# Nonzero trace in finite fields and proving irreducibility.

If we define trace to be $x+x^p+\cdots+x^{p^{n-1}}$. How do we know there is an element of nonzero trace? Clearly if $a\in F_p$ then its trace is zero as $a^{p^i}=a$ so $\operatorname{tr}(a)=a+a+\cdots+a=pa=0$. So we know this element has to come from $F_{p^n}$ and I reckon it needs to not be in any subfield, but I can't seem to show this.

Also, I am trying to show that $x^p-x-a$ which is in $F_{p^n}$ is either irreducible or factors completely. So far, I have seen that if the polynomial has some root, say $b$, then it also has roots $b+i$ for $i=1$ to $p-1$ so we found all the $p$ roots of the $p$ degree polynomial so it factors completely into linear factors if we can get one of its roots. But I don't see why it's that or irreducible. We can't the polynomial factor into two polynomials of lower degree that are irreducible?

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A question related to your first question, and your second question has been answered here. I am somewhat tempted to call this a duplicate, but cannot decide which :-) –  Jyrki Lahtonen Nov 28 '12 at 13:38

For your question about trace, note your reasoning is slightly wrong because that $\text{tr}(a) = na$ actually and not $pa$. However, to show an element of nonzero trace exists is a fairly simple argument. Suppose $x + x^p + ... + x^{p^{n-1}}$ vanishes everywhere in $F_{p^n}$. But then this polynomial has $p^n > p^{n-1}$ roots which is absurd, so it follows there must exist at least $p^n - p^{n-1}$ elements of nonzero trace in fact.
For your second question, note that $(x^p - x - a)' = -1$ where I am taking a derivative. It follows $x^p - x - a$ has no repeated factors since it is relatively prime with its derivative. Now let $x^p - x - a = p_1(x)p_2(x)...p_m(x)$ where $p_i(x)$ are distinct irreducible polynomials over $F_{p^n}$. Take roots $r_1,r_2,...,r_m$ from them. Then it is easy to see by your logic of $b$ being a root implying $b+i$ being a root that $F_p[r_1] = F_p[r_2] = ... = F_p[r_m]$ which implies all the irreducible polynomials have the same degree. But then $m|p$, implying $m=p$ or $m=1$ which gives the desired result.