If $\alpha$ is a root of $x^{p^n}-x+1$, then show that $\mathbb{F}_{p^n}\subset\mathbb{F}_p(\alpha)$

Question

Question

Let $\mathbb{F}_p$ be the field with $p$ elements, where $p$ is a prime. Let $f_{n,p}(x)=x^{p^n}-x+1$, and suppose that $f_{n,p}(x)$ is irreducible in $\mathbb{F}_p[x]$. Let $\alpha$ be a root of $f_{n,p}(x)$. Show that $\mathbb{F}_{p^n}\subset\mathbb{F}_p(\alpha)$ and $[\mathbb{F}_p(\alpha):\mathbb{F}_{p^n}]=p$.

Answer

I know that $\mathbb{F}_{p^n}$ consists of the roots of the polynomial $x^{p^n}-x$. Thus, if we can find a root, say $\gamma$, of this polynomial in $\mathbb{F}_p(\alpha)$, except $0$, in terms of $\alpha^i$'s, we are done as the other roots follow by $\gamma,\gamma+1,\dots,\gamma+p-1$. However, I couldn't find such a root. Is there any way I can follow to find it or do I need another approach to solve this question? Thanks in advance...

Answer 1

I need to apologize. I just noticed that even though we have handled non-irreducibility of this polynomial as well as its factorization over the prime field somewhat adequately, this time the question in the title is markedly different. I only saw what I expected to see, and failed to read closely.

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As stated, the claim in the title is actually false!

For a counterexample consider the case $p=2$, $n=3$. In the ring $\Bbb{F}_2[x]$ have the factorization into irreducibles

$$x^8-x+1=(x^2+x+1)(x^6+x^5+x^3+x^2+1).$$

And if $\alpha$ is a root of the quadratic factor $x^2+x+1$, then $\Bbb{F}_2(\alpha)=\Bbb{F}_4$, which is not a subfield of $\Bbb{F}_{2^n}=\Bbb{F}_8$.

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What I was keen on proving in the comments is that the splitting field of the polynomial $f(x)=x^{p^n}-x+1$ over the prime field $\Bbb{F}_{p}$ is $$ K=\Bbb{F}_{p^{pn}}, $$ that is, the unique extension of degree $pn$.

To see this we need to first show that if $\alpha$ is a zero of $f(x)$ from some extension field of $\Bbb{F}_p$, then $\alpha^{p^{pn}}=\alpha$, implying that $\alpha\in K$. While doing that we also observe that $\alpha^{p^n}=\alpha-1$, implying that we NEVER have $\alpha\in\Bbb{F}_{p^n}$. However, if $\alpha$ is any root of $f(x)$ and $z\in\Bbb{F}_{p^n}$ is arbitrary, we easily see that $\alpha+z$ is also a root of $f(x)$. Hence $z$ must be an element of the splitting field. Consequently $L=\Bbb{F}_{p^n}\subseteq K$.

As $[K:L]=p$ is a prime, there are no intermediate fields between $L$ and $K$. Hence $K$ must be the splitting field.

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We can also show that while we don't have the title claim for each and every root of $f(x)$, for a bit more carefully chosen root $\alpha$ of $f(x)$ we do have $K=\Bbb{F}_p(\alpha)$. This part of my answer requires familiarity with the properties of the trace map.

In the field $L$ there exists $p^{n-1}$ elements of trace $-1\in\Bbb{F}_p$. Among them we can find an element $z$ such that $L=\Bbb{F}_p(z)$. Reusing the calculation in the linked old answer of mine we see that $m_z(x)=x^p-x-z$ is a factor of $f(x)$ in the ring $L[x]$. So if $\alpha$ is a zero $m_z(x)$, then also $f(\alpha)=0$. Furthermore:

• $z=\alpha^p-\alpha\in \Bbb{F}_p(\alpha)$, and hence $L=\Bbb{F}_p(z)\subset \Bbb{F}_p(\alpha)$, but
• $\alpha\notin L$ because otherwise $z=\alpha^p-\alpha$ would have trace zero.

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The case $p=n=2$ is exceptional for the purposes of irreducibility of $f(x)$ simply because in that case $pn=4=p^n$.