Mariano Suárez-Alvarez
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Apr
23
comment The necessity of defining the stable equivalence in the construction of the Grothendieck group $K_0$
As I wrote, $P[A]/\sim$ has a decription in terms of f.g. projective modules, and I am using that. $A$ and $A\oplus A$ are f.g. projective modules, so they have classes in $P[A]/\sim$, etc.
Apr
23
comment The necessity of defining the stable equivalence in the construction of the Grothendieck group $K_0$
$P[A]/\sim$ is isomorphic to the set of finitely generated projective $A$-modules with the operation given by direct sum. If $A\oplus A\cong A$, then of course $A\oplus A\cong 0\oplus A$ and you cannot cancel the $A$.
Apr
23
comment The necessity of defining the stable equivalence in the construction of the Grothendieck group $K_0$
The Cuntz algebra $A$ is such that the is an isomorphism of modules $A\cong A^n$ for some $n\geq2$. Such an algebra provides an example.
Apr
23
comment If a finite group acts transitively on a set, does its center also acts transitively?
Since the center can certainly be trivial, of course no!
Apr
23
revised A semisimple Lie group has no character; Am I right?
added 37 characters in body
Apr
23
comment A semisimple Lie group has no character; Am I right?
The group has to be connected, for otherwise this is false. Take any compact $G$ and consider the Lie group $G\times\mathbb Z/7\mathbb Z$. It admits a non-trivial homomorphism to $S^1$.
Apr
23
answered A semisimple Lie group has no character; Am I right?
Apr
23
comment $L$ is algebraic curve. Why has $L$ finitely many singularities?($x,y\in \mathbb{R}$)
Once you have really done that case, do exactly the same thing with a polynomial in the variables and one of its derivatives.
Apr
23
comment $L$ is algebraic curve. Why has $L$ finitely many singularities?($x,y\in \mathbb{R}$)
Notice that you are not done. It is easier to let P be a irreducible common factor of f and f' and k be the maximal integer such that $p^k$ divides f. Then proceed more or less as you did.
Apr
23
comment $L$ is algebraic curve. Why has $L$ finitely many singularities?($x,y\in \mathbb{R}$)
Well, how did you solve it?
Apr
22
comment Is it true that the order of $ab$ is always equal to the order of $ba$?
Calling this «hand-waving» is quite misguided!
Apr
22
comment What is the Taylor series expansion of $z^{1/2}$ about origin.
If the function were analytic at zero, its restriction to the real line would have a derivative at zero. It doesn't. The polar form of the CR equation is not of any use at the origin, unless it is used with immense care — just avoid it.
Apr
22
comment What is the Taylor series expansion of $z^{1/2}$ about origin.
(In fact, it is not evn defined in a neighborhood of zero...)
Apr
22
comment What is the Taylor series expansion of $z^{1/2}$ about origin.
It cannot satisfy that equation if it does not even have derivatives at that point!
Apr
22
comment What is the Taylor series expansion of $z^{1/2}$ about origin.
For a function to have a Taylor series at a point it needs to be at the very least differentiable there.
Apr
21
comment Is the absolute Galois Group of $\Bbb Q$ countable?
@QiaochuYuan, there are uncountably many permutations of the complex numbers and yet the Galois group of $\mathbb C$ over $\mathbb R$ is finite!
Apr
21
comment Are all the numbers in this sequence a prime number?Sequence : $31 , 331, 3331, 33331$
You don't. ${}{}{}$
Apr
20
comment Sequences $(U_n)$ of neighborhoods of $0$ in a LCA group with $m(U_n)\to 0$
No, it is not always true, because you could have picked all $U_i$ equal to $G$!
Apr
20
comment For vector bundles $A\to E$, $B\to E$, is $\Gamma(A\oplus B)\cong \Gamma(A)\oplus\Gamma(B)$?
What have you tried?
Apr
20
comment Ring Extension: Mapping: $ \mathbb Q[\sqrt d] \rightarrow \mathbb Q$
It is certainly no more rigurous to write the ugly $\frac{1+0\sqrt d}{0+1\sqrt d}$ that the good ol' $1/\sqrt d$...