Determining when two biquadratic polynomials generate the same field Consider the family of monic biquadratic polynomials given by $f_{a,b}(x) = x^4 + 2ax^2 + b$ with $a,b$ integers. Let $K_{a,b}$ denote the isomorphism class of quartic fields obtained by adjoining any one of the roots of $f_{a,b}$. Is there a relatively nice list of criteria to decide whether $K_{a,b}$ is isomorphic to $K_{a',b'}$ when $(a,b) \ne (a',b')$? 
 A: The necessary and sufficient conditions are as follows:
1) Condition on the common quadratic subfield (as Watson Ladd's answer):
there exists $c\in\Bbb Q$ such that
$${a'}^2-b'=c^2(a^2-b)\;.$$
2) Condition on the discriminant: the polynomial discriminant being $256 b(a^2-b)^2$ we need for some $d\in\Bbb Q$
$$bb'=d^2$$
3) Standard algebraic computation in the field gives the additional
condition for some $e\in\Bbb Q$
$$\dfrac{aa'+c(a^2-b)\pm d}{2}=e^2$$
for a suitable sign $\pm$.
Example: $b=b'=1$, $a=2$, $a'=26$: $c=15$, $d=1$, $e=7$.
A: The Galois group of the splitting field is D4, the symmetries of a square. D4 has a unique normal cyclic subgroup of order 4, the rotations. Therefore we can write the Galois closure as a unique C4 extension of a quadratic field over $\mathbb{Q}$.
The quadratic field will be classified by its discriminant, which should be some function in $a$ and $b$. The C4 extension will be determined by class field theory. If we work over $\mathbb{Q}(i)$ instead the C4 extension will be represented by an element that can be written down in terms of $a$ and $b$.
The criterion for isomorphism then becomes the discriminant and the element being equal.
