Existence of splitting field of any polynomial in $F[X]$

Question

So I was reading some proofs from different books of existence of splitting field and I don't know why to bother so much about this. I think this can be done trivially.

Let $f(x)$ be a polynomial from $F[x]$. Suppose $a_1$, $a_2 ,\ldots,a_n$ be the roots of $f(x)$ which are outside F. Now if I take $ F(a_1,a_2,\ldots,a_n)$, this is the smallest field containing $F$ and the roots, hence $f(x)$ splits over this field and this $F(a_1,a_2,\ldots,a_n)$ is the smallest one containing $F$. Hence any polynomial has a splitting field.

Am I doing some mistake?

Answer 1

You are not making any mistake, but rather, not giving enough detail: You write $F(a_1, a_2, \ldots, a_n)$, but you do not define it: note that at that point in the proof one does not know whether there is a field extension of $F$ "containing the $a_k$". In fact, even the $a_k$ are not really being defined. The essence of the proof that a splitting field exists, is to make sense of "the roots of $f$ outside $F$", and the expression $F(a_1, \ldots, a_n)$.

Answer 2

I think Wikipedia is pretty good on this.

Factor $f$ into irreducible factors, and form the quotient by one of those nonlinear factors, thereby getting an extension field with a new root of $f$. You will need to do this at most $n$ times, where $n=\operatorname {deg}f$.