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I was wondering:

  1. If $a, b \in \mathbb{N}$ and obviously they both have a prime factorization, we know that $a+b \in \mathbb{N}$ and $a+b$ also has a prime factorization so $a$, $b$ and $a+b$ can be rewritten as $\prod_{i = 1}^{N}p_i^{n_i}$ with $n_i \in \mathbb{N}$.
  2. If $p, q \in \mathbb{Q}$ and obviously they both have a prime factorization, we know that $p+q \in \mathbb{Q}$ and $p+q$ also has a prime factorization so $p$, $q$ and $p+q$ can be rewritten as $\prod_{i = 1}^{N}p_i^{n_i}$ with $n_i \in \mathbb{Z}$.

But here it comes the question:

If I got two numbers $r$ and $s$ so they can be expressed as $\prod_{i = 1}^{N}p_i^{n_i}$ with $n_i \in \mathbb{Q}$ (so they are roots of any grade of positive rational numbers), can $r+s$ be also expressed as $\prod_{i = 1}^{N}p_i^{n_i}$ with $n_i \in \mathbb{Q}$ but we cannot find that factorization even when it exists? Or we just can't because the factorization doesn't exist? If not: why? Is it some kind of algebraic limitation?

Examples:

  • $2^2 + 2\cdot3 = 4 + 6 = 10 = 2\cdot5$
  • $2^{-1}+5^{-1} = \frac{1}{2}+\frac{1}{5} = \frac{7}{10} = 2^{-1}\cdot5^{-1}\cdot7$
  • $2^{\frac{1}{2}}+2^{\frac{1}{3}}\cdot3^{\frac{1}{3}} = \sqrt{2} + \sqrt[3]{6} = \textbf{???}$

Many thanks in advance!!!

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    a number which admits such a factoring is a root of a polynomial of the form $x^n-m$ for $n\in \mathbb N,m\in \mathbb Z$. It is easy to see that, for example $(1+\sqrt 2)$ is not the root of such a polynomial. – 2017-02-24
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    @lulu I'm not convinced, how can we prove there is no $P(x)(x^2-2x-1)=x^n-m$? or for any other minimal polynomial? – 2017-02-24
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    @i9Fn As I say, it's clear that $(1+\sqrt 2)$ is not the root of such a polynomial. Say it were. then $\sum \binom ni (\sqrt 2)^i=a$. but the left hand can be written as $s+r\sqrt 2$ for integers $s,r$ so this would tells us that $\sqrt 2\in \mathbb Q$. – 2017-02-24
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    @i9Fn for a more group theoretic proof, note that we can extend the $\mathbb Q$ homomorphism $\sqrt 2 \to -\sqrt 2$ to a $\mathbb Q$ automorphism of the splitting field of $1+\sqrt 2$. Thus $1-\sqrt 2$ would have to be a root of the same polynomial, which is absurd. – 2017-02-24

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Sadly, this is false. To see it, let's demonstrate that $\alpha = 1 +\sqrt 2$ has no such factoring.

Note that any number which admits such a factoring must be a root of a polynomial of the form $x^n-m$ for $n\in \mathbb N$ and $m\in \mathbb Z$. Indeed just take $n$ to be the least common multiple of the denominators of the $n_i$ in the factoring.

We'll now show that our $\alpha$ can not be the root of such a polynomial.

Suppose that it were. Then $(1+\sqrt 2)^n=m$ for suitable $n,m$. But by the binomial theorem we have $$(1+\sqrt 2)^n=\sum_{i=0}^n \binom ni \left(\sqrt 2\right)^i=m$$ But $$\sum_{i=0}^n \binom ni \left(\sqrt 2\right)^i=\sum_{i=2k}\binom ni 2^{k}+\sqrt 2\times \sum_{i=2k+1}\binom ni 2^{k}$$

As the two sums are clearly both integers we see that we have written $$r+s\sqrt 2= m$$ for integers $r,s$.

Inspection shows that $s>0$ (as is $r$ for that matter) so this would imply that $\sqrt 2 \in \mathbb Q$, which is our contradiction.

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    And does it hold for a sum of roots $r_1$ and $r_2$ so both of them are not equal to 1? What if $\alpha = \sqrt{2}+\sqrt{3}$? Don't we have another way to write $\sqrt{2}+\sqrt{3}$? – 2017-02-24
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    The same argument works in this case. Just write out $(\sqrt 2 +\sqrt 3)^n$ to show that there would exist non-zero integers $r,s$ with $r\sqrt 2 +s\sqrt 3\in \mathbb Z$. Easy to show that this is impossible. Just square! – 2017-02-24
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    So the sum of numbers $x$ so they belong to the set of "factorizable" numbers is an external operation when they are roots? – 2017-02-24
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    Sorry, that should have read: "non-zero integers $r,s,t$ such that $r\sqrt 2 +s\sqrt 3 +t \sqrt 6\in \mathbb Z$". Still easy to show that this can't happen. – 2017-02-24
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    I don't know what an external operation means. You might want to read about prime ideals in rings of algebraic integers...that's the kind of factoring which you can get. – 2017-02-24
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    External in the sense that if you add to numbers of the set, the result is inside the set (in this case, the set of factorizable numbers). So two natural numbers which can be factorized they can give a result which can be factorized. This holds for numbers in $\mathbb{Q}$, but not for roots, which can be factorized, but if you add them, then the result cannot be factorized, so it's out of the set. So... concerning the set of factorizable numbers, the summation of members of the set is closed respect to the set only for the subset of rational numbers. – 2017-02-24
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    As I say, look up the theory of prime ideals in rings of algebraic integers. There you can factoring of exactly the form you want. It's just that you can't get factoring into the usual sort of rational primes...but why should you expect to? – 2017-02-24
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    hi, I'm just thinking that what you've exposed is true with integer-valued polynomials, but would it hold if the polynomial in valued in roots of rational numbers? – 2017-03-29
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    Sorry, what statement do you want to be true? Like I said, there is a good notion of "prime" that works for number fields other than $\mathbb Q$ and with that notion, much (bot not all!) of the standard theory goes through. If you insist on trying to factor in usual primes, I don't think you'll get far. – 2017-03-29
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    Just to be clear: My argument for $1+\sqrt 2$ shows that it can't be the root of a polynomial of the form $x^n-m$ for $n\in \mathbb Z$ and $m\in \mathbb Q$, not just $m\in \mathbb Z$ ...nothing changes in the argument. – 2017-03-29
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    What if one were to allow *infinitely many* $n_i$ to be nonzero? In such a case, the least common multiple of the denominators may fail to exist, right? – 2018-02-07
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    @user76284 Not sure what sort of factoring you have in mind...for this question it is clear that the OP intended only finitely many $n_i$. – 2018-02-07
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    I was just wondering about the more general case. – 2018-02-07