## abstract algebra – flat ring extension preserves isomorphism in prime localizations

Let $$f:Arightarrow B$$ be a ring homomorphism and $$Arightarrow A’$$ be a flat ring homomorphism. We know that there exists a map $$g:A’rightarrow B’$$ by ring extension for $$B’:=Botimes_AA’$$. Now assume that the map $$A_{f^{-1}(mathfrak{q})}rightarrow B_{mathfrak{q}}$$ is an isomorphism for each prime ideal $$mathfrak{q}$$ of $$B$$, I was asked to show that $$A’_{g^{-1}(mathfrak{q}’)}rightarrow B’_{mathfrak{q}’}$$ for each $$mathfrak{q}’intext{Spec}(B’)$$.

I do not know how to apply the flatness of $$Arightarrow A’$$ here. Any hint or help will be appreciated.

## algebraic number theory – How can I prove this claim about splitting of prime ideals in real cyclotomic fields?

Let $$L_k = mathbb{Q}(zeta_{2^k} + zeta_{2^k}^{-1})$$ be the maximal real subfield of the cyclotomic field of conductor $$2^k, k ge 2$$ and $$f_k(x)$$ be the minimal polynomial of $$zeta_{2^k} + zeta_{2^k}^{-1}$$.

Define $$L = L_{k+1}, K = L_{k}$$ so $$L/K$$ has degree 2. Assume a prime ideal $$mathfrak{p} subset mathcal{O}_K$$ above $$p$$ splits in $$mathcal{O}_L$$ as $$mathfrak{p}mathcal{O}_L = mathfrak{p}_1 mathfrak{p}_2$$. Then I suspect that $$pmathcal{O}_L$$ totally splits (in which case $$pmathcal{O}_K$$ totally splits as well). How can I prove this?

I believe this is equivalent to showing that if $$f_{k+1}(x)$$ has 2 roots over $$mathbb{F}_p$$ then it factors completely over $$mathbb{F}_p$$. I’m not sure how that can be shown. Are there any general approaches to take here?

## Prove that if \$n + 1 | n! + 1\$ then \$n+1\$ is a prime number

How can we prove that if : $$n+1|n!+1$$ then $$n$$ is a prime number.
My try:
I try to do contrapositive: If $$n$$ is composite then $$n+1 nmid n!+1$$ and then to get that if $$n$$ is composite so $$n+1=ab$$ when $$2 le a le n$$ so that $$a|n+1$$ but here I don’t have idea how to continue, is this the good way?

## algebra precalculus – cardinality of the set of all prime factors of 120

need to find the cardinality of the set of all prime factors of 120

Will it be 16? Since the set of all prime factors of 120 will always have 16 elements i.e { 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 }

Or is this a trick question and requires a proof to answer?

## Binnomial congruence modulo prime number

Let $$1leq{j} with $$p$$ a prime number; is it true that for any positive integer $$n, nnotequiv{j}(mod{(p-1}))$$ , the congruence $$sum_{r>1}{nchoose rcdot(p-1)+j}cdot{r-1choose j}equiv 0(mod p)$$ is valid?

## nt.number theory – Simultaneous reductions of elliptic curves: same number of points \$|E(Bbb F_p)| = |E'(Bbb F_p)|\$ for some prime \$p\$?


Let $$E,E’$$ be two elliptic curves over $$Q$$. Is there at least one prime number $$p geq 5$$ of good reduction for $$E,E’$$ such that $$|E(F_p)| = |E'(F_p)|$$ ?

If $$E,E’$$ are isogenous over $$Q$$, this is clearly true. In any case, François Charles proved that there are infinitely many primes $$p$$ such that the reductions of $$E,E’$$ mod $$p$$ are isogenous over $$overline{F_p}$$. If the result holds over $$F_p$$, we are done.

For instance, for the curves $$y^2 = x^{3} – 3 x – 3, y^2 = x^{3} + x + 6$$, the smallest such prime is $$p = 3121$$.

One could ask if the above question can be generalized to arbitrary smooth projective irreducible curves $$C,C’$$ (not necessarily of the same genus!) over $$Q$$, or even over a number field $$K$$ (though the heuristic only works well if the primes $$mathfrak{p}$$ are totally split over $$Q$$, or at least have degree $$f leq 2$$).

Using ideas from there, one can show that if $$E,E’$$ are not isogenous over $$overline{Q}$$, then the set of such primes has density $$0$$. According to Lang–Trotter heuristics, one should expect the number of such primes $$p to be $$sim c sqrt{x} / log(x)$$ for some $$c geq 0$$ (if $$c neq 0$$, this should give infinitely many such primes).

Note: in his talk, he mentions that he doesn’t know what happens for three elliptic curves, nor what happens over $$F_p$$ instead of $$overline{F_p}$$.
For four (or more) elliptic curves, one should not expect infinitely many primes such that $$|E_i(F_p)|$$ are all equal when $$1 leq i leq 4$$.

## nt.number theory – Counting squares modulo \$p\$ that are also prime in an interval

What would be the best lower bound for the number of squares modulo $$p$$ in an interval $$[1,N]$$ with $$N that are prime?
Via the Burgess bound, I can find a lower bound for the number of squares modulo $$p$$ in $$[1,N]$$, but I would need a bound for the number of squares that are also prime. Since the size of $$N$$ matters, in my particular case I have
$$N=frac{sqrt{p}}{2}.$$
Thank you very much!

## prime factorization of homogeneous polynomials

Is prime factorization of homogeneous polynomials of n > 1 variables unique up to constant factors? Can anybody give an example of an ambiguous prime factorization of a homogeneous polynomial of several variables?

## prime numbers – Are there any papers about this observation of the distribution of the zeros of the zeta function?

Choose some $$x > 1$$. Then $$lim_{Ttoinfty} sum_{Im(rho) where $$rho$$ ranges over all zeros of the zeta function iff x is prime or the power of some prime. This implies that the zeta function’s zeros are in a sort of arithmetic progression, and are more likely to appear when $$cos(ln(x)rho)$$ is negative.

This can be (non-rigorously) obtained using the explicit formula for Chebyshev’s function: $$psi(x)=x-sum_{rho} frac{x^{rho}}{rho}-ln 2pi-ln(1-frac{1}{x^{2}})$$, so by taking the derivative, we get that $$psi'(x)=1-sum_{rho} x^{rho}+frac{2}{x-x^{3}}$$. Since $$psi'(x)$$ is $$infty$$ if $$x$$ is prime or the power of a prime, $$sum_{rho} x^{rho}$$ must be $$-infty$$. By using Euler’s formula, canceling sines, and dividing by $$sqrt{x}$$, we get that $$lim_{Ttoinfty}sum_{Im(rho) when $$x$$ is prime or a power of a prime.

I have also done computations for various values of $$x$$. If $$x=2$$ for example, $$sum_{Im(rho)<100}cos(ln(2)Im(rho)) = -7.078$$ (first 29 zeros summed), and $$sum_{Im(rho)<1000}cos(ln(2)Im(rho)) = -76.524$$ (first 649 zeros summed). The rate of divergence seems to be almost linear with respect to the number of zeros included in the summation, but it slows down very gradually.

Is this a known result, and is there a formal proof? Also, is there a good approximation for the rate at which this sum goes to $$-infty$$?

## appletv – Can I Watch Prime Video On Apple TV?

appletv – Can I Watch Prime Video On Apple TV? – Ask Different