Let $K$ be a field, $\alpha\in\bar{K}$, and $L/K$ a finite extension. How can we determine whether $\alpha\in L$, preferably in as much generality as possible?


Of course, there may be special cases where this is easy, e.g. $K\subset\mathbb{R}$ and $\alpha\in\mathbb{C}\setminus\mathbb{R}$. Another trick, using the field trace, is the described in [exercise 16 of Chapter 2](http://books.google.com/books?id=qMfGgJPCnVMC&lpg=PP1&dq=related%253AISBN0387902791&pg=PA41#v=onepage&q&f=false) of Marcus's *Number Fields*, though as far as I can tell this only works for the special case of radicals (since their traces are always 0).


There's only two general approaches I can think of at the moment: checking whether $\deg(\alpha)\mid [L:K]$, though this may not suffice; or somehow finding the minimal polynomial of $\alpha$ over $L$ (including proving that it is irreducible), which will be of degree 1 iff $\alpha\in L$ (not entirely sure how one would do this).


I'm wondering because I recently had the following messy situation: $K$ is the splitting field of a cubic $g\in\mathbb{Q}(t)[x]$, having $[K:\mathbb{Q}(t)]=3$, and $f\_1$, $f\_2\in K[x]$ are two cubics with splitting fields, $L\_1$ and $L\_2$ respectively, having $[L\_1:K]=[L\_2:K]=3$, and I wanted to know whether $L\_1=L\_2$. It would suffice to show $f\_1$ has a root in $L\_2$ or vice versa (since $L\_1=K($any root of $f\_1)$ and $L\_2=K$(any root of $f\_2)$, which led me to my question. Ultimately (and I still want to double-check my answer), I found that $L\_1=L\_2$, but I depended heavily on the specific properties of my $f\_1$, $f\_2$, and $g$.



Methods of showing an element is / is not in a field
Zev, I think in complete generality there is no good answer to the question. Just think about how to decide if a cubic is reducible. In principle it's easy: find a root! But unless the field has some special structure that makes it feasible to search for roots this may be easier said than done. This can be done on finite fields or ${\mathbf Q}$ (or other quotient rings of known basic UFDs), but on a totally general field what are you going to do?


Here are a few thoughts, but notice in each case they depend on knowing something meaty about the fields, so they're not "general" methods. 


1. If you can find a known Galois extension containing your two fields, then check if they have the same fixed subgroup. In your case it may be circular to "use" Galois theory to decide if $L\_1 = L\_2$ since figuring out the Galois group of $L\_1L\_2$ over $K$ requires you already know how much $L\_1$ and $L\_2$ overlap. (A useful example: if I wanted to show ${\mathbf Q}(3^{1/8})$ and ${\mathbf Q}(48^{1/8})$ are not isomorphic then I know that they both lie in the field ${\mathbf Q}(3^{1/8},\zeta\_8)$, and I can compute its Galois group over ${\mathbf Q}$ and then compute the subgroups corresponding to those original two fields. The subgroups turn out to be nonconjugate, hence the two fields are not isomorphic.)
2. Since your fields are coming to us (well, to you) as extensions of ${\mathbf Q}[t]$, exploit possible ring-theoretic properties related to ${\mathbf Q}[t]$ and its integral extensions in your fields, not just plain field theory. In other words, think instead about comparing the integral closures of ${\mathbf Q}[t]$ inside your two fields instead of the fields themselves. I will go out on a limb here and assume your polynomials have coefficients that are integral over ${\mathbf Q}[t]$. Then splitting properties of irreducibles from ${\mathbf Q}[t]$ in the two integral closures might quickly tell them apart. For comparison, if you want to distinguish two finite extensions of ${\mathbf Q}$, 
which have the same degree and the same number of real and complex embeddings, the next thing to look at is invariants connected to the integral closure of ${\mathbf Z}$ in the two fields: start comparing the splitting properties of 2, 3, 5, and so on. Usually 
when the two fields are not isomorphic you'll quickly find a prime which splits in the two rings of integers in different ways (practically, this means minimal polynomials cutting out the two fields will factor in different ways mod $p$, and separably, for some prime $p$ pretty quickly). This could be described also in terms of $p$-adic embeddings, which has a parallel in your setting of extensions of ${\mathbf Q}(t)$, e.g., maybe one the cubics has a root in ${\mathbf Q}((1/t))$ (assuming $K$ itself can be stuffed in there) and the other does not.
3. Exploit other structures on your specific fields. For example, ${\mathbf Q}(t)$ has a nontrivial derivation on it, which will extend to the larger finite extensions of it. 
If there were an intrinsic property of derivations which is not the same on the two fields, then they're not isomorphic.


In your case the two fields turned out to be the same. I don't know how you're going to succeed in proving that without directly showing they're equal (assuming it is circular to use Galois theory in this instance). All this invariant business will tell stuff apart, so I suppose if the discriminants are equal and then the first 30 irreducibles in ${\mathbf Q}[t]$ which you can write down turn out to factor in the same way in the integral closure of ${\mathbf Q}[t]$ in the two fields (yeah, like that'll be easy to do...) then you might suspect the fields are equal and should switch gears to try to show equality instead of nonequality. 


