Problem :

  1. Show that the characteristic of a finite field $K$ is a prime number.

  2. Show that $K$ is an extension of $\mathbb{Z}/p\mathbb{Z}$ and there exists $n$ such that Card($K$) $= p^n$.

  3. Show that every non zero elements of $K$ is a root of $t^{p^n -1}-1$.

  4. Show that $K$ is the splitting field of $t^{p^n}-t$.

I managed to prove the first and second problems. The first one by contradiction and the second by showing that $\mathbb{Z}/p\mathbb{Z}$ was isomorphic to a subfield of K (the image of the characteristic homomorphism).

However I have difficulties showing the third problem. I think it would equivalent to showing that $\forall x \in K : x^{p^n -1} = 1$, or maybe that $\forall x \in K : x^{p^n} = x$ and use Ferma's little theorem.


You're on the right track, but there are a couple of slips in your equivalences. The equation $\ x^{p^n-1} = 1\ $ doesn't hold for all $\ x\in K\ $, only for all non-zero $\ x\in K\ $, but that's precisely what it is that you're being asked to show in the third leg of your problem. And you can indeed use Lagrange's theorem (a generalisation of Fermat's little theorem) to show it, since the nonzero elements of $\ K\ $ form a multiplicative group of order $\ p^n-1\ $.

The other slip (just a typo, I presume) is in the equation $\ x^{p^n} = p\ $. This should be $\ x^{p^n} = x\ $.

  • $\begingroup$ Thank you! I was trying to figure out how to prove that for any $x \in K$, $x^{p^{n}-1} = 1$, and I totally forgot the group of inversible elements of $K$ was cyclic and precisely of order $p^{n} -1$. $\endgroup$ – NotAbelianGroup May 4 at 16:16
  • $\begingroup$ For the last question, can I conclude that since $t^{p^n}-t=t(t^{p^{n}-1}-1)$ and all the non zero roots of $(t^{p^{n}-1}-1)$ are in $K$ by the 3rd question, and the $0$ root is clearly in $K$, all the roots of $(t^{p^{n}-1}-1)$ are in $K$, thus $K$ is its splitting field? $\endgroup$ – NotAbelianGroup May 4 at 16:17
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    $\begingroup$ You need one more observation. A splitting field of a polynomial is the smallest extension of the field generated by its coefficients over which the polynomial splits into linear factors. That's trivially true in this case, because every member of $\ K\ $ is a root. $\endgroup$ – lonza leggiera May 5 at 2:37

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