Mariano Suárez-Alvarez
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12m
comment Diffeomorphism between Euclidean space
There is no need to restrict to a ball. Just consider the Jacobian at a point.
7h
comment A Quotient of the Euclidean Group
Please refer to the paper not only by a link but with author and title. If the URL stops working, for example, your question becomes incomplete, and we do not want that.
7h
comment Bad and good definitions in mathematics
Marking something as a soft question does not automatically turn it into a good match for this site. In this case, for example, it didn' t.
7h
comment Bad and good definitions in mathematics
@user4968912, no, this is is a question&answer site, not one where we discuss things, really.
7h
comment Bad and good definitions in mathematics
If there are other problems that we are not yet aware of and later we find about them, then we change the definitions again. That's what trial and error means. You should read about how science works, in general.
7h
comment Bad and good definitions in mathematics
But there is a method! Trial and error. This is what James's book documents, for example (and every other single history of science, in general)
7h
comment Bad and good definitions in mathematics
WHy would you possibly «hope it's not the case» that people use trial and error to figure out what the correct definitions are?! That's quite an absurd hope! What would you prefer they use? You seem to think that concepts can be captured in exactly one way in mathematical terms, and this is very weird given that you evidently know that this is not the case, as that is precisely what the evolution of definitions which you refer in what you wrote is!
1d
comment Algebraic methods to compute the cohomology ring of the complex topology of a variety?
@FrankScience, have you considered the case of curves?
1d
comment Algebraic methods to compute the cohomology ring of the complex topology of a variety?
When the variety is smooth (and in more general cases, but it becomes very complicated...) you can compute the de Rham cohomology of the underlying complex manifold as the cohomology of the "algebraic de Rham" compklex, built out of Kahler forms. You can find this explained in a great paper by Hartshorne. This is a purely algebraic method to compute.
2d
comment Show that $t^n-1 \mid t^m-1 \Leftrightarrow n\mid m$
What I meant was, it is not that you can assume that $P(t)$ is in $F[t]$, it is there!
Jul
31
revised Show that $t^n-1 \mid t^m-1 \Leftrightarrow n\mid m$
added 480 characters in body
Jul
31
comment Show that $t^n-1 \mid t^m-1 \Leftrightarrow n\mid m$
@user26857, it doesn't really matter!
Jul
31
comment Show that $t^n-1 \mid t^m-1 \Leftrightarrow n\mid m$
Why would $P$ be in $k[t,t^{-1}]$?
Jul
31
answered Show that $t^n-1 \mid t^m-1 \Leftrightarrow n\mid m$
Jul
31
comment Motivate why $a^{-n}$ equals to $\frac{1}{a^n}$
Well, if $n$ is positive, $a^{-n}$ is defined to be equal to $1/a^n$. At most, then, what you can do is explain what motivates this definition!
Jul
31
comment What is the Krull dimension of $A[x,y,z]/\langle xy,xz \rangle$
That $(x)$ be prime depends on $A$, no?
Jul
31
comment What is the Krull dimension of $A[x,y,z]/\langle xy,xz \rangle$
If $B$ is your ring, then $B/(x)=A[y,z]$, so $\dim B\geq\dim A+2$, and eequality holds when $A$ is a field.
Jul
31
comment Implement some actions in Gap
This is almost offtopic, really :-/
Jul
30
comment Decomposition of a homogeneous polynomial
This has the following nice generalization: If we had a multigraded domain $A=\bigoplus_{\alpha\in\mathbb Z^n}A_\alpha$, we could let $w(a)$ be the convex hull in $\mathbb R^n$ of the set of multiudegrees of its nonzerocomponents. This "width" is also multiplicative, with the sum being the Minkowski sum of convex sets.
Jul
30
comment Decomposition of a homogeneous polynomial
One way to say this is as follows: the width of a non-zero element $a$ of a graded domain is the difference $d_{\max}(a)-d_{\min}(a)$ of the degrees of the topmost nonzero component of $a$ and that of the bottommost (?) component. Then $w$ is multiplicative: $w(a)+w(b)=w(ab)$. The result follows formally from this, just as for one-variable polynomials and the usual degree.