# Finding the degree of minimal polynomials

Let a number $x = \sqrt[a_1]{p_1} + \sqrt[a_2]{p_2} + \ .. \ + \sqrt[a_n]{p_n}$ be a number such that all $a_n$ are integers and all $p_n$ are rational. I've been noticing that for every number x, the degree of its minimal polynomial is seemingly always equal to $\prod_{1}^n \ a_n$.

Is that valid for all values of $a_n$? If so, is there a proof?

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On a related topic, see exercises 18--22 on pp. 290--291 of Lang's Undergraduate Algebra (3rd edition), especially the remark after exercise 18 for context. If you want to find this on Google books, search for the phrase "most people" in the book, which is part of the remark. Those exercises concern the field degree of a field extension obtained by adjoining to a field $F$ several $n$-th roots of elements of $F$, and in practice one often finds a "random" sum of numbers algebraic over a field is a primitive element for the extension generated by all of those numbers. –  KConrad Apr 17 '11 at 3:57
The question at mathoverflow.net/questions/26832/… is relevant to the question here. –  KConrad Apr 17 '11 at 4:02

No. Some conditions are needed on the $a_i$ and $p_i$. For instance, take n=2, $a_1 = a_2 = 2$, $p_1 = p_2 = 2$. Then $x = 2 \sqrt{2}$, which has minimal polynomial $x^2 - 8$. As an even simpler example, n=1, $a_1 = 2$, $p_1 = 4$, then $x$ is rational.

For a less trivial example, take $a_1= 4$, $a_2 = 6$, $p_1=p_2=2$. Check that this has a polynomial of degree 12. In fact, this isn't really true at all.

One can, however, prove that the degree of the minimal polynomial is at most $\prod a_n$, which is an easy exercise in field theory. Any graduate algebra textbook covering Galois theory will be more than sufficient to prove this; just remember the degree of the minimal polynomial is the same as the dimension of the extension field viewed as a vector space over the base field.

EDIT:

After much miscommunication on my part, we've reached the following results:

Suppose $a_1,\ldots,a_n$ are pairwise relatively prime positive integers, $p_1, \ldots, p_n$ integers such that $\sqrt[a_i]{p_i}$ is of degree $a_i$ for each i. Then $\sqrt[a_1]{p_1} + \cdots + \sqrt[a_n]{p_n}$ is of degree $\displaystyle \prod_{i=1}^n a_i$.

The condition that each $\sqrt[a_i]{p_i}$ is met (by Eisenstein Criterion) should there be a prime $q_i$ such that $q_i | p_i$ and $q_i^2 \not{|} p_i$ for each i.

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What if $gcd(a_n, p_n) = 1$, and no pairs ${a_n, p_n}$ are equal? then is it satisfied? –  Victor Apr 17 '11 at 2:47
I'm not totally sure what having $gcd(a_n,p_n)=1$ gives you (or conditions regarding pairs $(a_n,p_n)$), but if you have $gcd(a_1,...,a_n)=1$ then your statement holds. If $gcd(p_1, \ldots ,p_n)=1$ and $a_1= \cdots = a_n$, it should also hold. I'm not sure about the general case for $gcd(p_1, \ldots ,p_n)=1$, though I wouldn't be surprised if your statement held then. –  Logan Maingi Apr 17 '11 at 2:58
There are a good number of cases where your statement will hold, but it's not true in general. If the $a_i$ and the $p_i$ aren't related in any obvious way, it's probably true for small values of n, but I'd still check. Wolfram alpha (www.wolframalpha.com) can compute the minimal polynomials in sufficiently small examples, and most computational algebra engines can do it for arbitrary numbers of the form you want. –  Logan Maingi Apr 17 '11 at 3:00
Scratch what I said above; it's not true unless each $p_i^{1/a_i}$ has minimal polynomial of degree $a_i$. This holds in the case that some prime q divides $p_i$, but $q^2$ doesn't divide $p_i$, by the Eisenstein Criterion. With more advanced arguments, a little bit stonger statements can be made. It fails in the case where we take $\sqrt{4}$. Another similar case is that the kth root of unity $e^{i2 \pi /k}$ satisfies the polynomial $a^{k−1}+\cdots+x+1$, which is of degree k−1. But when $p_i^{1/a_i}$ is of degree $a_i$ over $\mathbb{Q}$, the above should hold. –  Logan Maingi Apr 17 '11 at 6:57
Dear Logan, your statement after "...not true unless..." is not correct: the sum $\sqrt[2]{2.3}+ \sqrt[2]{2.3^3}$ does not have degree 4 although all your hypotheses are verified: the numbers $\sqrt[2]{2.3}$ and $\sqrt[2]{2.3^3}$ have minimal polynomials of degree 2 and the prime $q=2$ divides $2.3$ and $2.3^2$ whereas $q^2=2^2$ divides neither. In view of this and your "scratch what I said above", could you please sum up your claim in a new statement and provide a proof of that new statement or a reference ? You are welcome to use "more advanced arguments". –  Georges Elencwajg Apr 17 '11 at 17:23

Besicovitch has proved the following related interesting result:

Consider an integer $n\gt 1$ and distinct prime numbers $p_1,p_2,\ldots ,p_k.$ Then the field $F=\mathbb Q (\sqrt[n]{p_1},\ldots ,\sqrt[n]{p_k})$ has dimension $n^k$ over $\mathbb Q$ .
More precisely, a $\mathbb Q$-basis of that field $F$ is given by the radicals $$\sqrt[n]{p_1^{m_1}\ldots p_i^{m_i} \ldots p_k^{m_k} } \quad (\; 0\leq m_i \lt n \quad , \quad 1\leq i\leq k )$$

(The case $n=2$ is a classical chestnut in Galois theory.)
This does not answer the OP's question but at least assures us that, for example, $$\sqrt[3]{900}+\sqrt[3]{36}+ \sqrt[3]{15}+\sqrt[3]{150} \notin \mathbb Q$$
which is not so simple to check directly.

I have the pessimistic feeling that there is no very satisfactory general answer to the question "when does the sum $\sqrt[n_1]{a_1}+ \sqrt[n_2]{a_2}+...+\sqrt[n_k]{a_k}$ have degree $n_1 n_2 ...n_k$", but I'd love to be shown wrong.

Bibliography: Besicovich's original article is: Abram S. Besicovitch, "On the linear independence of fractional powers of integers", Journal of the London Mathematical Society 15 (1940), 3-6.

Here is a more recent and accessible proof : Ian Richards, "An application of Galois theory to elementary arithmetic", Advances in Mathematics 13 (1974), 268-273. 13 (1974), 268-273.

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