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 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 $g$.

  • $\begingroup$ If I were trying to test whether two irreducible cubics $f_1$ and $f_2$ in $K[x]$ give isomorphic cubic extensions of $K$, I think the first thing I'd do is see if their discriminants are the same (up to multiplication by squares in $K$). But of course a yes answer is inconclusive. $\endgroup$ Jul 1, 2010 at 2:31
  • 2
    $\begingroup$ Zev, that trick in Marcus is due to lack of technique which he has at that point in the book. The better way to solve that problem is to use splitting properties: an obvious prime ramifies in the quadratic field but not in the quartic field, hence you can't embed the quadratic field in the quartic field. I remember thinking that trace trick in Marcus was interesting when I first did that exercise, but in the long run it's kind of a dead end. The trace is still crucial for other things (e.g., discriminant, different), so I am not saying the trace function is a dead end! $\endgroup$
    – KConrad
    Jul 1, 2010 at 3:31

3 Answers 3


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.


See section 5.4 "The Subfield Problem and Applications" in Henri Cohen's book "A Course in Computational Algebraic Number Theory". This is an excellent reference for computational algebraic number theory and the first place one should look when investigating such topics.


As long as you are considering low-degree extensions, you can probably do your calculations of field inclusions in a straightforward way using the primary decomposition functions in a computer algebra system.

For example, in the particular question that you mentioned, you assert that $K = \mathbf{Q}(t)[x]/(g(x))$ and $$ L_1 = K[u]/(f_1(u)) = \mathbf{Q}(t)[x,u]/(g(x),f_1(u)) $$ are fields and ask whether $f_2(v)$ is irreducible in $L_1[v]$. In other words, you want to know whether $$ \mathbf{Q}(t)[x,u,v]/(g(x),f_1(u),f_2(v)) $$ is a field.

You can get the answer by asking a CAS to find the primary decomposition of the ideal $(g(x),f_1(u),f_2(v))$ in the ring $\mathbf{Q}(t)[x,u,v]$.

For this kind of computation, you might try Singular, which is a free system that has many tools for primary decomposition. Since your ideal is 0-dimensional and the generators form a Groebner basis (for many term orderings), you may be able to use some quick routines such as Singular's zerodec. For what it's worth, you also know already that your ideal is radical, and so it suffices to ask a computer for the minimal associated primes.

If the calculations over $\mathbf{Q}(t)$ are too slow, you may try clearing denominators and computing associated primes over $\mathbf{Q}[t]$. That may be faster because of better implementations of basic algorithms. You can recover the associated primes over $\mathbf{Q}(t)$ from those over $\mathbf{Q}[t]$ by discarding the associated primes that contain polynomials in $t$ of positive degree.


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