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Apr
6
comment Understanding limsup
@kiwifruit The supremum is not the largest value of the sequence. That would be maximum. However, maxima need not exist. In the case of the sequence $3-1/n$, the supremum is $3$ but there is no maximum because the elements in the sequence keep getting bigger and approach to $3$ in the long run.
Apr
6
comment Understanding limsup
For the sequence $3-1/n$, the supremum is $3$. Yes, as $n$ gets larger, the sequence values get larger, but the supremum stays the same. The supremum is not the thing as maximum, so supremum doesn't need to come from the set itself (indeed, $3$ does not belong to any of the elements in your sequence $3-1/n$).
Mar
31
revised Order of elements in a disjoint cycle
Latex
Mar
31
comment Order of elements in a disjoint cycle
The order of $(12)(34)$ is $2$. And same for $(13)(24)$ and $(14)(23)$. Your answer for $3$-cycles is right. All the $3$-cycles have order $3$.
Mar
31
comment Order of elements in a disjoint cycle
You can take the element $x\in A_{4}$, and then look at $x, x^2, x^3, …$ until you reach $x^{n}=1$ (the identity element). The smallest such $n$ is the order of $x$.
Mar
27
comment $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?
I think in a situation like this, one would like to show that $x^{3}-5$ is irreducible over the field $\mathbb{Q}(\sqrt[3]{2})$. But I don't know if that is any easier (Perhaps using some irreducibility criterion… For example, Eisenstein criterion, if only you can show $\sqrt[3]{5}$ is prime in $\mathbb{Q}[\sqrt[3]{2}]$).
Mar
27
accepted Computing a field extension by hand
Mar
27
comment Computing a field extension by hand
@ABC: That's true. But I want to upvote and accept your answer as a form of gratitude.
Mar
27
comment Computing a field extension by hand
@ABC: That's almost ridiculous :P (this is a compliment!) Please post it as an answer.
Mar
27
comment Computing a field extension by hand
@ABC: Oh I see (I don't have much experience with expanding rational functions in series, other than geometric series $1/(1-x)$ :)). What do you think about the statement in block quotes?
Mar
27
comment Computing a field extension by hand
@ABC: Ah second comment is very nice. Thank you :)
Mar
27
comment Computing a field extension by hand
@ABC: For your first comment, I understand how to gather the terms for polynomials… How to do it for rational functions? Edit: Oh expansion in series… like infinite series.
Mar
27
asked Computing a field extension by hand
Mar
12
comment Convergence of $\sum_{n\leq x} \frac{\chi(n) \Lambda(n)}{n}$
@EricNaslund: Good point. I suspect that the convergence of the series above is an easier task than showing that it, in fact, converges to $0$. (In fact, Apostol proves the convergence of (1) at least 3 chapters before he proves the Prime Number Theorem, so this somehow doesn't feel as hard as PNT).
Mar
11
awarded  Nice Question
Mar
11
comment Are there nonisomorphic fields with isomorphic multiplicative groups?
I remember asking a similar question question before.
Mar
11
accepted How do $x^{p}+y^{p}=z^{p}$ and $x\equiv y \pmod{p}$ together imply $x\equiv -z \pmod{p}$?
Mar
11
comment How do $x^{p}+y^{p}=z^{p}$ and $x\equiv y \pmod{p}$ together imply $x\equiv -z \pmod{p}$?
Just for the record, the exercise is that if $x+\omega y=u\alpha^{p}\pmod{p}$ where $\alpha\in\mathbb{Z}[\omega]$ and $u$ is a unit, then $x\equiv y \pmod{p}$. Which is more or less what you said.
Mar
11
comment How do $x^{p}+y^{p}=z^{p}$ and $x\equiv y \pmod{p}$ together imply $x\equiv -z \pmod{p}$?
@StellaBiderman: I should have included the screenshot in full in the beginning. My apologies! :(
Mar
11
comment How do $x^{p}+y^{p}=z^{p}$ and $x\equiv y \pmod{p}$ together imply $x\equiv -z \pmod{p}$?
@CalvinLin: Reading your comments make sense. Thanks! So he is reaching contradiction in a different way (using the result of exercises).