Michael Zhao
Reputation
763
Top tag
Next privilege 1,000 Rep.
Create new tags
 Apr 11 comment Exact sequence of free abelian groups, $\sum_{i=0}^n(-1)^i\text{rank}(F_i)=0$. A similar statement holds true when you have vector spaces in place of your groups. The idea behind the proof there is to use the rank-nullity theorem, so I guess you can do something similar here. Since a free abelian group is a free $\mathbb{Z}$-module, if you tensor with the flat $\mathbb{Z}$-module $\mathbb{Q}$, you'll get an exact sequence of $\mathbb{Q}$ vector spaces with the dimensions equal to the number of $\mathbb{Z}$ summands (i.e. the rank). Mar 30 awarded Nice Answer Mar 2 asked Extension of scalars by a submodule, which is also a ring Jan 3 accepted Matrix Algebra over Algebraically Closed Field Jan 3 asked Matrix Algebra over Algebraically Closed Field Sep 6 awarded Yearling Jun 1 comment Show that the ring of integers $A$ of the cubic field $\mathbb Q[x]$ with $x^3=2$ is principal. (continued:) We also have $(2) \subset \xi$ and $(2) = I^3 \subset I$. Since $I$ is maximal, if the ideal $\xi \neq I$, then $\xi + I$ is an ideal larger than $I$ containing $I$, a contradiction. So we must have $\xi = I$. So every ideal class contains the principal ideal $I$, which means the class group is trivial. Jun 1 comment Show that the ring of integers $A$ of the cubic field $\mathbb Q[x]$ with $x^3=2$ is principal. I am trying to unravel the last paragraph; could you see if I am reasoning this out correctly? The only ideals that aren't principal must have norm $\leq$ the Minkowski bound $N_0$ since otherwise, take an ideal $I$ with $N(I) > N_0$. Then the class $I \cdot F(A)$ has an ideal $I_0$ with $N(I_0) < N_0$. But $N(I) \leq N(I_0)$ since ideal norm is multiplicative. We know $I = (\sqrt[3]{2})$ is prime, and hence maximal, since $\mathbb{Z}[\sqrt[3]{2}]/I \cong \mathbb{Z}$, which is an integral domain. The norm is 2 since the other cosets contain $\sqrt[3]{4} + I$ and $1 + I$. Feb 5 awarded Nice Answer Jan 6 awarded Popular Question Sep 6 awarded Yearling Feb 14 answered please help me understand the lecture note? heat equation and fourier series Sep 6 awarded Yearling Apr 22 comment how to show that a function$f$ is contained in all natural numbers? For the proof, that's not necessary to know. It's sufficient to show that the expression for $f(a,b)$ is always a positive integer, for any positive integers $a$ and $b$, which is what we did. Apr 22 answered how to show that a function$f$ is contained in all natural numbers? Apr 18 comment I have some questions for double integrals, please help… I really need help If you had $d\theta$ instead of $dx$ or $dy$ respectively, you could, unless the function inside had any $\theta$'s involved. Apr 17 revised I have some questions for double integrals, please help… I really need help type-setted it in latex Apr 17 comment I have some questions for double integrals, please help… I really need help No, that is not acceptable since the bounds of the integral tell you that $x$ runs from 0 to $2\pi$ (respectively, that y runs from 0 to $\pi/2$). Just because you see $2\pi$ or $\pi/2$, doesn't mean it has anything to do with the variable $\theta,$ which I think is the mistake you're making. Apr 17 suggested approved edit on I have some questions for double integrals, please help… I really need help Mar 25 accepted Concurrency of A'L, B'M, C'N.