Roots of a monic polynomial $f(X) \in k[X]$ in splitting field distinct, form a field, then char $k = p$ and $f(X) = X^{p^n} - X$ This is Lang's Algebra 3rd edition chapter 5 exercise 13
If the roots of a monic polynomial $f(X) \in k[X]$ in some splitting field are distinct, and form a field, then char $k = p$ and $f(X) = X^{p^n} - X$ for some $n \geq 1$.
I know that $f(x)= (x - \alpha_1)\cdots(x - \alpha_n)$ for $\alpha_i \in F$ where $F$ is the splitting field. Not sure why it must have finite characteristics or why $f$ takes on the form given in the problem. Some help would be greatly appreciated!
 A: This is a nice exercise!
Let $F$ be a splitting field for $f$. It's a little vague to say that the roots of $f$ form a field, so I will make things a bit more precise towards what I believe Lang intended: if $K \subset F$ denotes the set of roots of $F$, then $K$ is a subfield of $F$, meaning that the addition and multiplication operations on $K$ are consistent with those of $F$.
First, note that $K$ is finite, since every polynomial over a field has finitely many roots. Hence, $K$ has characteristic $p$ for some prime $p$, as every finite field has prime characteristic. Since $K$ is a subfield of $F$, we have $\operatorname{char}(F) = \operatorname{char}(K) = p$. Similarly, because $k$ is a subfield of $F$, we have $\operatorname{char}(k) = \operatorname{char}(F) = p$. This establishes the first claim.
Next, note that $|K| = p^{n}$ for some integer $n \geq 1$. Indeed, since $\operatorname{char}(K) = p$, $K$ contains $\mathbb{Z}/p\mathbb{Z}$ as its prime subfield, and so in particular is a $\mathbb{Z}/p\mathbb{Z}$-vector space. Its dimension over $\mathbb{Z}/p\mathbb{Z}$ must be finite, because $K$ is itself finite; if the dimension of $K$ as a $\mathbb{Z}/p\mathbb{Z}$-vector space is $n$, we must have $|K| = p^{n}$, as desired.
Finally, observe that every element of $K$ is a root of $g(X) = X^{p^{n}}-X$. Clearly, $0$ is a root of $g$. Since $K^{\times}$ is a group of order $p^{n}-1$, every element of $K^{\times}$ has order dividing $p^{n}-1$ by Lagrange. In particular, every element $\alpha \in K^{\times}$ satisfies $\alpha^{p^{n}-1} = 1$, and so satisfies $\alpha^{p^{n}} = \alpha$, which shows that $\alpha$ is a root of $g$. We have exhibited $p^{n}$ distinct roots of $g$, and so it follows that $g(X) = \prod_{\alpha \in K} (X-\alpha)$. But this is precisely the factorization of $f$ in $F$, since $f$ is monic and each root has multiplicity one, so we have $f = g$. This establishes the second claim.
