
Why $Z_p$ is closed. - Mathematics Stack Exchange
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What differences are there between $\\mathbb Z_p$ and …
If we talk about the finite cyclic group $\Bbb Z_p$ then we are still visualising the numbers $\{0,1 ...
Why $\\mathbb{Z}_p^*$ is a cyclic group? - Mathematics Stack …
$\begingroup$ 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. Otherwise, it may or may not be. More generally, any finite subgroup of the multiplicative subgroup of a field is cyclic.
order of elements in $\\mathbb{Z}_p^*$ with $p$ prime
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Is $\\mathbb Z _p^*=\\{ 1, 2, 3, ... , p-1 \\}$ a cyclic group?
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abstract algebra - Ring of $p$-adic integers $\mathbb Z_p ...
Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
ring theory - $Z_p[i]$ is a field? - Mathematics Stack Exchange
It is a field if $x^2 + 1$ is irreducible in $\mathbb{Z}/p\mathbb{Z}$. Because the degree of $x^2 + 1$ is two, this is exactly the case when $x^2 \equiv -1 \bmod{p ...
Show that if $p$ is prime then $\\Bbb Z_p$ is a field
2016年9月26日 · $$[a]\cdot\Bbb Z_p=\Bbb Z_p\implies \exists [b]\in\Bbb Z_p:[a][b]=[1]$$ In other words: every element of $\Bbb Z_p$ but $[p]$ have a multiplicative inverse as stated above. $\Box$ Because $\Bbb Z_p$ is a commutative ring with $[1]\neq[0]$ and have multiplicative inverse for all their elements but $[p]=[0]$ then $\Bbb Z_p$ is a field. $\Box$
Topology on $\\mathbf{Z}_p$ - Mathematics Stack Exchange
2015年1月22日 · The operations (sum and product) are continuous in the various $\mathbf{Z}/p^n \mathbf{Z}$ as these rings have the discrete topology, so that the "compositions" of operations of $\mathbf{Z}_p$ with the projections $\mathbf{Z}_p \to \mathbf{Z}/p^n \mathbf{Z}$ are all continuous, and by definition of the product topology, this means that the ...
Prove that $\\mathbb Z _p$ is compact - Mathematics Stack …
2014年1月20日 · $\begingroup$ There are hints in the book #A Course in the Arithmetic# of J.P Serre. $\mathbb Z_p$ inherits a topology of $\prod \mathbb Z/p^n\mathbb Z$ if we define the discrete topology of $\prod \mathbb Z/p^n\mathbb Z$. We just need to prove that $\prod \mathbb Z/p^n\mathbb Z$ is compact and $\mathbb Z_p$ is closed. $\endgroup$ –