Is it possible to recover the degree of a field extension from a list of elements and the ground field?

Question

I'm interested to know if there is anything known about recovering the degree of a field extension, $E/k$, given $E=k(\alpha_1,\ldots, \alpha_n)$ (here I'm assuming that the extension is of finite degree. Obviously there are some silly examples anyone could eyeball, like $\sqrt[m]{d}$ for $n=1$ and $\alpha_1=d\in k$ or the case $E=\mathbb{F}_q$ and $k=\mathbb{F}_p$ where we can just divide orders (intermediate fields are clearly equally trivial). Where, to give the necessary bit of care, we assume this is a nontrivial extension. If $E/k$ is Galois and we can appeal to other bits of theory, we might also get the degree by calculation of the Galois group('s order). Is there anything known about more general extensions? It is conceivable given that $E/k$ is an extension of algebraic number fields that the theory of ideals might give an insight, especially given the (IMO) rather fascinating fact that $\mathcal{O}_E$ is finitely generated as a $\mathbb{Z}$-module is an equivalent statement, and using the machinery of algebraic number theory, or some other extra structure, but I'm principally concerned with the more general theory if any exists or if just anything is known about this problem.

Answer 1

To put this one to rest, I will answer the more precise question that, after much prodding, we got Adam to formulate in the comments. I am merely paraphrasing a comment of Qiaochu.

If you are given the $\alpha_i$ as roots of irreducible polynomials, then the degree is not a function of the $\alpha_i$. Of course, when you adjoin only one root, the degree of the extension is just the degree of the minimal polynomial. But as soon as you adjoin two roots, you cannot recover the degree. Linear independence over the base field doesn't help either: let $k=\mathbb{Q}$, let $\alpha_1$, $\alpha_2$ be two distinct roots of $x^9-2$. Then they can generate $\mathbb{Q}(\sqrt[9]{2},\mu_3)$ or $\mathbb{Q}(\sqrt[9]{2},\mu_9)$ for some 9-th roots $\sqrt[9]{2}$ of 2, depending on whether $\alpha_1$ and $\alpha_2$ differ by a 3-rd or by a 9-th root of unity. Accordingly, the degree will be either 18 or 54. In either case, the roots will be linearly independent over $\mathbb{Q}$, so they satisfy your conditions.

Adjoining roots of distinct polynomials won't help either, since you can just take some other element of the top field whose minimal polynomial has some roots over the bottom field. Now, if instead you adjoin roots of polynomials, whose splitting fields are disjoint over the base field, then the degree is just the product of the degrees of the minimal polynomials.