8

Let $p$ be a prime.

a) Show that $f$ has no roots in $\Bbb{F}_p$.

Let $F^*$ be the multiplicative group of $\Bbb{F}_p$. Then, by lagrange's thoerem for all nonzero $\alpha \in \Bbb{F}_p$, $\alpha^{p-1} = 1 \implies \alpha^p=\alpha \implies \alpha^p-\alpha=0$. Of course $0^p=0$, so this is true for all elements of $F$ and not just the nonzero ones. But then $\alpha^p - \alpha - 1 = -1$ for all $\alpha \in \Bbb{F}_p$ and so it must have no roots in $\Bbb{F}_p$. I could have also done this using the Frobenius automorphism, right?

b) Let $\alpha$ be a root of $f$ (in some algebraic closure of $\Bbb{F}_p$). Show that $\alpha + s$ is also a root for all $s \in \Bbb{F}_p$.

Let $\alpha^p - \alpha -1 =0$. Let $E$ be an algebraic closure of $\Bbb{F}_p$. Since $E$ has characteristic $p$, $(\alpha + s)^p = \alpha^p + s^p$. So we have,

$$(\alpha + s)^p - (\alpha + s) -1 = \alpha^p + s^p - \alpha -s -1 = s^p - s = 0.$$

c) Conclude that $f$ is irreducible over $\Bbb{F}_p$, for every $p$.

By b) and the fact that $\Bbb{F}_p$ has $p$ distinct elements, we know that the roots of $f$ are $\alpha, \alpha+1, ... , \alpha + p-1$. So if $K$ is a splitting field, we have $$x^p - x -1 = (x-\alpha)(x-(\alpha+1))...(x-(\alpha+p-1)).$$

Now let's assume that $f$ is reducible over $\Bbb{F}_p$. Then $f=gh$ for some $g$ and $h$ with degrees less than that of $f$. So $g$ and $h$ must be of the form $(x-(\alpha+s_1))...(x-(\alpha+s_k))$ where k is less than n. Let's say that g has degree 2, because the other cases are similar.

So $g =(x-(\alpha + s_i))(x-(\alpha + s_j))$ and the constant term for $g$ is,

$$(\alpha + s_i)(\alpha + s_j) = \alpha^2 + s_is_j\alpha + s_is_j.$$

Since $\Bbb{F}_p$ is a field, if $\alpha s_is_j$ is an element of $\Bbb{F}_p$ , then so is $((\frac{1}{s_is_j})(\alpha s_is_j) = \alpha$, a contradiction.

We can show by induction that if we multiply $(x-(\alpha+s_1))...(x-(\alpha+s_k)$, we get a term that looks like $s_1s_2...s_k\alpha$. So this is also true for $k>2$.

Do you think that my answer is correct?

Thank you in advance

Community
  • 1

4 Answers4

7

a) and b) are fine. I haven't checked c) but you could also note that if the polynomial splits into $h \cdot g$ with $h,g \in \mathbb{F}_p[x]$ then, say $$ h(x) = \prod_{i \in S} (x - (\alpha + i)) \in \mathbb{F}_p[x] $$

with $S$ proper subset of $\{0,1,\ldots,p-1\}$. Therefore $$ \sum_{i \in S} (\alpha + i) \in \mathbb{F}_p $$ hence $|S| \alpha \in \mathbb{F}_p$ but since $0 < |S| < p$ this implies that $\alpha \in \mathbb{F}_p$, but then $$ f(x) = \prod_{i} (x - i) = x^p - x $$ which is not true.

blabler
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5

I like your idea very much, but I didn't understand how you managed to ignore the $\alpha^2$ term from your constant term.

I have written up an answer to this question with a similar idea, where I look at the next to highest degree term of $g(x)$ (=the term of degree $k-1$). In your degree two example, this would be the linear term. Its coefficient is $2\alpha+(s_i+s_j).$ As $s_i,s_j$ are in the prime field, and $2$ is invertible, we can, as in your argument, conclude that $\alpha\in\mathbb{F}_p$, which is a contradiction.

By studying the lowest degree term, you get a lot of clutter from powers of $\alpha$. The degree $k-1$ term is IMHO easier to manage.

