Frank Science
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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