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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.

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    $\begingroup$ Silly examples anyone could eyeball, huh? What do you think about d = -4, m = 4? $\endgroup$ Nov 7, 2010 at 21:41
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    $\begingroup$ I thought I had a valid formalization of this problem, but I did not. How are we given k, how are we given E, and how are we given the alpha_i? $\endgroup$ Nov 8, 2010 at 1:35
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    $\begingroup$ This question doesn't seem well-posed to me. Qiaochu is trying to prod you towards stating it properly. What information are you given and in what format? What do you mean by "given $E=k(\alpha_1,\ldots\alpha_n)$"? $\endgroup$
    – Alex B.
    Nov 8, 2010 at 2:39
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    $\begingroup$ @Adam: the degree of E isn't uniquely determined by radical expressions (since there is no canonical way to distinguish which root you're taking). For example [Q(a, b) : Q] where a, b both satisfy a^3 = 2 could be equal to 3 or 6 depending on whether a and b are equal. For the special case of k a number field we can distinguish elements alpha_i by explicitly identifying them in C but in general there's no canonical way to do this. So, as Alex says, the problem as it stands is not well-posed. $\endgroup$ Nov 8, 2010 at 16:18
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    $\begingroup$ Adam, frankly, your unwillingness to think about the comments and to inspect your assumptions and your thinking critically is not only detrimental to your aim of getting a good answer (and more generally to doing mathematics), but is also getting frustrating for those who are trying to help. Surely, only a few seconds' thought would have revealed to you, that Qiaochu's example generalises in a way that you haven't ruled out: two linearly independent roots of $x^9−2$ can either generate a degree 18 or a degree 54 extension of Q, depending on whether they differ by a 3rd or a 9th root of 1. $\endgroup$
    – Alex B.
    Nov 9, 2010 at 1:58

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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.

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  • $\begingroup$ But is it easy to calculate the polynomial? That was my original thought on the subject, but it didn't seem feasible to calculate in general. $\endgroup$ Nov 9, 2010 at 2:30
  • $\begingroup$ I was trying to be too general/ambitious in my original post it seems. That really hurt the communication of what I was really looking for, and I apologize for the frustration this has caused the patient posters. I've thought more about what I'm looking for, and I think it's this: Given $\{\alpha_i\}_{i=1}^n\subseteq\mathbb{C}$ which are $\mathbb{Q}$-linearly and algebraically independent, and for which each $\alpha_i$ is algebraic, can we recover $[E:\mathbb{Q}]$ for $E=\mathbb{Q}(\alpha_1,\ldots,\alpha_n)$ in an algorithmic [and efficient if possible] way? $\endgroup$ Nov 9, 2010 at 3:00
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    $\begingroup$ Adam, if the $\alpha_i$ are algebraic, then they aren't algebraically independent. More rethinking required, I think. $\endgroup$ Nov 9, 2010 at 3:05
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    $\begingroup$ @Adam: there are at least two problems with the reformulation in your last comment. First, as Gerry says, you can't possibly want the $\alpha_i$'s to be algebraically independent over $\mathbb{Q}$, for then $[E:\mathbb{Q}]$ is infinite. Second, you say "given" the set, but you don't specify how you're being given it. If it's being given in a reasonable way -- e.g., by minpolys for the $\alpha_i$'s together with enough approximate information to distinguish $\alpha_i$ from other roots of its minpoly, then sure, standard computer algebra packages already have such algorithms implemented. $\endgroup$ Nov 9, 2010 at 3:35
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    $\begingroup$ Okay, thanks Pete and Gerry, and I apologize to all involved for the fiasco that seems to have ensued from my poor handling of this topic. $\endgroup$ Nov 9, 2010 at 3:56

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