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Jan
31
comment Why $V_{\mathbb{C}} = V_{1,0}\oplus V_{0,1}$?
@MarianoSuárez-Alvarez And I should clarify that I was an undergrad, and abstract algebra (including Galois theory) was taught before complex geometry, and it seemed to me natural to study general Galois extensions, and the decomposition is in fact in accordance to the action of Galois group, and it now seems to me that the introduction of $\otimes_\mathbb R\mathbb C$ shares the same motivation with Galois descent, not as random as «this is in Euler's works somewhere».
Jan
31
comment Why $V_{\mathbb{C}} = V_{1,0}\oplus V_{0,1}$?
@MarianoSuárez-Alvarez Sorry, I'm not mature enough to judge math, but the first time I learnt these linear algebra on the complex geometry class (text: Huybretch's introduction), I felt that the manipulation of tensors is quite messy, and appearance of some constants was quite annoying. My intuition was that we can do everything properly for a general Galois extension instead of $\mathbb C/\mathbb R$. After class, I did them on my own. I got into a theory where I didn't know how to advance. Recently, I discovered Galois descent and I realized that it was what I wanted.
Jan
30
comment Why $V_{\mathbb{C}} = V_{1,0}\oplus V_{0,1}$?
Note that $\mathbb C/\mathbb R$ is a finite Galois extension. A very general framework for this is Galois descent.
Jan
28
comment Functional Equation: When $f(x+y)=f(x)+f(y)-(xy-1)^2$
Conceptually, first let $y=1$ and solve the equation for $f\vert_{\mathbb N}$, then restrict the original equation to $x,y\in\mathbb N$, which is strong enough to lead to a contradiction.
Jan
28
awarded  Custodian
Jan
28
awarded  Yearling
Jan
24
awarded  Nice Question
Jan
20
comment $OABCD$ tetrahedron with $OA ⊥ OB ⊥ OC ⊥ OA$
Denote $H'$ the orthogonal projection of $O$ onto the plane $ABC$, and show that $AH'\perp BC$ via Theorem of Three Perps, then deduce that $H'=H$.
Jan
20
comment Minimal cyclotomic field containing a given quadratic field?
@JyrkiLahtonen I don't know whether it's easy to compute the determinant of a general cyclotomic polynomial. In fact, it's not hard to enumerate ALL quadratic subfields of a cyclotomic field, which is equivalent to determine all index-2 subgroups of an abelian group, or nontrivial homomorphisms from an abelian group to $\mathbb Z/2\mathbb Z$, which leads to the answer. I have no time to spell it out here.
Jan
18
asked Minimal cyclotomic field containing a given quadratic field?
Jan
13
comment about the the tensor product
@David The point is that MSE is not only for the OPs but also for others, and for reference, not like a chat room. You needn't translate it for me, but if you're pleased, translate for audience, since English is the lingua franca. What about deleting? Well, apparently it will reduce the number of beneficiaries, so no need to do that.
Jan
13
revised about the the tensor product
deleted 14 characters in body; edited title
Jan
13
comment about the the tensor product
@David I don't understand why you start to speak French here.
Dec
23
comment Any left ideal of $M_n(\mathbb{F})$ is principal
@TomOldfield For the first one, you need to define $T(Ax)=Bx$ on the image $A(V)$. I give you a more chance to think throughly of the second point (the first point is needed here).
Dec
23
comment $\tan (x+yi)=\alpha +\beta i \Rightarrow \tan(x-yi)=\alpha-\beta i$?
$\tan z$ has a Taylor series at $0$ with real coefficients.
Dec
23
comment If $R$ is a Noetherian ring, why is $R[[x]]$ also Noetherian?
Another proof: consider Artin-Rees lemma
Dec
23
answered Any left ideal of $M_n(\mathbb{F})$ is principal
Dec
23
awarded  Socratic
Dec
22
comment An exercise from Stein's Fourier analysis about wave equation
I think on Stein's book, it contains a method to solve wave equation (nameyly, via Fourier transform).
Dec
22
revised (Co)different ideal is divisorial?
added 25 characters in body