Theorem 5.1. from the book of Algebra by Aluffi

Question

The following is from the book of Algebra by Aluffi: My questions :

1- $\mathbb{F_p} \subset F$ is a field extension. $F$ includes $\mathbb{F_p}$ but $E$ doesn't include $\mathbb{F_p}$: Let $a \in E$ and $f \in \mathbb{F_p}$ then $fa \notin E$. So how $E=F$? (Obviously, $E \subset F$ and not equality.)

2- Number of roots in the splitting field of $f(x)$ with degree $d$ over $\mathbb{Z}$ or $\mathbb{C}$ is $d$. I haven't seen a proof that 'similar' holds for a finite field like $\mathbb{F_p}$: Is it true that the number of roots of $x^q − x$ over $\mathbb{F_p}$ is $q$ (proof)?

Answer 1

1. $E$ does include $\mathbb F_p$. Every element $x \in \mathbb F_p$ obeys $x^p = x$ (by Fermat's little theorem), and therefore, also obeys $x^q = x$ for $q = p^n$.

2. The polynomial $f(x)$ certainly factorises a product $(x_1 - a_1) \dots (x_n - a_n) $ with $a_1, \dots , a_n \in F$, but the question is whether the $a_i$ are distinct. For $f(x) = x^q - x$, the answer is yes: the fact that $(f(x), f'(x)) = 1$ implies that $f$ has no repeated roots.

Answer 2

1. Obviously, $E \subset F$, as you said. Also, $E$ is a field, as the proof shows. ($E$ definitely contains $\Bbb{F}_p$, since any $x\in \Bbb{F}_p$ satisfies $x^{p^d}=x$ for any $d \in \Bbb{N}$.) Now, $F$ is defined as the smallest field that splits $x^q-x$. Therefore, F is a subset of any field that splits $x^q-x$. Since $E$ is a field containing all the roots of $x^q-x$, this means that $F \subset E$. Thus, since $E \subset F$ and $F\subset E$, $E=F$.

2. The definition of splitting field of a polynomial says that the polynomial splits into linear factors in that field. Thus, in $F$, $x^q-x$ can be written as the product of a bunch of linear factors. Since multiplying $n$ linear factors leads to a polynomial of degree $n$, if we split $x^q-x$ into linear factors, then we will get $q$ linear factors. Also, each linear factor $x-a$ implies that $a$ is a root of $x^q-x$. Furthermore, since $x^q-x$ is separable, each root must be distinct. Therefore, $x^q-x$ has $q$ distinct roots, one for each of its linear factors.

Answer 3

1. The elements of $\mathbf F_p$ satisfy the equation $x^p=x$ (that's Fermat's theorem), hence by an easy induction $x^{p^k}=x$ for any $k\ge 1$. Taking $k=d$, you get the inclusion $\mathbf F_p\subset E$.
2. Over any field, a polynomial of degree $q$ has at most $q$ distinct roots, and has actually $q$ roots if you take into account the multiplicity of the roots. This has not nothing to do with the field being finite or infinite – it just relies on the fact that an element $ \alpha$ is a root of $f(x)$ if and only if $f(x)$ is divisible by $x-\alpha$, and the multiplicity of a roots is determined by the successive derivatives having or not $\alpha$ as a root.

Now, in the case of $f(x)=x^{p^d}-x$, one has $f'(x)=-1$ (we're in characteristic $p$), so all roots of $f(x)$ are simple, and it has exactly $p^d$ roots.