that group, assuming you view the set as a subset of Z_p, is guaranteed to be cyclic if p is prime. Otherwise, it may or may not be. More generally, any finite subgroup of the multiplicative …

I read some books about finite fields, sometimes the author refers to the finite field $\mathbb {F}_p$ and sometimes to the finite cyclic group $\mathbb Z_p$. What is the difference …

2013年10月20日 · With the usual addition, R R cannot be a vector space over Z/pZ Z / p Z, because every vector space over this field is a torsion group, precisely an elementary abelian …

2018年5月5日 · I have a small question, is there a theorem about the order of the elements in the multiplicative group Z∗ p Z p ∗ when p p is prime? I'm looking into the reduction of order finding …

2015年1月22日 · The operations (sum and product) are continuous in the various Z/pnZ Z / p n Z as these rings have the discrete topology, so that the "compositions" of operations of Zp Z p …

According to wikipedia, the Pontryagin dual of a Prüfer group is isomorphic to a group of p-adic integers. Where can I find a proof for it on the internet?

There are a few ways to define the p p -adic numbers. If one defines the ring of p p -adic integers Zp Z p as the inverse limit of the sequence (An,ϕn) (A n, ϕ n) with An:= Z/pnZ A n:= Z / p n Z …

2025年1月9日 · No, the completion of a totally ramified Zp -extension does not need to be perfectoid in general. The extension needs to be ramified in a stronger sense, ensuring that …

2015年7月28日 · show that g = 2 g = 2 is a generator of group Z∗19 Z 19 ∗ Can anyone explain me how i can show in this example and generally that an element is a generator in a group?

It is a field if x2 + 1 x 2 + 1 is irreducible in Z/pZ Z / p Z. Because the degree of x2 + 1 x 2 + 1 is two, this is exactly the case when x2 ≡ −1 mod p x 2 ≡ − 1 mod p has no solution.