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 subgroup of a field is cyclic. The proof relies on the following characterization of finite cyclic groups: A finite group is cyclic iff for every divisor d of the order of the group there are at most …

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 between them?

Oct 20, 2013 · 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 p p -group, that is, a direct sum of copies of Z/pZ Z / p Z. Since R …

May 5, 2018 · 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 to factoring. could it be that the order of all elements x ∈ Z∗ p x ∈ Z p ∗ are odd.

Jan 22, 2015 · 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 with the projections Zp →Z/pnZ Z p → Z / p n Z are all continuous, and by definition of the product topology, this means that the operations of Zp Z p are continuous.

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 and ϕn: An → An−1 ϕ n: A n → A n − 1 (like in Serre's book), how to prove that Zp Z p is the same as

Jan 9, 2025 · 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 the p th-power map on OK∞/p is surjective. By contrast, many classical Lubin-Tate or π 1 p∞ -type extensions do have that stronger ramification property and thus yield perfectoid completions.

Jul 28, 2015 · 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.