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!
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.
Here are two generalizations:
What happens if we only ask for the roots of $f(x)$ to be closed under addition?
What happens if we only ask for the roots of $f(x)$ to be closed under multiplication?
We'll continue to ask for $f$ to be monic and for the roots to be distinct. Let $F$ be a splitting field, and let me write the base field as $K$ because I'll be using $k$ as an index.
The multiplication case turns out to be arguably less interesting so let's do that one first. First, $f(x)$ could have $0$ as a root, but that isn't so interesting, so let's assume WLOG that the roots are nonzero by dividing out by $x$ if necessary.
Next, if $\alpha \in F$ is a root then so is every power $\alpha^2, \alpha^3, \dots$ so in order for $f(x)$ to have finitely many roots, every root must have finite order, so the roots of $f(x)$ contain $1$ and are closed under inverses, hence form a finite subgroup of the multiplicative group of $F$ of order $n = \deg f$. By Lagrange's theorem every root satisfies $x^n = 1$, so $x^n - 1$ is a monic polynomial of degree $\deg f$ with the same roots as $f$, which gives $\boxed{ f(x) = x^n - 1 }$. Since we possibly divided by $x$ we can also have $\boxed{ f(x) = x^{n+1} - x }$. Either way, the nonzero roots of $f$ must be the $n^{th}$ roots of unity.
(The previous argument here used the fact that finite subgroups of the multiplicative group of a field are cyclic, which turns out to be overkill, if not circular. In fact this argument is the beginning of most (?) proofs of that result.
Now for addition. This one is fun and new to me! If $\alpha \in F$ is a root then so is every multiple $2 \alpha, 3 \alpha, \dots $ and these are all distinct unless $F$ (and therefore $K$) has positive characteristic $p$. Next, since the roots of $f(x)$ are closed under addition we must have $f(x + \alpha) = f(x)$ for all roots $\alpha$. Taking formal derivatives gives $f'(x + \alpha) = f'(x)$, so $f'(0) = f'(\alpha)$ for all roots of $f(x)$. But there are $\deg f$ of these and $f'(x) - f'(0)$ has degree $\deg f - 1$, so this is only possible if $f'(x) - f'(0) = 0$ and hence $f'(x)$ is a constant. This means $f(x)$ has the form
$$f(x) = g(x^p) + f'(0) x$$
for some $g(x) \in k[x]$. The translation condition $f(x + \alpha) = f(x)$ then gives
$$g(x^p + \alpha^p) + f'(0) (x + \alpha) = g(x^p) + f'(0) x$$
which gives $g(x^p + \alpha^p) - g(x^p) = f'(0) \alpha$, hence $g(y + \alpha^p) - g(y) = f'(0) \alpha$. Taking formal derivatives again gives $g'(y + \alpha^p) - g'(y) = 0$, and the same argument as above (we need to observe that the Frobenius is injective over a field so the $\alpha^p$ are still distinct) gives that $g'(x)$ is a constant, so $g$ can be written in the same form as $f$ above.
Iterating this argument we conclude that $f(x)$ is a "polynomial in Frobenius"
$$\boxed{ f(x) = \sum_{k \ge 0} f_k x^{p^k} }.$$
These are exactly the polynomials in characteristic $p$ which are additive in the sense that $f(x + y) = f(x) + f(y)$, so $f(x)$ defines an additive homomorphism $F \to F$ and its set of roots, which is the kernel of this additive homomorphism, is closed under addition as desired.
Now when we ask for the roots of $f(x)$ to be closed under both addition and multiplication we get the intersection of these results: $f(x)$ must both have the form $x^{n+1} - x$ (since $0$ must be a root) and be a polynomial in Frobenius, so the only possibility is $f(x) = x^{p^k} - x$. Of course this is a roundabout way to get the desired result and Alex's nice argument is more direct, but I like seeing the two generalizations and how they intersect.
Edit: Now that we know the conclusion of the additive case here is an alternative argument. Suppose the roots of $f(x)$ are closed under addition, and consider
$$g(y) = f(x + y) - f(x) \in F(x)[y]$$
regarded as a polynomial in $y$ with coefficients in $F(x)$. This is a monic polynomial of degree $\deg f$ with $\deg f$ roots, namely the roots $\alpha \in F$ of $f$. But this means that $g(y) = f(y)$! So $f(x + y) = f(x) + f(y)$ and $f$ is an additive polynomial.
This gives an alternate route where we proceed by classifying additive polynomials, which can be done without taking formal derivatives by just expanding all the powers out and comparing term-by-term (it's not as bad as it sounds); see e.g. this blog post, ctrl+F for "additive polynomial."
Edit #2: This also suggests a slightly different alternative approach to the multiplicative case where we avoid appealing to Lagrange's theorem. First, as above we can assume WLOG that $f(0) \neq 0$ by dividing by $x$ if necessary.
The roots of $f(x)$ are closed under multiplication iff $f(\alpha x) = f(x)$ for all roots $\alpha$; note that we don't need to normalize the leading terms because the two polynomials have the same roots and the same nonzero constant term. By comparing the leading terms, it follows that if $\deg f = n$ then $\alpha^n = 1$ (without Lagrange's theorem! This is a lightly disguised version of a nice proof of Lagrange's theorem for finite abelian groups), hence that $x^n - 1$ is a monic polynomial of the same degree as $f$ with the same roots as $f$, which gives $f(x) = x^n - 1$.
