This question originates from Pinter's Abstract Algebra Chapter 31 Exercise C5.
Prove that if $p(x) = x^4 +ax^2 + b$ is irreducible in $F[x]$, then $F[x]/\langle p(x)\rangle$ is the root field of $p(x)$ over $F$.
Let $c$ be a root of $p(x)$, so $\displaystyle F(c)\cong \frac{F[x]}{\langle p(x)\rangle}$ where $[F(c):F]=4$. \begin{align*} p(x) &= x^4 + ax^2 +b \\ &= \left(x^2+\frac{a}{2}\right)^2 + \left(b-\frac{a^2}{4}\right) \end{align*}
Let $\displaystyle y=x^2+\frac{a}{2}$, \begin{align*} p(y) &= y^2 + \left(b-\frac{a^2}{4}\right) \end{align*}
Let $d_1, d_2$ be the roots of $p(y)$, so $[F(d_1):F]=2$. As every extension of degree 2 is a root field, $d_1,d_2\in F(d_1) = F(d_2)$. Note the values of $x$ that satisfies $\displaystyle d_1=x^2+\frac{a}{2}$ are exactly the roots of the irreducible polynomial $\displaystyle x^2+\frac{a}{2}-d_1$ over $F(d_1)$.
Let $e_{11},e_{12}$ be the roots of $\displaystyle x^2+\frac{a}{2}-d_1$, so $[F(d_1,e_{11}):F(d_1)]=2$. This implies $e_{11},e_{12}\in F(d_1,e_{11})$ is a root field of $\displaystyle x^2+\frac{a}{2}-d_1$ over $F(d_1)$. Similarly, let $e_{21},e_{22}$ be the roots of $\displaystyle x^2+\frac{a}{2}-d_2$, so $e_{21},e_{22}\in F(d_2,e_{21}) = F(d_2,e_{22})$.
Note
- $F(d_1,e_{11})$ and $F(d_1,e_{21})$ are both finite extensions of $F(d_1)$;
- $F(d_1,e_{11})$ is a root field;
- We can construct an isomorphism $h: F(d_1,e_{11})\rightarrow F(d_1,e_{21})$ which fixes $F(d_1)$ with $h(e_{11})=e_{21}$ and $h(e_{12})=e_{22}$. This implies $F(d_1,e_{11}) = F(d_1,e_{21})$, which therefore contains all the roots of $p(x)$ over $F$.
Note $F(d_1,e_{11})$ is a minimal field extension of $F$, and $c\in F(d_1,e_{11})$. This implies $F(d_1,e_{11})\subseteq F(c)$. Conversely, $F(c)\subseteq F(d_1,e_{11})$. Hence $F(c) = F(d_1,e_{11})$, which is the root field of $p(x)$ isomorphic to $\displaystyle \frac{F[x]}{\langle p(x)\rangle}$.
Does this look reasonable?