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I want to show this: $p$ prime, $p \equiv 2 \bmod 3 \implies t^{3}-a$ reducible in $\mathbf{F}_p[t]$ for all $a\in \mathbf{F}_{p}$. By using Fermat's little theorem, but I don't know how.

If somebody would show me, I'd be very glad.

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2 Answers 2

up vote 6 down vote accepted

We will prove a more general statement:

If $p$ is a prime and $n$ is relatively prime to $p-1$, then the map $x \mapsto x^{n}$ is a bijection from $\mathbf F_p$ to itself. In other words, every element has a unique $n^{\text{th}}$ root in $\mathbf F_p$.

As a simple corollary, every polynomial of the form $t^n - a$ is reducible, since it is divisible by $t-b$ where $b$ is a $n^{\text{th}}$ root of $a$.

Now, even though the above lemma is stated for the field $\mathbf F_p$, it is convenient and more enlightening to pass to the multiplicative group $\mathbf F_p^{\times}$: the additive structure is irrelevant and the zero element is best regarded separately here. As an exercise, verify that our first lemma follows as a consequence of the following.

If $G$ is a finite group and $n$ is relatively prime to $|G|$, then the homomorphism $\varphi : G \to G : x \mapsto x^{n}$ is an automorphism of $G$. That is, every element in $G$ has a unique $n^{\text{th}}$ root.

Proof. Since $G$ is finite, establishing one of injectivity and surjectivity suffices, but we show both for fun. We need the following two facts:

  • $a^{|G|} = 1_G$ for all $a \in G$.

  • Since $\gcd(n, |G|)= 1$, by Bézout's lemma, there exist integers $u, v$ such that $u n + v|G| = 1$.

Armed with these facts, we are ready to show

  • Surjectivity: Given any $a \in G$, $$ \left( a^{u} \right)^n = a^{un} = a^{un} \cdot a^{v|G|} = a^{un + v |G|} = a, $$ so every element of $G$ is in the image of $\varphi$.

  • Injectivity: Since $\varphi$ is a homomorphism, it suffices to check that its kernel is trivial. Let $a \in G$ such that $a^n = 1_G$. Then $$ a = a^{u n + v|G|} = \left(a^{n} \right)^u \cdot a^{v|G|} = 1_G \cdot 1_G = 1_G, $$ and hence we are done.

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Automorphism is a strange word to use. It's not a field automorphism. All you really need is that the map is a bijection. –  Thomas Andrews Dec 22 '11 at 17:17
    
@Thomas Good point. It's an automorphism of the group $\mathbf F_p^*$ though, right? –  Dylan Moreland Dec 22 '11 at 17:19
1  
@ThomasAndrews Yikes! I'm wondering what I was thinking then. :-) Thanks for your comment. –  Srivatsan Dec 22 '11 at 17:20
    
I'd also say that it is easier to prove surjection rather than injection. Let $nx + (p-1)y = 1$. Then for any $a\neq 0$ in $\mathbb F_p$, $(a^x)^n = a$. Also, clearly $0$ is in the image, so the map is surjective. And the part $(a^x)^n = a$ uses Fermat's Little Theorem. –  Thomas Andrews Dec 22 '11 at 17:20
    
@DylanMoreland Yes, it is an automorphism of $\mathbb F_p^\times$ –  Thomas Andrews Dec 22 '11 at 17:23

You need to show that $t^3 - a$ has a root in $\mathbf F_p$. By little Fermat we know that $a^p = a$. As $p \equiv -1\ \bmod 3$, we also know that $3$ divides $p + 1$. You can put these two facts together to find a cube root $b$ of $a^2$. Now, what is the cube of $a/b$?

[Maybe it's better to think about it this way: $\mathbf F_p^*$ is a cyclic group of order $p - 1$, and this order is coprime to $3$. Once again, thanks to Srivatsan for correcting an awful blunder of mine in the original version.]

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@ Dylan Moreland: Thanks!!! –  Tashi Dec 22 '11 at 19:21
    
Since $2-p \equiv 0 \mod 3$ and $a^{2-p} = a$, we can also write the cube root of $a$ directly as $a^{\frac{2-p}{3}}$. (Another equivalent expression is $a^{\frac{2p-1}{3}}$.) –  Srivatsan Dec 23 '11 at 14:29

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