# Every finite field is perfect

In the book Contemporary Abstract Algebra by Gallian, I'm reading the following proof of the theorem that every finite field is perfect: But what seems questionable to me is the part where the author is arguing that $(a+b)^p= a^p + b^p$. Consider $\mathbb{Z}_3=\mathbb{Z}/3\mathbb{Z}$, which is a field. Then $(a+b)^2=a^2+2ab+b^2$. But does $3$ divide $2$? If so then $2=3n=0\mod 3$, for some integer $n$, which is obviously not true. Would someone please clarify this for me?

• You should take $3$ in your example. This property only holds for the characteristic $p$ of that field, where the characteristic is the smallest natural number $p$ such that $p \cdot 1 = 0$. In your example, the characteristic is $3$ instead of $2$. – Student Mar 14 '17 at 18:53
• @RCT why is $\phi$ onto? – Babai Jan 31 '18 at 15:22

## 1 Answer

At the beginning of the proof, $p$ is defined as the characteristic of the field. The claim is that the Freshman's Dream holds for that power, not just any power.

• Can you please explain why is $\phi$ onto? – Babai Jan 31 '18 at 15:23
• @Babai $\phi$ is a map between finite sets of the same size, so it is surjective if and only if it is injective. – RCT Feb 1 '18 at 16:58