I am going to take as an assumption that your statement, that when $a_m\neq 0$, all roots have multiplicity $ 1 $, is true. It certainly sounds correct, but I'm no expert on that part.
However, you can easily show the case for $a_m=0$ from that point, as follows:
Suppose that the smallest index that arises is $p^e$. That is,
$$
f(x) = \sum_{i=0}^{m-e} a_ix^{p^{m-i}}
$$
where $a_0=1$ and $a_e\neq 0$. Now, let $y=x^{p^e}$. This allows us to write
$$
f(x) = \sum_{i=0}^{m-e} a_i y^{p^{m-i}/p^e} = \sum_{i=0}^{m-e} a_i y^{p^{m-e-i}/p^e}
$$
Let $M=m-e$, so we have
$$
f(x) = \sum_{i=0}^M a_i y^{p^{M-i}} = g(y)
$$
Now, $g(y)$ has, from the assumed statement, $p^M$ roots of multiplicity $ 1 $.
Now is where we bring in the characteristic being $p$, which means that $(x+y)^p = x^p+y^p$. So we seek solutions to $x^p=a$. If $x_1^p=x_2^p=a$, then $x_1^p-x_2^p = (x_1-x_2)^p=0$, and so $x_1=x_2$. So $x^p-a=0$ has one root of multiplicity $p$, and by iterating this process, we find that $x^{p^e}-a=0$ has one root of multiplicity $p^e$.
But our roots of $g(y)=f(x)$ take the form $y=x^{p^e}=a$, so each root of $g(y)$ is associated uniquely with a root of $f(x)$, and has multiplicity $p^e$.
I am almost positive that your hunch is correct. The extra factor $x^2+1$ is there simply to make sure that we can think of the splitting field as an extension of $K$.
A convenient way of including $\sqrt{-1}$.
My copy of Jacobson's Basic Algebra I is in my office (IIRC published after Lectures in Abstract Algebra), so I cannot check whether he later edited the proof.
An alternative way of organizing the proof, based on exact same ideas, would be to take an irreducible polynomial $g(x)\in K[x]$. Then consider the polynomial $f(x)=g(x)\overline{g}(x)\in F[x]$, where $z\mapsto\overline{z}$ is the obvious $F$-automorphism of $K$. Then proceed along the same route:
- Let $L$ be the splitting field of $f$ over $F$.
- Because $f$ is separable $L/F$ is Galois. Let $G$ be the Galois group, and let $P\le G$ be a Sylow $2$-subgroup.
- Let $M$ be the fixed field of $P$. Because $M/F$ is simple and $[M:F]$ is odd, we can conclude that we must have $M=F$ and, consequently $G=P$.
- Let $P_m=\{1\}\unlhd P_{m-1}\unlhd\cdots\unlhd P_2\unlhd P_1\unlhd P_0=P$ be the decomposition series. By basic properties of $p$-groups $[P_{i-1}:P_i]=2$ for all $i$.
- The fixed field of $P_1$ is a quadratic extension of $F$, and the quadratic formula shows that it is isomorphic to $K$. So we can identify it with $K$.
- The earlier lemma showed that $K$ has no quadratic extensions so $P_2$ cannot exist, implying that $L\simeq_F K$.
The way the above outline reintroduces $K$ as the fixed field of $P_1$ is not very elegant. We should justify that this reintroduction doesn't meddle with the polynomial we started with! Proving that $[L:F]=2$ is one way, and there are probably alternative ways of making the desired conclusions, and I may have missed the simplest way. But having that extra factor $(x^2+1)$ takes care of such issues.
Best Answer
Let us fix a splitting field $K$ for $f$. Let $a \in K$ be a root of $f$. Then every root of $f$ is a root of $f(X+a)$, and $f$ is separable so $f|f(X+a)$. Since both polynomials are monic of the same degree, $f(X+a)=f$.
Now, consider $g(Y)=f(X+Y)-f(X) \in F(X)[Y]$. Then $g$ is monic with the degree of $f$ and if $y \in K$ is a root of $f$, then $g(y)=0$. As $f$ is separable, $f(Y)|g(Y)$, and thus $f(X+Y)=f(X)+f(Y)$.
So now, we show that a polynomial $f$ over a field $F$ of characteristic $p$ such that $f(X+Y)=f(X)+f(Y)$ must be a linear combination of the $X^{p^i}$.
Indeed, if it isn’t the case, we can subtract from $f$ a linear combination of the $X^{p^i}$ so that $f(X)=\alpha X^t+O(X^{t-1})$ with $\alpha \neq 0$ and $t$ not a power of $p$.
Then $0=f(X+Y)-f(X)-f(Y)=\alpha\sum_{k=1}^{t-1}{\binom{t}{k}X^kY^{t-k}}+\sum_{k=0}^{t-1}{O(X^kY^{t-1-k})}$. Thus for each $1 \leq k < t$, $p|\binom{t}{k}$. This forces $t$ to be a power of $p$.