mathbf{Z}_p$ hyperbolic.
I am pretty sure that there is a sense of hyperbolicity attached to $\mathbf{Z}_p$ as the boundary of hyperbolic groups converge to a cantor set, in turn homeomorphic to $\mathbf{Z}_p$ @Mike
Why should that be a bad thing? I doubt you're interested in all the questions I am, or would be even after you broadened your horizons a bit more. There's no accounting for taste.
Hi everyone, I have a question about three algebraic expressions. Let's say that a<b and both are whole numbers. The first question leads: "is a/b more, less or equal to a/b*a/b".
So it's more, yes. But I wrote this: a/b = a^2/a^b + a/b^2.
@r9m Well, I don't pretend to be a perfect person, I often make mistakes, sometimes I got upset, annoyed, but I never ever find a pleasure in trying to harm someone!
Even there, he upvoted much more. Our only hope is that he upvoted in order to downvote later on
@Skull on the other hand mr. Upvote downvoted quite a lot
I am sensing a pattern in here
Mike, mind hinting me in the right direction in here? I am throwing fists at all theorems and nothing helps. I wanna show that a sylow-2 subgroup of $D_60$ is (isomorphic to) $D_4$. I managed to simplify it to showing it is non-abelian, and if we take an element of order 4, than $H-<a>$ has an element of order 2. I also managed to show it isn't cyclic (and thus doesn't have an element of order $8$.
Don't care either way.
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Given a commutative Ring $R$ of ordered pairs $(x,y)$ of reals $x,y$ with addition and multiplication defined in the following way.
$$(x,y) + (u,v) = (x+u,y+v)$$
$$(x,y).(u,v) = (xu-yv,xv + yu)$$
I already showed that $R$ is an integral domain , now i need to show to prove $R$ is a field or no...
