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Denote the integers modulo $p$, $\mathbb{Z}$ mod $P$, as $\mathbb{Z}_P$. Denote the set of integers equivalent to $n$ mod $P$ - the equivalence class of $n$ as $\overline{n}$.

We know that for any prime $p$, $\mathbb{Z}_P$ is a field. As a finite field contains a finite number of elements, and $\mathbb{Z}_P$ has elements $\overline{0}, \overline{1}, \ldots, \overline{p-1}$, which is obviously a finite set. So is $\mathbb{Z}_P$ a finite field?

(Could it be the case that $\mathbb{Z}_P$ is not a finite field, because though $\mathbb{Z}_P$ contains only $p$ elements, every one of these elements is an infinite set?)

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    $\begingroup$ Yes, it is finite. There are only $p$ elements. $\endgroup$
    – M.B.
    Commented Nov 19, 2013 at 22:25
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    $\begingroup$ Yes, it's a finite field. That the elements of $\mathbb{Z}_p$ are themselves infinite sets doesn't matter. $\endgroup$ Commented Nov 19, 2013 at 22:26
  • $\begingroup$ In the sense of equivalence relations, even a "number" can be regarded as an inifinite set. What we commonly regard as "$2$" could be interpreted as the equivalence class $\{2,\frac{4}{2}, \frac{6}{3},\dots \}$. $\endgroup$
    – Doc
    Commented Nov 19, 2013 at 23:42
  • $\begingroup$ @Doc that's true for $\mathbb{Q}$ and supersets thereof, but actually not true for $\mathbb{Z}$ itself - for example, there is no $\frac42$ in $\mathbb{Z}$. :-) $\endgroup$
    – Newb
    Commented Nov 20, 2013 at 0:41

3 Answers 3

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Yes, it is a finite field. It doesn't matter that the elements of that field are themselves (infinite) sets, the "element of" relation is not transitive [in general, there are sets with the property that $x\in y \in T \Rightarrow x\in T$, but this is not one of them].

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For any ring $R$ and ideal $I \leqslant R$, $R/I$ is a field if and only if $I$ is maximal. Let $R = \Bbb{Z}$ in your problem. Show that every prime ideal is maximal in $\Bbb{Z}$.

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Yes it is a field, as proved. I answer just to stress something :

Noting $\mathbf{Z}_p$ for $\mathbf{Z} / p \mathbf{Z}$ is a bad idea, even if it may customary at some levels/countries (although I don't think it may be) for $\mathbf{Z}_p$ is a universally standard notation for the ring of $p$-adic integers, completion of $\mathbf{Z}$ for a $p$-adic absolute value on $\mathbf{Z}$. There is a universally accepted notation for $\mathbf{Z} / p \mathbf{Z}$ when $p$ is prime, and it is $\mathbf{F}_p$. I don't think there is one for a non prime $p$.

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    $\begingroup$ $\mathbf{Z}_n$ or $\mathbb{Z}_n$ is also a widely used standard notation for the ring of integers modulo $n$. Usually, if $\mathbf{Z}_p$ is used, it is clear from the context whether the ring of integers modulo $p$ or the $p$-adic integers are meant. If not, one needs to explicitly say which is meant. $\endgroup$ Commented Dec 19, 2014 at 16:21
  • $\begingroup$ $\mathbb{Z}$, $\mathbb{F}$ etc should never appear elswhere than on a blackboard where it is boring to quickly write bold letters. That's why $\mathbb{Z}$, $\mathbb{F}$ writing style appeared. In book, and when you have tex, it's $\mathbf{Z}$, $\mathbf{F}$. My point was therefore not of the way to write $\mathbf{Z}$ like this or like $\mathbb{Z}$, but on putting an index. By the, $\mathbf{Z}_n$, the ring of $n$-adic integers, exists also for non-prime $n$'s. $\endgroup$
    – Olórin
    Commented Dec 19, 2014 at 16:24
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    $\begingroup$ Many people prefer blackboard bold also on paper, there's nothing wrong with that. I understood your point was the subscript, but my point was, the subscript notation is also standard for the ring of integers modulo $n$ or $p$, nothing wrong with that either. $\endgroup$ Commented Dec 19, 2014 at 17:17

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