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I'm a student of mathematics in Germany.


Jan
17
comment Connected manifold with disconnected boundary?
Or just $[0..1]$. (I don’t get that cylinder-example, though @ThomasAndrews.)
Jan
17
comment Kernel of a ring homomorphism involving group rings over the integers
Because $ℤ[ℤ/nℤ]$ is, as an abelian group, the free abelian group generated by the set $ℤ/nℤ$, so any other additive group structure you can imagine $ℤ/nℤ$ to have doesn’t matter. The “$+$” in $ℤ[ℤ/nℤ]$ doesn’t care.
Jan
17
comment Kernel of a ring homomorphism involving group rings over the integers
You just proved it to be false. because $1, t, … t^{n-1}$ are all different in $ℤ/nℤ$, hence they are linearly independent in $ℤ[ℤ/nℤ]$ so, both $t-1$ and $1 + t + … + t^{n-1}$ are nonzero. This works for any group $G$ with an nonidentity element of finite order.
Jan
17
comment Kernel of a ring homomorphism involving group rings over the integers
Are you implicitly assuming that $ℤ[ℤ/nℤ]$ has no zero divisors? I still don’t see how it follows that $1+t+…+t^{n-1}$ is sent to null.
Jan
17
comment Kernel of a ring homomorphism involving group rings over the integers
How do you conclude that $1+t+…+t^{n-1}$ is mapped to $0$ through the map given by $t ↦ t$ (where $t^n ↦ 1$)?
Jan
17
comment My proof that $G(x)\to G / G_x$ is injective
@user174981 Let $y ∈ Gx$. From the surjectivity of $G → Gx$ or the definition of $Gx$, there is a $g' ∈ G$ such that $g'x = y$. So $[g'] ∈ G/G_x$ maps to $y$ as well under $G/G_x → Gx,~[g] ↦ gx$, proving the surjectivity of the map.
Jan
17
comment Homeomorphisms of an open set onto itself.
What open set $U$ are you talking about? Is $U ⊂ ℝ^n$ where $ℝ^n$ is given euclidean topology?
Jan
17
comment My proof that $G(x)\to G / G_x$ is injective
@user174981 It means that the map $G/G_x → Gx,~[g] ↦ gx$ coming from the surjective map $G → Gx,~g ↦ gx$ is still surjective.
Jan
17
answered My proof that $G(x)\to G / G_x$ is injective
Jan
16
comment nilpotent elements of $M_2(\mathbb{R})$, $M_2(\mathbb{Z}/4\mathbb{Z})$
People want to close your question (not me). Counter that by providing a bit more context, telling us whether this is homework and adding a short note on where you’re stuck.
Jan
16
comment Cardinality of set of all bijections $\mathbb{N}\to\mathbb{N}$; is my proof correct?
@EricTroy If $F(f)$ is not surjective, it is not an element of $S$, the set of bijective functions $ℕ → ℕ$.
Jan
16
comment Cardinality of set of all bijections $\mathbb{N}\to\mathbb{N}$; is my proof correct?
Your $F$ is not well-defined, because for $f = 1_ℕ$, either observe that $F(f)(0) = 1 > 0$ or $F(f)(1) = 3 > 1$ (depending on whether $0 ∈ ℕ$ or not (it’s not)), and since $F(f)(x) ≥ x$, one concludes that $F(f)$ is not surjective.
Jan
16
answered Ideal contained in annihilator problem
Jan
16
comment Is $A$ diagonalisable if its unitary?
@learnmore I misunderstood your question.
Jan
16
comment Is $A$ diagonalisable if its unitary?
Furthermore, unitary matrices are invertible, so from $A^2 = A$, you already can conlude $A = I$.
Jan
15
comment A fan, a horn, and a snowflake - unusual math terms
“The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.” what.
Jan
15
comment Finding a maximal ideal in the set of continuous real-valued functions on $\mathbb{R}$
The quotient $X/I = \{f + I;~f ∈ X\}$ is a factor ring, not $X \setminus I = \{f ∈ X;~f \notin I\}$.
Jan
15
comment A fan, a horn, and a snowflake - unusual math terms
When I once told someone of the hairy ball theorem he found it funny that it was discovered by someone named “Harry Ball”.
Jan
14
comment A fan, a horn, and a snowflake - unusual math terms
@Axoren How is “cleavage” ever a risky click?
Jan
14
comment A fan, a horn, and a snowflake - unusual math terms
General abstract nonsense.