1,053 reputation
319
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location Hangzhou, China
age 22
visits member for 2 years, 1 month
seen yesterday

I'm just a guy learning and enjoying in math. My name in MO is yizhongxunhuan. Which in Chinese means a recycle.


1d
accepted To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$.
Nov
18
awarded  Yearling
Sep
28
comment Tensor product of domains is a domain
In your answer you said 'the good news is that for a perfect field k (in particular for an algebrically closed field) every algebra is separable and every extension field is primary'......
Sep
28
comment Prove Poincare duality theorem with Morse theory.
Thanks for your answer. The orientation of unstable manifold can be chosen without any assumption?
Sep
28
accepted Prove Poincare duality theorem with Morse theory.
Sep
28
comment Tensor product of domains is a domain
According to your cited theorem, C tensor C over R is a domain(as C is a domain and R is perfect). But it's commutative 4-dim R algebra, hence is not a domain. Is there something wrong?
Sep
24
awarded  Autobiographer
Aug
8
awarded  Altruist
Aug
7
awarded  Investor
Aug
7
comment Norm map over Galois extension
I can prove it if F is a finite field. But in general I guess it's not right and should have some counterexample. And I just can't find one...
Aug
7
comment if $K/F$ is a Galois extension, show that any intermediate field $L$ is generated by the traces of elements from $K$ over $L$.
@Alex Then you just check when is the trace of a normal basis generator, call it $\beta$, is fixed by an element in Gal(K/F). The answer is iff the element is actually in Gal(K/L). Then by the Fundamental Theorem in Galois Theory, we may conclude that F($\beta$)=L.
Aug
7
comment if $K/F$ is a Galois extension, show that any intermediate field $L$ is generated by the traces of elements from $K$ over $L$.
I think there is a typo. L should be generated by the traces of K over L with coefficients in F. So @MattE 's trick probably won't work here. As the case L=F is trivial...
Aug
5
comment To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$.
Sorry, but the inequality doesn't hold can't apply we have a inverse inequality. So I don't think you really solved this problem.
Aug
5
comment A proof of the normal basis theorem of a cyclic extension field
@MakotoKato Could you tell me what's the proof that applies to both cases? I was asked to prove the normal basis theorem as a homework, but I really don't have any idea... Thanks a lot.
Jul
31
revised To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$.
added 2 characters in body
Jul
31
asked To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$.
Jul
2
awarded  Curious
May
26
awarded  Revival
May
20
comment Reference request-What is the prerequisite of S.S.Chern's proof of the generalised Gauss-Bonnet theorem?
Oh, BTW, the 2-dimensional case can be proved by Chern-Weil theory, as it is a well-known fact that all 2-real-dim'l Riemannian manifolds are Riemann surfaces.
May
20
comment Reference request-What is the prerequisite of S.S.Chern's proof of the generalised Gauss-Bonnet theorem?
I only know that the Chern-Weil theory will give the G-B-C theorem for complex manifolds. The way I know to get Pfaffian is that from Mathai-Quillen's geometric construction of Thom classes.