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I have been trying to understand the following portion of this answer:

First, any field of order $p^n$ will have characteristic $p$, so the underlying additive structure of the group is necessarily $(\mathbb{Z}_p)^n$.

I hadn't heard of the term "characteristic" before. According to Wikipedia:

In mathematics, the characteristic of a ring $R$, often denoted $\text{char}(R)$, is defined to be the smallest number of times one must use the ring's multiplicative identity $(1)$ in a sum to get the additive identity $(0)$ if the sum does indeed eventually attain $0$. If this sum never reaches the additive identity the ring is said to have characteristic zero.

However, from this alone, I'm not being able to deduce how a field of order $p^{n}$ must have characteristic $p$. Also, how are we able to conclude that the underlying additive structure of the group is necessarily $(\Bbb Z_{p})^n$?

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  • $\begingroup$ Relevant: math.stackexchange.com/questions/72856/… $\endgroup$ – Robert Wolfe May 24 '18 at 5:19
  • $\begingroup$ This is the structure of a vector space (of dimension $n$) over $\mathbb Z/p$. So you should prove that you have such a vector space. $\endgroup$ – GEdgar May 24 '18 at 12:43
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Short answer: the distributive law.

Long Answer: The additive group has order $p^n$, so it must have some nonzero element $s$ of additive order $p$. Then $1$ plus itself $p$ times multiplied by $s$ is zero, since it's the same thing as $s$ added to itself $p$ times. But this means $1$ plus itself $p$ times must be $0$ since $s\ne 0$. But then any field element $z$ added to itself $p$ times be zero, since that's the same as $z$ times $1$ added to itself $p$ times. So all nonzero elements of the additive group have order $p$.

But the structure for finite abelian group says that this must be $(\Bbb{Z}/(p))^n$, since a finite abelian group must be a direct product of cyclic groups, and the cyclic factors must all have order $p$ lest there be a nonzero element of some other order in their direct product.

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  • $\begingroup$ Why must it have some nonzero element $s$ of additive order $p$? That is not clear to me $\endgroup$ – user563280 May 24 '18 at 3:36
  • $\begingroup$ @user48929: Cauchy's theorem. $\endgroup$ – Nick Matteo May 24 '18 at 3:41
  • $\begingroup$ @Kundor Thanks, makes sense $\endgroup$ – user563280 May 24 '18 at 3:55
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Given a field $F$, one may define a ring homomorphism $\varphi : \Bbb Z \to F$ by $n \mapsto n \cdot 1_F$ (where $1_F$ is the multiplicative identity of $F$). If the field has characteristic $0$ then $\ker \varphi = \lbrace 0 \rbrace$ so that $F$ has a copy of $\Bbb Z$ sitting inside it and hence a copy of $\Bbb Q$ (since all non-zero elements must have multiplicative inverses in a field).

Now suppose $F$ has characteristic $p$. Then $\ker \varphi = p\Bbb Z$ and so $F$ has a copy of $\Bbb Z/p\Bbb Z$ sitting inside it. Thus $p \cdot 1_F = 0_F$. Note that I made no assumption on the order of $F$, just on its characteristic. One can show that if $F$ has characteristic $p$ then $F = \Bbb F_{p^n}$, and so $F$ is an $n$-dimensional $\Bbb Z /p\Bbb Z$-vector space. From this it follows that $F \cong (\Bbb Z/p\Bbb Z)^n$ as $\Bbb Z/p\Bbb Z$-vector spaces (since vector spaces are uniquely determined by their dimension), so as an additive group $F$ is isomorphic to $(\Bbb Z/p\Bbb Z)^ n$.

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You are correct that this is not obvious from the definition! The first question, about why a field of order $p^n$ has characteristic $p$ is straightforward. The key observation here is that any field of finite characteristic must have prime characteristic. To see this, suppose for contradiction that $\mathbb{F}$ is a field of characteristic $n = pq$ where neither $p$ nor $q$ is 1. Then, $\sum_{i=0}^{pq} 1 = 0$, hence $\sum_{i=1}^n p= 0$ where $p \in \mathbb{F}$ is the element formed by adding 1 to itself $p$ times. Distributivity shows then that $pq = 0$ in $\mathbb{F}$. But neither $p$ nor $q$ are $0$ in $\mathbb{F}$ since they are formed by adding the additive identity to itself strictly fewer times than $pq$. So $p$ and $q$ are zero divisors, which is impossible since $\mathbb{F}$ is a field, hence an integral domain. This shows that any field of prime finite characteristic must have prime characteristic, hence a field of order $p^n$ has characteristic $p$.

Answering your second question takes some basic Galois theory. The form of this fact I've typically seen is in fact stronger: the only finite fields that exist are exactly $(\mathbb{Z}_p)^n$ up to both additive and multiplicative structure. Wikipedia has a nice treatment of the proof (https://en.wikipedia.org/wiki/Finite_field#Existence_and_uniqueness). If you haven't seen Galois theory before though, the article isn't particularly approachable. The big idea that makes this true is that any field of order $p^n$ must be a splitting field of the polynomial $x^{p^q} - x$ over $\mathbb{F}_p$. Then, the fact from Galois theory that all splitting fields of a polynomial over the same base field are isomorphic implies the result.

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