# Proof that a sum of the first period of powers of integer roots of rationals is irrational

I'm looking for a simple proof or a reference to any proof that

For $j \in ℤ$, $0<j<m$, when each $k^j \notin ℚ$ and each $d_j \in ℚ$, $d_j \neq 0$, then $\sum_{j=1}^{m-1} d_j k^j \notin ℚ$

My searches have turned up a few papers that appear to make use of this property, but none that prove it, refer to a source, or even give it a name.

EDIT:

I'm working with $k$ restricted to positive integer roots of rational numbers: $k=r^{1/m}=(a/b)^{1/m}$ with $r$ and $m$ chosen so that $m$ is the least positive integer for which $k^m \in ℚ$ (so when $k \in ℚ$, $m=1$).

The set of positive integer roots of rational numbers (set $𝕂$) is not closed under addition (see the $\sqrt{2}-1$ case in comments) Given $l$, $h$ members of $𝕂$, $\left(l\pm h\right) = l \left(1\pm{h\over l}\right) = l \left(1+k\right) \notin 𝕂$ occurs when $\left(1+k\right) \notin 𝕂$.

The sum before the edit is a rearrangement of the binomial expansion of $\left(1+k\right)^n$ where a different choice of $n$ changes the values of $d_j$, and excluding the rational $j=0$ term. If the sum is irrational, then $\left(1+k\right)^n \notin ℚ$ for any $n$, $\left(1+k\right) \notin 𝕂$, and $\left(l\pm h\right) \notin 𝕂$ occurs when $\pm {h\over l} = k \notin ℚ$.

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More conditions are necessary for this to be true: $d_1 = 2$, $d_1 = -1$, $k^1 = \sqrt{2}$, $k^2 = 2 \sqrt{2}$, but $d_1 k_1 + d_2 k_2 = 0 \in \mathbb{Q}$. – copper.hat Jun 8 '12 at 22:03
Do you think that it holds for $k=\sqrt{2}-1$ and $m=3$? – WimC Jun 8 '12 at 22:04
There is something missing here, because you can set the final polynomial equal to zero and use this as the definition of $k$. This will give a counterexample if you choose an irreducible polynomial. – Phira Jun 8 '12 at 22:04
I don't understand. If $m=3$, $k=\varphi=\frac{1+\sqrt{5}}{2}$ and $d_j=(-1)^j$ then $k^j \notin ℚ$ for $0 \le j$, $d_j \in ℚ$, $d_j \neq 0$ but $\sum_{j=1}^{m-1} d_j k^j = \varphi^2 - \varphi = 1 \in ℚ$. – Henry Jun 8 '12 at 22:06
This was answered (counterexample) in a very recent answer to a similar question you asked. – André Nicolas Jun 8 '12 at 23:30

The result that you want can be found in a fair number of places. The one I know is Lang's Algebra. Here is a link. It starts at page $297$, in the Galois theory chapter.

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That link gives me a page that is "unavailable for viewing," and my library doesn't seem to have that book. I imagine I could find it in another book with a chapter on Galois theory, is there any subheading I should look for? – Polyergic Jun 9 '12 at 15:07
Strange, I just tested link again, worked fine for me. Unfortunately can't lend you my copy of the book! – André Nicolas Jun 9 '12 at 15:12
Tried again and I can see it this time. It's over my head enough I can't just read it and understand, but it does look like what I was looking for. Thanks! – Polyergic Jun 10 '12 at 2:23

The current question is as follows:

Let $k=(a/b)^{1/m}$, where $a,b,m$ are integers and $m$ is minimal for such a representation of $k$. (This implies that $k^j$ is not rational, for $0<j<m$.)

$k$ is a root of the polynomial $bx^m-a$.

For most values of $a$ and $b$, this polynomial is irreducible, so there cannot be a polynomial of smaller degree such that $k$ is a root which implies your conclusion.

(In particular, if $a$ is square-free, and different from $0,\pm 1$, then the Eisenstein criterion guarantees that the polynomial is irreducible.)

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This is very close, but for what values of $a$ and $b$ would the polynomial be reducible? $a$ is not necessarily square-free, for example, if $k = \left({2^4\over 3}\right)^{1\over5}$ – Polyergic Jun 9 '12 at 2:18