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I'm studying the concept of field applied to modular aritmetic. Is it correct to say that, if the dimension is a prime number $p$ then field properties are satisfied for the integers $\bmod p$ ? And that, if the dimension is a power of a prime number $p$ (that is $p^{n}$), not all integers $\bmod p^{n}$ form a field, but only a set of $p^{n}$ integers ?

Bernard
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AleWolf
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1 Answers1

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The integers modulo $p$ for a prime $p$ does form a field.

The integers modulo $p^n$ for $n>1$ doesn't form a field. And there isn't a subset of that which forms a field either.

There are fields with $p^n$ elements for any prime $p$ and natural number $n$. However, for $n>1$ this field does not coincide with the ring of integers modulo $p^n$. For instance, the additive group of such a field is $(\Bbb Z_p)^n$.

Arthur
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  • Thanks so much. I would like to study your assertions on a book or websites, do you have some links? – AleWolf May 13 '19 at 19:30
  • @AleQuercia Any introductory book in abstract algebra (as long as it covers fields at all) should do. – Arthur May 13 '19 at 19:34
  • Thanks so much. – AleWolf May 13 '19 at 19:38
  • However, take a look at Theorem 4.1 here – AleWolf May 13 '19 at 19:59
  • @AleQuercia That seems to agree with me: The integers modulo $p$ is a field iff $p$ is prime. That's what theorem 4.1 says too. – Arthur May 13 '19 at 20:00
  • Yes yes, you're right :) – AleWolf May 13 '19 at 20:01
  • What means that for a prime $p$ and $n>1$ there exists only one field with order $p^{n}$ ? Isn't obvious that there exist only one field ? – AleWolf May 14 '19 at 09:59
  • @AleQuercia First of all, it should say "There is only one field up to isomorphism" (there are different fields, but they are isomorphic). Second, no, it's not at all obvious that all fields of order $27$ are isomorphic. Or that all fields of order $625$ are. And so on. But they are isomorphic, and that's what that statement means. – Arthur May 14 '19 at 10:05
  • If there are different fields, in what are different ? – AleWolf May 14 '19 at 10:17
  • @AleQuercia When two things are different, but isomorphic, the only difference between them is basically the names of the elements. For instance, you can have the field with elements ${0, 1, a, a+1}$, where $x + x = 0$ for any $x$, and $a\cdot a = a+1$. This would be a field with $4$ elements. You can also have a field with elements ${p, q, r, p+q}$ with the rules that for any $x$ in the field, $x + r = x$, $x + x = r$ and $p\cdot x = x$, while $q^2 = p + q$. This is also a field of $4$ elements. They are not the same field. But they are isomorphic. – Arthur May 14 '19 at 10:27
  • Is it correct to say that if we consider the field with order $p^{n}$ , where p is prime , then the element $p^{n}$ doesn't belong to the field since its inverse doesn't exist ? – AleWolf May 14 '19 at 11:44
  • @AleQuercia What do you mean by "the element $p^n$"? And what do you mean by "exist"? If you add $1$ to itself $p$ times in such a field, then that gives you $0$. So in a sense $p = 0$. And you can raise $0$ to the $n$th power no problem. Inverses aren't requied for powitive exponents to make sense. – Arthur May 14 '19 at 11:47