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Jan
22
answered $f(x) \in R[X]$ irreducible $\Rightarrow (f(x))$ ideal?
Jan
22
comment $f(x) \in R[X]$ irreducible $\Rightarrow (f(x))$ ideal?
So you’re asking if the ideal generated by $f$ is (A) an ideal that is (B) generated by one element?
Jan
22
comment $\mathbb{A}^n$ with the Zariski Topology is Quasi-Compact.
Ok, I’m confused now: Does or doesn’t one need the axiom of choice? @KevinCarlson
Jan
21
revised $\mathbb{A}^n$ with the Zariski Topology is Quasi-Compact.
edited body
Jan
21
comment $\mathbb{A}^n$ with the Zariski Topology is Quasi-Compact.
@KevinCarlson But $U_1 ⊂ U_1 ⊂ U_1 ⊂ …$ might be a chain with an upper bound which is not all of $\mathbb A^n$? Don’t you need something guaranteeing the existence of a chain running through suffiently many open sets? Oh, one can try to construct strictly increasing chains and fail to do so, right?
Jan
21
answered $\mathbb{A}^n$ with the Zariski Topology is Quasi-Compact.
Jan
21
comment $\mathbb{A}^n$ with the Zariski Topology is Quasi-Compact.
What’s your definition of noetherian? Ascending sequences of open sets terminate?
Jan
20
comment Incorrect Chain Rule Proof
It already says why: Consider constant $g$ as an extreme case. Do you see the deathly sin secretly commited in the proof?
Jan
20
revised What is the difference between coordinates transformation and change of coordinates?
added 9 characters in body; edited tags
Jan
20
revised What is the difference between coordinates transformation and change of coordinates?
added 9 characters in body; edited tags
Jan
20
comment Question about cyclic subgroup of a non-abelian group of order $8$.
@Libertron Yes, of course I did.
Jan
20
revised Question about cyclic subgroup of a non-abelian group of order $8$.
deleted 3 characters in body
Jan
19
answered Question about cyclic subgroup of a non-abelian group of order $8$.
Jan
18
comment is complex number under absolute value a group?
@jeniffer Not sure, why it’s that long: For $a, b, c ∈ ℂ$: $$\begin{align*} (a*b)*c &= |ab|*c = ||ab|·c| = |ab|·|c|\\ &= |abc|\\ &= |a|·|bc| = |a·|bc|| = a*|bc| = a*(b*c) \end{align*}$$
Jan
18
comment is complex number under absolute value a group?
@jeniffer Yes, that’s what I mean. The notion of an inverse (for $a ∈ ℂ$ some element $a' ∈ ℂ$ such that $a*a' = e$ and $a'*a = e$, the identity) doesn’t make any sense if there isn’t an identity to begin with. Also note, that with the same argument the operation doesn’t even yield a group (/monoid) when restricted to $ℝ$.
Jan
18
revised is complex number under absolute value a group?
added 17 characters in body
Jan
18
comment is complex number under absolute value a group?
No, $1$ is no neutral element. Maybe I did misunderstand your answer, though?
Jan
18
answered is complex number under absolute value a group?
Jan
18
comment Extension Fields, complex numbers.
On understanding the argument: Have you heard of the omni-important isomorphism theorem (for rings) yet? If not, look at it. As soon as you got that, take as a first example the ring homomorphism $ℝ[X] → ℂ,~X ↦ \mathrm i$ and determine its kernel.
Jan
18
comment $A\subset f^{-1}(f(A))$ with equality if and only $f$ is injective.
@idm What do you consider your proof? And of which statement exactly?