Easy criteria to determine isomorphism of fields? Let $K$ be a field and $f,g$ irreducible polynomials in $K[X]$, is there a nice iff condition for $K[X]/(f)\cong K[X]/(g)$?
($\cong$ denotes an isomorphism that is the identity on restriction to $K$).
Thoughts: It is sufficient that they are $K^\times$ multiples of each other. I'd hoped this was necessary but it isn't as $\mathbb{Q}[X]/(X^2-2)\cong\mathbb{Q}(\sqrt2)=\mathbb{Q}(\sqrt2+1)\cong\mathbb{Q}[X]/(X^2-2X-1)$ with the middle two fields viewed as subfields of $\mathbb{C}$. It is necessary that they have the same degree. Please let me know if there are any other simple necessary conditions. Thanks!
 A: If $f,g$ are quadratic polynomials (and $K$ is any field of characteristic not 2), then by the quadratic formula the isomorphism classes are classified by the discriminant "$b^2 - 4ac$" (if $f = ax^2 + bx + c$), modulo squares. Ie, the isomorphism classes are in bijection with $K^\times/(K^\times)^2$.
In any other situation things become a lot more difficult.
If $K$ is $\mathbb{Q}$, then $\mathbb{Q}[X]/(f)\cong\mathbb{Q}[X]/(g)$ if and only if the action of $G_\mathbb{Q} := \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on the roots of $f$ is isomorphic to the action of $G_\mathbb{Q}$ on the roots of $g$. Ie, if $X_f,X_g$ denote the sets of roots of $f,g$, then $\mathbb{Q}[X]/(f)\cong\mathbb{Q}[X]/(g)$ if and only if there is a bijection $\phi : X_f\stackrel{\sim}{\rightarrow}X_g$ such that $\phi(\sigma x) = \sigma\phi(x)$ for all $x\in X_f$, $\sigma\in G_\mathbb{Q}$. 
This is a consequence of the Galois correspondence, which says that the association sending any finite extension $L := \mathbb{Q}[X]/(h)$ of $\mathbb{Q}$ (with $h$ irreducible) to $X_h$ (as a set with $G_\mathbb{Q}$-action) gives an equivalence of categories between the category of finite field extensions of $\mathbb{Q}$ and the category of finite sets with a transitive $G_\mathbb{Q}$-action.
The result follows from the fact that two finite extensions of $\mathbb{Q}$ are isomorphic as fields if and only if they are isomorphic as extensions of $\mathbb{Q}$ (any abstract isomorphism between them must fix their prime subfields).
The same result will also be true if $K$ is replaced by any $\mathbb{F}_p$ (though this situation is trivial since finite extensions of $\mathbb{F}_p$ are uniquely determined by degree). With suitable modifications the result is also true when $K$ is any finite extension of $\mathbb{Q}$, and with more care, even true when $K$ is an algebraic extension of $\mathbb{Q}$.
If $K$ is not an algebraic extension over its prime subfield, then things can get weird, as we move into the world of arithmetic geometry. For example, for any field $k$, you can set $K := k(t)$, then $K = K[X]/(X) = k(t)$ is isomorphic to $K[X]/(X^2-t)\cong k(\sqrt{t})$.
If $K$ is finite over $\mathbb{Q}$, then in certain cases you may also be able to use class field theory.
Though, I should mention that in practice it's rare that you would care about isomorphisms between fields (as abstract fields). You generally will want to restrict yourself to "nice" extensions of a fixed base field, in which case a number of the conditions above can be relaxed.
