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4h
revised Expected genus of a function field over a finite field
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7h
comment What probability mass, density or distribution function might corresponds to this moment generating function?
BCLC: The point I was also trying to make elsewhere is that you seem to have a tendency to self-answer and accept way before "the case is closed". At least that is the impression I get, when you continue with the same exercise in another thread. I am concerned, and I think Did has a valid point here.
7h
answered Expected genus of a function field over a finite field
8h
comment Positivity of the alternating sum associated to at most five subspaces
A question/suggestion. If I find the time, I will work on it and turn it into a question. An alternating sum of dimensions just smells of homological algebra. Strongly.
14h
comment Discrete Fourier Transform DFT
They both only need the existence of primitive roots of unity of the prescribed order $N$. In $\Bbb{C}$ we have those for all $N$. In $\Bbb{F}_q$ we need $N$ to be a factor of $q-1$. Both are about sequences with period $N$. They are called discrete, because it it about sequences rather than functions with a continuous domain.
1d
revised Show that if p is an odd prime, with p $\equiv 3 \pmod 4$ then $(\mathbb{Z}^*_{p})^4$ = $(\mathbb{Z}^*_{p})^2$.
added 2 characters in body
1d
comment Show that if p is an odd prime, with p $\equiv 3 \pmod 4$ then $(\mathbb{Z}^*_{p})^4$ = $(\mathbb{Z}^*_{p})^2$.
BTW, many users expect askers to discuss their thoughts, or show some partial work. Here it is not easy to show partial progress, but can you try and apply the hint in my answer to the case $p=7$, please? Then you may see what's going on (using that $a^6=1$ for all $a\in\Bbb{Z}_7^*$)!
1d
comment Show that if p is an odd prime, with p $\equiv 3 \pmod 4$ then $(\mathbb{Z}^*_{p})^4$ = $(\mathbb{Z}^*_{p})^2$.
I deleted my first suggestion, because I realized that you don't need cyclicity of $\Bbb{Z}_p^*$. I don't think CRT helps you this time.
1d
answered Show that if p is an odd prime, with p $\equiv 3 \pmod 4$ then $(\mathbb{Z}^*_{p})^4$ = $(\mathbb{Z}^*_{p})^2$.
1d
comment Prove that: $(x_1+…+x_k)^2\leq 2(x_1^2+…+x_k^2)$.
The best you can do is $(x_1+\cdots+x_k)^2\le k(x_1^2+\cdots+x_k^2)$. You get this from Cauchy-Schwarz using the all ones vector as the other vector. C-S is tight in the known cases, so....
1d
comment Positivity of the alternating sum associated to at most five subspaces
So, there is no nice description of the homology of the obvious chain complex $\alpha$ is the Euler characteristic of?
1d
comment Number of points satisfying a quadratic equation over $GF(q)$
Somewhat related. Not sure what you think of as elementary.
1d
comment Does the complex modulus satisfy the power identity $|z^r|= |z|^r$?
@samjoe: For all positive real numbers $x$ we have $$x=e^{\ln x}.$$ Going via the complex logarithm is standard in the definition of complex powers.
1d
comment Does the complex modulus satisfy the power identity $|z^r|= |z|^r$?
We really need a "mother" question about these finer points of complex powers, refer all askers to it, and close all the other questions as duplicates. I'm fairly sure there is a suitable duplicate target to this question, but I don't frequent the tag, so I don't remember one.
1d
answered Does the complex modulus satisfy the power identity $|z^r|= |z|^r$?
1d
comment Why is Gaussian matrix full rank?
A non-zero polynomial (here the determinant) in $n^2$ variables vanishes only in a set $N$ of measure zero. The probability of determinant vanishing is the integral of the continuous function (=a product of $n^2$ i.i.d. gaussians in distinct variables) over $N$. What do you know about integrals over sets of measure zero?
2d
comment Group theory: Intuition as to what a group is
I would say the intuition will come when you work with groups enough. Not sure I have a strong intuition about it. yet :-).
2d
comment Spliting subspaces and fields
No need for $W$ to be a subring. After all, for any non-zero $b\in K$ the space $bS$ satisfies the requirements. But, it is easy to show that the set $$L=\{\alpha\in K\mid W\alpha\subseteq W\}$$ is a field. I just don't see right away why we should have $L=S$. A cool question!
2d
comment To find a field of $p^{2}$ elements ,where $p$ is prime
Show that if $p>2$ there exists a quadratic non-residue $a$ modulo $p$. Then show that $x^2-a$ is irreducible.
2d
awarded  Good Question