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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.