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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.
Mar
19
asked Concurrency of A'L, B'M, C'N.
Jan
30
comment Funny thing. Multiplying both the sides by 0?
Yes, I know, so I'm saying that the "usual way" doesn't really work in that you supposedly address the zero value of $\cos \theta = 0$ in the same step as you address non-zero values of $\cos\theta$, i.e. the "usual way" that the OP talks about overlooks a case. Sure, it might be trivial, but it still is overlooked.
Jan
28
answered Funny thing. Multiplying both the sides by 0?
Oct
31
answered Expected Value of a Binomial distribution?
Oct
22
comment Expected Number of Successes in a Sample
@Austin Mohr: Thanks. At any rate, it depends on whether you're sampling with replacement or without replacement (are the probabilities of being broken independent from one calculator to the next?). If you're sampling with replacement (probabilities are independent), then you still use the fact that if $X \sim B(n,p)$, then $E[X] = np = 10\cdot 20/200 = 1$, as below.
Oct
21
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