@Wojowu My mistake. I thought you meant (3) follows from (1), but you actually said it follows from (2), which is right.

@Wojowu You are right, (3) is superfluous, but it doesn't always follow from (1). If $K={\mathbb F}_q$, where $q$ is a prime power, $q\equiv 3\pmod 4$, then $-1$ is not a square in $K$ so for every $a\in K^*$ exactly one of $\pm a$ is a square. If $q\equiv 1$ then for half of the values of $a\in K^*$ both $\pm a$ are squares and for the other half neither of them is, so (1) fails. In all cases though, (3) follows from (2). If $P=X^q-X+1$, with odd degree $q$, then for every $a\in K$ we have $P(a)=1$ so $P$ has no roots in $K$ so (2) fails.

@Wojowu I forgot to mention that $F$ is not of characteristic $2$. In characteristic $2$ we have equations of the type $X^2+X+a=0$, which have roots in the so called Artin-Schreier extensions.

We have the following statement: Every polynomial over a field $K$ has a root in $K(\sqrt{-1})$. Then FTA says the statement is true for $K=\mathbb R$. Obviously, it's not true for every field $K$, we need some properties. A minimal set of properties would be: (1) $X^4-a^2$ has a root in $K$ for every $a\in K$, (2) Every polynomial in $K[X]$ of odd degree has a root in $K$, (3) $K$ is not finite. (Note that (1) simply states that either $a$ or $-a$ has a square root in $K$.)

Completely without analysis you can't. Otherwise you could prove that every equation with rational coefficients has solutions in ${\mathbb Q}(i)$. The very definition of $\mathbb R$ involves some analysis. The one "most algebraic" proof I know uses two facts that require some analysis: (1) If $a>0$ then the equation $X^2-a=0$ has a real root. (2) Every polynomial of odd degree has a real root. If we write the degree of $P\in{\mathbb R}[X]$ as $n=2^km$, with $m$ odd, then the proof is done by induction on $k$.

The general case of Cayley-Hamilton can actually be proved by "horrendous calculations". (Only done in a smart way.) Basicaly, if $A=(a_{i,j}$ and $P_A(X)=c_nX^n+\cdots +c_0$ , then you write each $c_i$ and the $n^2$ entries of each $A^i$ as explicit polynomials in $a_{i,j}$ and you check that all $n^2$ entries of $P_A(A)=c_nA^n+\cdots +c_0I_n$ are $0$. It's just playing with sums of monomials and proving that all of them reduce at the end. The proof in the 3-4/2014 issue of Gazeta Matematica, Seria A, pag. 32-36. Online at: ssmr.ro/gazeta/gma/2014/gma3-4-2014-continut.pdf

Thank you for your answer. I'll have to look into this Poincaré-Birkhoff-Witt theorem. It's a subject I'm not familiar with. Do you believe it could provide a short proof of the result? My proof is quite technical. I only used elementary properties of the tensor products. At some point I also use the fact that the symmetric group $S_n$ is generated by the transpositions $\sigma_i=(i,i+1)$ with $1\leq i\leq n-1$ and the relations between the generators are $\sigma_i^2=1$, $\sigma_i\sigma_j=\sigma_j\sigma_i$ if $|i-j|\geq 2$ and $(\sigma_i\sigma_{i+1})^3=1$ for $1\leq i\leq n-2$.

* If you saw this or something equivalent somewhere, then please let me know. (By mistake, I wrote "I you saw...".)

I edited my post with the precise definition of the bimodule $M(V)$ and of the short exact sequence $0\to M(V)\to T(V)\to S(V)\to 0$. I you saw this or something equivalent somewhere, then please let me know.