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?



Finding the degree of minimal polynomials
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.


