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comment What notions are used but not clearly defined in modern mathematics?
When I read this answer my first thought was "it just means $d\omega = 0$, so where's the ambiguity?" Then I read the comments...
2d
revised Are the elliptic curve discrete log problem and the elliptic curve Diffie-Hellman problem equivalent?
added 78 characters in body
May
22
comment A constrained positive polynomial
If $n \equiv 4 \bmod 5$ and $n > 4$ then $Q(x) = x^n + x^3+x^2+x+1$ decomposes in $\mathbf Z[x]$ with factor $Q_+(x) = x^4+x^3+x^2+x+1$ and $Q(1) = Q_+(1) = 5$, a prime that is $1 \bmod 4$, and necessarily $Q_m(1) = 1$. This doesn't fit your condition that $a > 1$, however, since the prime value occurs at $a = 1$. A similar construction could be made with $5$ replaced by any prime number (my previous family of examples was for the prime value $3$).
May
22
comment A constrained positive polynomial
And, for a family of examples, $Q(x) = x^n + x + 1$ for $n \equiv 2 \bmod 3$ with $n > 2$ decomposes in $\mathbf Z[x]$ with factor $Q_+(x) = x^2+x+1$ and $Q(1) = Q_+(1) = 3$, so $Q_m(1) = 1$.
May
22
comment A constrained positive polynomial
Obviously $a = 1$.
May
22
comment A constrained positive polynomial
Could you please give some context, e.g., why the interest in a prime value that is $1 \bmod 4$?
May
21
comment Enumerating cosets of the modular group
For coset representatives you could use matrices $(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ in the modular group where $0\leq b < |a|$ if $a \not= 0$ and $0 \leq d < |c|$ if $a=0$.
May
21
comment Enumerating cosets of the modular group
Good. It would be helpful to explain where this question comes from. The title is just about enumerating cosets (no algorithm), and it's possible to give a concrete list of coset representatives without dealing with your word task.
May
19
comment Non-standard Gauss sums
@Liss, if all you test is a finite set of primes $p$, especially what you call "small" $p$, then of course you'll find a positive lower bound on them. But is there any reason to expect over all $p$ that the sums don't include values that get arbitrarily small? When you're dealing with a family of exponential sums (in your case, varying over all primes), even if individually they are never $0$ it doesn't mean particular values couldn't get arbitrarily close to $0$.
May
18
comment Proof of Cauchy's Theorem from Group Theory - Generalizable?
I don't think it is realistic to expect a proof exploiting commutativity to carry over to the non-abelian case. When I first learned about groups, they were all abelian (unit groups mod $m$). For the results (i) order of each $g \in G$ divides $|G|$ and (ii) $g^{|G|} = e$ for all $g \in G$, I had learned a proof of (ii) that exploited commutativity and derived (i) from (ii). To prove the same two results for general finite groups, you prove (i) first via cosets and then derive (ii) from (i). At the end of the day you have the same two theorems, but without commutativity new ideas are needed.
May
18
comment Non-standard Gauss sums
Some Kloosterman sums involving a quadratic character are zero. I'm not saying yours might be zero, but if you are looking for some off-the-shelf result that will prove character sums $S$ in some family do not equal $0$ then you will be disappointed. The standard task for exponential sums is to get sharp upper bounds, not lower bounds away from $0$. Saying there is a lower bound that is a negative number is not really the point of these bounds which is to say $|S|$ is less than or equal to some sharp value.
May
18
comment What are the most misleading alternate definitions in taught mathematics?
And then how would you prove the commutator subgroup is normal for students who learn about that subgroup for the first time? Or prove the cosets by a normal subgroup form a group? As Milne would say, introduce multiple points of view at the start rather than worry that one is "the" definition.
May
17
comment TVS with null topological dual space
You sure have the most appropriate username to answer this question.
May
17
comment Non-standard Gauss sums
Whoops, my previous comment on Jacobi sums was incorrect since the sum is not just $\sum_{k \in \mathbf Z/(p)} (\frac{k(k+1)}{p})$ but has the factor $\omega_p^{kl}$ in there too. Replacing $k$ with $k/l$ makes the sum $\sum_{k \in \mathbf Z/(p)} (\frac{k(k+l)}{p})\omega_p^k$. I agree with Will that it is hopeless to expect an exact formula for this but it gets a bound like $2\sqrt{p}$ from the Weil conjectures.
May
16
comment Non-standard Gauss sums
Are you trying to count how often consecutive numbers $k$ and $k+1$ in $\mathbf Z/(p)$ are perfect squares?
May
16
comment Non-standard Gauss sums
@Liss, first of all you are right that $(\frac{l}{p})^{-1}$ comes out, but the character is quadratic so the exponent doesn't matter: $a^{-1} = a$ if $a = \pm 1$. Concerning a sum with $(\frac{k^2+k}{p}) = (\frac{k(k+1)}{p})$ in it, replacing $k$ with $-k$ makes that $(\frac{-1}{p})(\frac{k(1-k)}{p})$ and then you basically have a Jacobi sum, which you can look up elsewhere. This is not really a research-level question. I suggest if you have similar questions that you ask them on math.stackexchange.
May
16
comment Non-standard Gauss sums
The missing term at $k = 0$ is $1$, so your sum is $\sum_{k \in \mathbf Z/(p)} (\frac{k+1}{p})\omega_p^{kl} - 1$. Now you're summing over an additive group, so replacing $k$ with $k-1$ makes it $\sum_{k \in \mathbf Z/(p)} (\frac{k}{p})\omega_p^{(k-1)l} - 1 = \omega_p^{-l}\sum_{k \in \mathbf Z/(p)} (\frac{k}{p})\omega_p^{kl} - 1$, and omitting the $k=0$ term (which is $0$) and then doing a multiplicative change of variable you get $\omega_p^{-l}(\frac{l}{p})\sum_{k \not\equiv 0 \bmod p} (\frac{k}{p})\omega_p^k - 1$. This last sum is a standard Gauss sum of the Legendre symbol.
May
16
comment Examples of TVS with no non-trivial open convex subsets
For what it's worth, this construction of TVS with dual space $\{0\}$ works for $L^p(X)$ where $X$ is any measure space with no atoms and $0 < p < 1$. In this way the sequence spaces and functions on $[0,1]$ that you describe are special instances of a more general construction. See math.uconn.edu/~kconrad/blurbs/analysis/lpspace.pdf.
May
16
comment Integers $d$ for which the Negative Pell equation is soluble for both $d$ and $2d$?
The number 25 is in your list of values of $q$, so that is the first missing value.
May
15
comment Existence of Euler product on critical line for $L(\chi,s) L(\overline{\chi},1-s)$?
Now in your case you are dealing with two products at once, so in your sum there is a certain amount of cancellation, but I don't see why that should make the task of proving convergence more feasible in any sense.