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I'm studying Fraleigh's Abstract Algebra, and I'm completely new to fields. I'm studying Theorem 33.10, which states:

A finite field of order $p^n$ exists for every prime power $p^n.$

The proof goes by considering $\mathbb{Z}_p$ and its algebraic closure $\overline{\mathbb{Z}}_p$. We let $K\subseteq \overline{\mathbb{Z}}_p$ denote the set of the distinct zeros of the polynomial $x^{p^n} - x$. Then, we show that $K$ is closed under addition, multiplication, has additive/multiplicative identities and inverses. From there we deduce that $K$ is a subfield of $\overline{\mathbb{Z}}_p$, with order $p^n$, as desired.

Well, here's what confuses me: Fraleigh states that $K$ is a subfield of $\overline{\mathbb{Z}}_p$ that contains $\mathbb{Z}_p$. I don't really understand why $K$ contains $\mathbb{Z}_p$. The reason must be fairly simple, because there's really no explanation, but please understand I am a complete beginner. Would appreciate some help.

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  • $\begingroup$ $x^p = x$ for all $x \in \mathbb{F}_p$. So $x^{p^2} = (x^p)^p = x^p = x$, and so on. $\endgroup$
    – orangeskid
    Commented Jul 14, 2022 at 16:01
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    $\begingroup$ A subfield contains the unit $1$. Therefore it contains $1+1+\dots+1$ for any number of terms. These make up $\mathbb{Z}_p$. $\endgroup$
    – GEdgar
    Commented Jul 14, 2022 at 16:01
  • $\begingroup$ @orangeskid Oh, so the elements of $\mathbb{Z}_p$ are all zeros of the polynomial? Wow, I should've just checked the definition of $K$. Thanks. $\endgroup$
    – Ssay
    Commented Jul 14, 2022 at 16:04
  • $\begingroup$ @GEdgar Thank you very much Edgar. $\endgroup$
    – Ssay
    Commented Jul 14, 2022 at 16:06

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Another different approach that maybe you coul find easier is this constructive argument:

Take a finite field $K$ with $p$ elements, and $f\in K[t]$ irreducible of order $n$. Let $m$ the ideal generated by $f$. We know that the quotient $K[t]/m$ is a field with $p^n$ elements.

An example for $5^3$. Take $K=\mathbb{Z}_{5}$ and $f=t^3+t+1$

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    $\begingroup$ You need to prove that an irreducible $f$ of degree $n$ exists. $\endgroup$
    – lhf
    Commented Jul 14, 2022 at 16:06
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    $\begingroup$ Hi Guillerminho77, thanks for the answer, but I think the existence of such an irreducible $f$ is proven after my theorem in the question is proven. Thanks anyways. $\endgroup$
    – Ssay
    Commented Jul 14, 2022 at 16:09

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