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 Dec24 accepted To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$. Nov18 awarded Yearling Sep28 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'...... Sep28 comment Prove Poincare duality theorem with Morse theory. Thanks for your answer. The orientation of unstable manifold can be chosen without any assumption? Sep28 accepted Prove Poincare duality theorem with Morse theory. Sep28 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? Sep24 awarded Autobiographer Aug8 awarded Altruist Aug7 awarded Investor Aug7 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... Aug7 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. Aug7 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... Aug5 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. Aug5 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. Jul31 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 Jul31 asked To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$. Jul2 awarded Curious May26 awarded Revival May20 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. May20 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.