Root Field of an Irreducible Bi-quadradratic Polynomial Charles Pinter A Book of Abstract Algebra chapter 31 problem C5:
Prove:
If $p(x) = x^4 + ax^2 + b$ is irreducible in F[x], then F[x]/$<p(x)>$ is the root field of p(x) over F.
Discussion:
As the hint in the back of the book states, F[x]/$<p(x)>$ = F(c) where c is a root of p(x).  The hint then goes on to rewrite p(x) as $(x^2 +a/2)^2 – (a^2/4 – b)$ (actually it has $(x^2 - a/2)^2 – (a^2/4 – b)$, but that’s a typo).  The hint then says to let $d_1$ be a root of $x^2 – [(a^2/4) – b]$ and $d_2$ be a root of $x^2 – (a/2 +d_1)$.  It says then that F($d_1$, $d_2$) = F(c) and that F($d_1$, $d_2$) contains all the roots of p(x).
$d_1 = {±\sqrt{a^{2}/4-b}}$
$d_2 = {±\sqrt{a/2±\sqrt{a^{2}/4-b}}}$
Note that $d_2$ isn’t necessarily a member of F[x].  I had considered that p(x) may have all its roots in F(c) because it has at least one root in the root field of $x^2 – (a/2 +d_1)$, but because $x^2 – (a/2 +d_1)$ isn't in F[x], that isn't applicable.
The roots of p(x) are ${±\sqrt{a/2±\sqrt{a^{2}/4-b}}}$, the same as $d_2$.
So if I select ${\sqrt{a/2+\sqrt{a^{2}/4-b}}}$ as c, then it’s clear that F($d_1$, $d_2$) = F(c).  Since $d_1 = d_2^2 – a/2$, the $d_1$ is extraneous, and F($d_1$, $d_2$) = F($d_2$) = F(c).
But for F(c) to be the root field, it would also have to contain  ${\sqrt{a/2-\sqrt{a^{2}/4-b}}}$.
Does there exist some set of $k_0$ to $k_n$ member of F such that $k_0 + k_1c + k_2c^2 + … + k_nc^n = {\sqrt{a/2-\sqrt{a^{2}/4-b}}}$?
The only members of F specified are 0, 1, a and b.  The roots of p(x) are specifically excluded.  There could be other members, but nothing I can assume.
Letting c = ${\sqrt{a/2+\sqrt{a^{2}/4-b}}}$ and d = ${\sqrt{a/2-\sqrt{a^{2}/4-b}}}$, then F(c, d) would contain all the roots of p(x).  In addition, F[x]/$<p(x)>$ is isomorphic to F(c) is isomorphic to F(d), but that’s not the same as F(c, d).
What am I missing here?
 A: I don't have the book with me; I assume that by root field the author means what is more commonly called a splitting field, i.e. the field extension obtained by adjoining all the roots of $p(x)$.
With this interpretation, the statement is false; $F(c)$ need not be the splitting field. As indicated in this answer, the statement will be false precisely when both $b$ and $b(a^2-4b)$ are not squares in $F$ (e.g. take $F = \mathbb{Q}$, $a=1, b=2$).
To see that this is the case, note that the field $F(c, d)$ contains the element $\sqrt{b} = cd$ and the element $\sqrt{a^2 - 4b} = c^2 - d^2$. Hence it also contains $\sqrt{b(a^2 - 4b)} = \sqrt{b}\sqrt{a^2 - 4b}$, whereby $F' = F(\sqrt{b(a^2-4b)})$ is a subfield of $F(c, d)$. But the polynomial $p(x)$ is still irreducible over $F'$, as $F'$ does not contain the square root of the discriminant $a^2 - 4b$. It follows that $F(c, d)$ has degree at least $4$ over $F'$, whereby $F(c, d)$ has degree at least $8$ over $F$. Thus we cannot have $F(c, d) = F(c)$.
