1

By Fermat's little theorem, we know if $p$ is a prime number, we have: $$a^p\equiv a\,\,\,(\text{mod }p).$$ We know in a finite field $F$ of order $p$ when $p$ is prime, characteristic of $F$ is $p$. So, using Fermat's little theorem $a^p=xp+a$ for some $x$. This means $a^p=a$ in $F$. This shows that all elements of $F$ satisfy the polynomial $X^p-X=0$. As this equation has $p$ roots, we conclude all the elements of $F$ are roots of this equation or conversely all the roots of this equation are elements of $F$.

At this point, I appreciate any corrections if I have any mistakes in the previous claims.

Now, a more general claim is that for every field $F$ of order $p^n$ when $p$ is prime, every element of $F$ satisfies the polynomial $X^{p^n}-X=0$. But I can't understand how this is concluded. Because, here $p^n$ is not a prime number. Any explanation of why this is correct is apprecaited!

Micheal Brozovic
  • 21
  • This follows from Lagrange's theorem for groups, which is also a nice way to prove Fermat's Little Theorem in the first place. – Izaak van Dongen Oct 10 '23 at 19:27
  • How much group theory do you know? (i.e. Would you understand the explanation: the multiplicative group of units of $F$ has order $p^n-1$, so every element of that group has order dividing $p^n-1$.) – Daniel Schepler Oct 10 '23 at 19:27
  • Hint: $0$ is a root of $x^{p^n}-x,$ and $F\setminus{0}$ is a group under multiplication of size $p^n-1,$ so $x^{p^n-1}-1=0$ for all non-zero $x.$ – Thomas Andrews Oct 10 '23 at 19:30
  • @DanielSchepler I haven't studied abstract algebra, I'm studying linear algebra recently. So I don't know a lot about groups (I can say I know nothing). Actually I need to prove this to use in my other question. I just know about fields and its axioms, some properties and characteristic. – Micheal Brozovic Oct 10 '23 at 19:33
  • OK, then I guess you could generalize a certain proof of Fermat's little theorem: if $\alpha \ne 0$, then multiplication by $\alpha$ permutes all the nonzero elements of the field, so $\prod_{x \in F \setminus { 0 }} (\alpha x) = \prod_{x\in F \setminus { 0 }} (x)$. But the first is also equal to $\alpha^{p^n-1} \prod_{x\in F \setminus { 0 }} (x)$, from which you can conclude $\alpha^{p^n-1} = 1$. – Daniel Schepler Oct 10 '23 at 19:36
  • @ThomasAndrews I don't know a lot about groups. Can you give me a resource that I can read about what you said briefly? – Micheal Brozovic Oct 10 '23 at 19:36
  • @DanielSchepler Awesome, I understand. But I have a little confusion in the part you said multiplication by $\alpha$ permutes all the non-zero elements of the field. At first, what is $\alpha$? What set does it belong to? And next question is why it permutes all the non-zero elements? Do you mean $\alpha x$ is equal to another element of the field and multiplying them is equal to multiplying the element of the field? If so, how do we say that all the elements are covered by multiplying them by $\alpha$? Actually how can we say it permutes the non-zero elements? – Micheal Brozovic Oct 10 '23 at 19:48
  • Literally any introduction to abstract algebra will cover this. @MichealBrozovic – Thomas Andrews Oct 10 '23 at 20:12

1 Answers1

2

Since you basically asked for a group theory free proof, here is one. (Of course, it's just group theory in disguise.)

Lemma. Let $F$ be a finite field with $q$ elements. Then $a^q=a$ holds for all $a \in F$.

Proof: For $a=0$ this is clear. So let $a \neq 0$. We consider the product of all the non-zero elements in $F$. $$s := \prod_{\large 0 \neq b \in F} b.$$ This means that when $F = \{b_1,b_2,b_2,\dotsc,b_q\}$ and $b_1 = 0$, we let $s = b_2 \cdot b_3 \cdots b_q$. Since $F$ has no zero divisors, $s \neq 0$. Now, the map $b \mapsto a \cdot b$ is a bijection from the set of non-zero elements of $F$ to itself. In fact, $b \mapsto a^{-1} \cdot b$ is inverse to this map. This means that we can make a substitution in the product. We get $$s = \prod_{\large 0 \neq b \in F} a \cdot b$$ There are $q-1$ factors, in each one we can take out $a$. So we get $$s = a^{q-1} \cdot \prod_{\large 0 \neq b \in F} b = a^{q-1} \cdot s$$ Since $s \neq 0$ and $F$ has no zero divisors, we get $1 = a^{q-1}$. Hence, $a = a^q$. $~~~\square$

Martin Brandenburg
  • 163,620
  • I get what you said completely except of the part that you say the map $b \mapsto a \cdot b$ is a bijection from the set of non-zero elements of $F$ to itself. As I asked in the comments, is $a$ an element of $F$? Then how do we conclude this map is a bijection? – Micheal Brozovic Oct 10 '23 at 20:03
  • 1
    @MichealBrozovic The map $b\mapsto a^{-1}\cdot b$ is an inverse of that map. – Jyrki Lahtonen Oct 10 '23 at 20:07
  • Yes, $a \in F$ by assumption. It is the same $a$ as in the claim of the Lemma. 2) A map is bijective iff it has an inverse map. (One can argue that this should be the definition of a bijection, by the way. In any case, don't forget this!) I specifed the inverse map.
    • – Martin Brandenburg Oct 10 '23 at 20:07
  • @MartinBrandenburg Yes now I got it. One last question just for assurance. When we say a relation is a map? Saying every element in domain has only one correspondence in range is enough? Is this the correct definition or this is for functions? Then what is a map? – Micheal Brozovic Oct 10 '23 at 20:23
  • https://en.wikipedia.org/wiki/Function_(mathematics) – Martin Brandenburg Oct 10 '23 at 20:25
  • @MartinBrandenburg So a function is also called a map. Thanks a lot. Now I got everything. – Micheal Brozovic Oct 10 '23 at 20:28