Jyrki Lahtonen
  • 133,153
  • I don't know...maybe my answer is wrong. I wasn't looking at $\alpha^2$, I was just looking at the term $s_is_j\alpha$. –  Jul 23 '13 at 21:07
  • There is a problem with that step, sure. You could try the induction on the $x^{k-1}$ term. You made it so far that it would be a shame to abandon this general approach. – Jyrki Lahtonen Jul 23 '13 at 21:13
4

(This is an alternate approach that doesn't use the outlined argument.)

In general, if $h(x)|x^{p^n}-x$ in $\mathbb F_p[x]$ then $h(x)$ is the product of distinct prime polynomials, each of degree equal to a factor of $n$.

If you know this, a fun way to prove this theorem (but skipping the outlined steps in the question) is to prove that $x^p-x-1$ is a divisor of $x^{p^p}-x$. Since $x^p-x-1$ has no roots in $\mathbb F_p$, this would mean that it has to be a prime polynomial.

You show that it is a factor by noting that:

$$x^{p^p}-x = \sum_{k=0}^{p-1} \left(x^{p^{k+1}}-x^{p^k}\right) = \sum_{k=0}^{p-1} (x^p-x)^{p^k} = \\\sum_{k=0}^{p-1}((x^p-x-1)+1)^{p^k} \equiv 0\pmod{x^p-x-1}$$

Thomas Andrews
  • 177,126
-1

Another approach:

Suppose $x^p-x -1 = gh$ for $g, h \in \mathbb F_p[x]$ with degrees $\geq 1$. Then taking a (formal) derivative, we have $$-1 = g'h + gh'$$ since $px^{p-1} = 0$ in $\mathbb F_p[x]$. Note that $g'$ and $h'$ cannot both be $0$, otherwise, $-1 = 0$. This implies the degree of the right hand side is at least $1$ (verify!). But the degree of $-1$ is $0$. Contradiction.

cat
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  • An interesting idea! I really want it to work! But, you lost me at the verification step. How do you arrive at this contradiction? Observe that $x^p-x-1$ does factor over the field $\Bbb{F}_{p^p}$, it splits into linear factors over that field. So any argument needs to use a property specific to the field $\Bbb{F}_p$. Characteristic $p$ is not enough, because the polynomial factors over a bigger field with the same characteristic. – Jyrki Lahtonen Sep 18 '16 at 15:27
  • Your first equation may be wrong as there may be cancellations (and there will be cancellations in extension field situation I described). The formula $$\deg (a(x)+b(x))=\max{\deg a(x),\deg b(x)}$$ only holds, when the polynomials $a(x)$ and $b(x)$ have distinct degrees. Otherwise their leading coefficients may cancel each other. – Jyrki Lahtonen Sep 18 '16 at 18:22
  • To address your second claim, the argument I state uses the fact that for any polynomial $p(x) \in \mathbb F_p[x]$, its derivative $p'(x)$ is also in $\mathbb F_p[x]$. So if I assume $x^p-x -1 = gh$ in $\mathbb F_p[x]$, then taking derivatives gives us an equation still n $\mathbb F_p[x]$. – cat Sep 18 '16 at 18:22
  • I see my error now, thank you. – cat Sep 18 '16 at 18:23
  • For example, $f(x)=x^2+x+1$ factors as $f(x)=(x+\alpha)(x+\alpha+1)$ over $\Bbb{F}_4$. Here $\alpha$ is zero of $f(x)$. With $g(x)=x+\alpha$, $h(x)=x+\alpha+1$ we have, indeed, $$g'h+gh'=1\cdot(x+\alpha+1)+(x+\alpha)\cdot1=1=-1.$$ – Jyrki Lahtonen Sep 18 '16 at 18:25
  • Hmm. Both $\deg gh'$ and $\deg g'h$ are equal to $r+s-1$. To reiterate: ANY proof of this result cannot work for all fields of characteristic $p$ because the claim is then false. If the argument does not use the fact that the field is specifically $\Bbb{F}_p$ (rather than its extension field) it will be wrong. – Jyrki Lahtonen Sep 18 '16 at 19:58
  • oh doh, my bad. And I see what you are saying, thank you. – cat Sep 18 '16 at 20:02