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By definition: If $F$ is a field and $K\subset F$ is the smallest field contained in $F$, we call $K$ the prime subfield of $F$. Denote the prime subfield of $F$ by $P(F)$ (hope you don't mind my introducing this notation).

We know that for any field $F$ with characteristic $p$, $P(F)=\{a\cdot b^{-1}, \text{where } a\in\{0,1,...,p-1\},b\in\{1,2,...,p-1\}\}$. I'd like to know if $P(F)$ is a set of integers, but it's not clear to me that it should be so (e.g. $2(p-1)^{-1}$ an integer?)

For some context: I am trying to prove that if the characteristic of $F$ is $p$ for some $p$ prime, $P(F)$ is isomorphic to $\mathbb{F}_p$. I'm convinced that $\sigma:P(F)\to \mathbb{Z}_p$ with $\sigma(f)=\overline{f}\equiv f$ mod $p$ will give a homomorphism, and hence $\sigma$ is an isomorphism between $P(F)$ and $P(\mathbb{Z}_p)=\mathbb{Z}_p$. [I haven't yet verified that $P(\mathbb{Z}_p)=\mathbb{Z}_p$; seems $\mathbb{Z}_2\subset \mathbb{Z}_p$, but $\mathbb{Z}_2$ is not a field with the addition inherited by $\mathbb{Z}_p$?] I know that the desired homomorphism properties - $\sigma(a+b)=\sigma(a)+\sigma(b); \sigma(ab)=\sigma(a)+\sigma(b); \sigma(1)=1$ - hold if $a,b$ are integers, hence the initial question of the post.

If my wishful thinking is off and $P(F)$ is not a set of integers, I was thinking $P(F)$ is isomorphic to $\{0,1,...,M\}$ for some $M\le p^p$, and then we can define the above homomorphism from $\{0,1,...,M\} \to \mathbb{Z}_p$. But we no longer have the guarantee that $\{0,1,...,M\}$ is a field. So I'm not sure where I would go from there.

Thanks a bunch in advance! And please let me know if I should filter my posts more before I answer questions. I usually don't include much "thought-process" text - maybe I should not do so.

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  • $\begingroup$ Have you seen the first isomorphism theorem for rings or Bézout’s identity? $\endgroup$ – k.stm Aug 30 '16 at 19:35
  • $\begingroup$ I have not. In fact, no discussion of rings yet. Ot seems rings are also useful for understanding polynomials (or vice versa? Both?), the latter of which features in my HW assignment (and the former not at all). I will read on rings and look out for this theorem. Thanks for the tip! $\endgroup$ – manofbear Aug 30 '16 at 19:41
  • $\begingroup$ Rings are extremely useful and absolutely foundational to abstract algebra. In fact, I’d recommend studying some basic ring theory before or alongside field theory. $\endgroup$ – k.stm Aug 30 '16 at 19:42
  • $\begingroup$ Excellent, thanks for that insight. I'm going through Artin Ch 10 (rings) as we speak. Any particular references you'd recommend as nice introductions without much in the way of background? $\endgroup$ – manofbear Aug 30 '16 at 19:43
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    $\begingroup$ Sorry, I’m not too familiar with English textbooks on basic algebra/linear algebra, but there are plenty of questions about recommendations here, just search for questions with the reference-request tag and one of the abstract-algebra or linear-algebra tags. Lang’s Algebra is often cited I think. From what I can recall, it’s rather encyclopedic. I learnt my basic algebra from first-year introductory lectures, Bosch’s Linear Algebra and Algebra (in German) and a lot of Wikipedia. $\endgroup$ – k.stm Aug 30 '16 at 19:49
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If my wishful thinking is off and P(F) is not a set of integers,

You're right to question your wishful thinking. The prime subfield is not a set of integers.

Think about the $p$-element field $\mathbb{Z}_p$, which is its own prime subfield. It has $p$ elements that it's convenient to name with the names $0, 1, \ldots , p-1$ of the first $p$ nonnegative integers, but it doesn't contain those integers, since their arithmetic isn't ordinary integer arithmetic, it's arithmetic mod $p$. Your question suggests that you sort of understand this.

In answer to your last paragraph: you should think a question through as best you can before asking it here - but you need not arrive at perfect clarity. If you could, you'd probably have the answer.

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  • $\begingroup$ Ethan, the voice from the past says that I like this answer very much. $\endgroup$ – Lubin Sep 11 '16 at 23:45
  • $\begingroup$ @Lubin Voices from the past are always welcome (and comforting). $\endgroup$ – Ethan Bolker Sep 11 '16 at 23:48
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In characteristic $p$, the prime subfield is just $\mathbf Z/p\mathbf Z$. This is because for any commutative ring $R$, there's a canonical ring homorphism which maps each $n\in \mathbf Z$ onto $n\cdot 1_R$, and if it is not injective, its kernel is generated by an integer $a>0$, whence by the 1st isomorphism theorem an injective ring homomorphism $\mathbf Z/a\mathbf Z\hookrightarrow R$.

Now if $R$ is an integral domain (e.g. a field $F$), the kernel is a prime ideal, generated by a prime number $p$. You now have your injection from $\mathbf Z/p\mathbf Z$ into the field $F$.

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