Let $K$ be a finite extension of $F$. $K$ is a splitting field over F iff every irreducible with a root in $K$ splits over $K[x]$
This question is taken from the third edition of Abstract Algebra by Dummit and Foote (chapter 13.5 question 5). Could someone please verify my attempt? This solution uses theorems 8 and 27 from chapter 13 of the text.
$\implies$
Since $K$ is a splitting field over $F$, there must be $f \in F[x]$ for which $K$ is the splitting field of $f$. Now, let $g \in F[x]$ be irreducible and have $\alpha_1 \in K$ as a root. Then let $\beta_2$ be another root of $g$, then $F(\alpha_1) \simeq F(\beta_2)$ by the homomorphism $\psi$ which takes $\alpha_1$ to $\beta_2$ (theorem 8). Note that this isomorphism, when restricted to $F$, is the identity. Now, consider the splitting field of $f$ over $F(\alpha_1)$ - which is $K$ since $K$ is a superset of $F(\alpha_1)$ - and $F(\beta_2)$, which we will call $E_2$. Now I will paraphrase theorem 27 in the text:
If $F' \simeq F$ and $f \in F[x]$ and $f'$ is the associated polynomial in $F'[x]$ of $f$, then the splitting field $E$, and $E'$ of $f$ and $f'$ respectively are isomorphic.
Using this theorem, we deduce that $K \simeq E_2$ and so the element in $K$ which is associated with $\beta_2$ in $K$, which we will call $\alpha_2$, must be a root of $g$ since $g(x) = (x-\beta_2)h(x)$ in $F(\beta_2)$ for some $h \in F(\beta_2)[x]$.
Now let us define $l()t(x)$ for any polynomial $t$ and homomorphism $l$ to be the polynomial obtained by $l$ evaluating all of the coefficients of $t$.
Now since $g \in F[x]$ and $\psi$ fixes all elements in $F$, we must have that $g(x) = \psi()g(x) = (x- \alpha_2)\psi()h(x)$. We see that $\alpha_2$ is a root of $g$ since $g(\alpha_2) = \psi()g(\alpha_2) = (\alpha_2 - \alpha_2)\psi()h(x)$ = 0. If $n = deg(g)$, we may continue this construction to find the rest of the roots of $g$, $\alpha_3 ,..., \alpha_n$, in $K$. Since $K$ contains $deg(g)$ roots of $g$, the polynomial must split in $K$.
$\impliedby$
By assumption $[K:F] = n < \infty$. Since finite extensions are algebraic, take $a_1 \in K\backslash F$ which is algebraic over F and let $m_1$ be its minimal polynomial. Consider $F(a_1)$ which is a subset of $K$. If $[K:F(a_1)] = 1$, then $m_1$ splits by hypothesis since it has a root in $K$ and also since $K = F(a_1)$ , any proper subset of $K$ would deprive $m_1$ of $a_1$, by the definition of $F(a_1)$, and so $K$ is a splitting field. If $[K:F(a_1)] > 1$, then take $a_2 \in K \backslash F(a_1)$ and do the same procedure again. Surely each time $[K:F(a_{i + 1})] < [K:F(a_{i})]$ so we may continue this process and terminate it as we have a descending chain of positive integers. Then, $F(a_1, a_2, ..., a_m) = K$ and the product of the minimal polynomials $a_1, a_2, ..., a_m$ splits over $K$ as each individual minimal polynomial splits and again if we consider any proper subset of $K$, then we'd remove one of the $a_i$s and remove a root of the product of polynomials so $K$ is a splitting field.
Thank you!