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In the proof that a finite field has a power of $p$ elements (where $p$ is a prime), we identify $\mathbb{Z}/p\mathbb{Z}$ in our field $F$, and think of $F$ as a finite dimensional vector space over $\mathbb{Z}/p\mathbb{Z}$. We then let $n$ denote the dimension of our vector space and let $v_1,\dots,v_n$ be our basis. The next step then assumes that any element $v\in{F}$ can be written as $$a_1v_1+\dots+a_nv_n$$ where $a_i\in\mathbb{Z}/p\mathbb{Z}$.

My question is, why can we assume that this setup will generate all of $F$? In other words, how do we know that there does not exists a $u\in{F}$ that cannot be written in the form $$a_1v_1+\dots+a_nv_n?$$ Maybe the better question is, why are we allowed to make this assumption? In other words, why are we allowed to identify $F$ with a vector space over $\mathbb{Z}/p\mathbb{Z}$?

Seth
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    By definition, every element of a vector space is a linear combination of basis elements. We can identify $F$ with a vector space because it is a vector space: it satisfies all the axioms. In general, if $K$ and $L$ are fields, and $K$ is a subfield of $L$, then $L$ is a vector space over $K$. – Arturo Magidin Dec 09 '21 at 06:01
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    "Why do we identify $F$ with a vector space over $\mathbb Z/p\mathbb Z$." This is important to know: if a field $K$ is inside a field $L$ then $L$ can be viewed as a vector space over $K$. Indeed, we can add and subtract elements of $L$ and we can multiply elements of $L$ by elements of $K$, all axioms of vector spaces hold true. – markvs Dec 09 '21 at 06:18
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    If $F$ had such a $u$, then $v_1,...,v_n$ would not be a basis. We can build a basis iteratively starting from the empty set by taking an element of $F$ not in the span of the current set and adding it to the set, until there is no such element. Since $F$ is finite, we end up with a finite basis. – Karl Dec 09 '21 at 06:25
  • Please choose one question to ask. – Vincent Dec 10 '21 at 10:36
  • Dear @ArturoMagidin, in your last sentence, what happens for example with $\mathbb{Q}$ over $\mathbb{R}$? Is there a basis that can generate all of $\mathbb{R}$? Or is specific to finite fields? – Seth Dec 11 '21 at 07:06
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    @Seth: It's $\mathbb{R}$ over $\mathbb{Q}$. Whether it has a basis depends on your Set Theory. Assuming the Axiom of Choice, a basis exists. But whether there is a basis or notdoes not matter for that: it is still a vector space over $\mathbb{Q}$. And for finite fields, the existence of a basis can be proven as outilined by Karl. – Arturo Magidin Dec 11 '21 at 11:44

1 Answers1

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(1). Preliminary. For any field $F$, the characteristic $\chi (F),$ sometimes written $\chi_F,$ is the least $n\in\Bbb Z^+$ such that $\sum_{j=1}^n1=0,$ if such $n$ exists. If no such $n$ exists it is customary to define $\chi(F)=0.$ The definition of a field forbids $1=0$ so $\chi(F)\ne 1.$

Let $S(m)=\sum_{j=1}^m1\in F$ for any $m\in\Bbb Z^+.$ If $F$ is finite then $\{S(m):m\in\Bbb Z^+\}$ is finite so there exist $m,n\in \Bbb Z+$ with $S(m)=S(m+n),$ so $S(n)=S(m+n)-S(m)=0.$ So for a finite field $F$ we have $0<\chi(F),$ so $2\le \chi(F).$

Now if $2\le n=\chi(F)$ then $n$ is prime. For if $n=n'n''$ with $n',n''\in\Bbb Z^+$ then $0=S(n)=S(n')S(n'')$ but the minimality of $n$ (from the definition of $ \chi (F)\,$) implies $S(n')\ne 0\ne S(n'')$ unless $n'\ge n$ or $n''\ge n.$

(2). Let $F$ be a finite field with $\chi(F)=p$ where $p$ is prime. We can define $\Bbb Z/p\Bbb Z=\{j\in \Bbb Z: 0\le j<p\}$ with arithmetic modulo $p$. For $0\ne j\in \Bbb Z/p\Bbb Z$ and $f\in F,$ define $jf=S(j)f\in F.$ And if $0_p$ is the additive identity of $\Bbb Z/p\Bbb Z$ then define $0_pf=S(p)f=0f=0\in F$. You can confirm that $(F,\Bbb Z/p\Bbb Z)$ satisfies all the axioms for the definition of a vector space over $\Bbb Z/p\Bbb Z. $ And since $F$ is finite, a vector-space basis $B$ for $F$ must be finite.

DanielWainfleet
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