Posted by: holdenlee | April 18, 2010

## pth roots are linearly independent

In this post (actually in the pdf) we will prove by semi-elementary means (though with some language about fields) that for prime p, the pth roots of rational numbers are linearly independent over $\mathbb{Q}$. This means that we can’t multiply “different” pth roots by rational numbers and sum them to give zero. By “different” we mean not differing by a pth power- $\sqrt{2}$ and $\sqrt{8}$ would not be considered “different”. Before we jump in the proof, we need a few concepts.

• A field is a number system where addition and multiplication satisfy all the familiar laws of arithmetic (commutative, associative, distributive, inverses), for example the rational numbers $\mathbb{Q}$ and the complex numbers $\mathbb{C}$. Here we only care about fields included in the complex numbers.
• A number is algebraic (in $\mathbb{Q}$) if it is the root of a polynomial equation (with coefficients in $\mathbb{Q}$).
• The minimal/ irreducible polynomial of an algebraic number $\alpha$ in a field $F$ is an irreducible monic polynomial with coefficients in $F$ that has $\alpha$ as a root. It is unique because if there were two, we could take the greatest common divisor of the two polynomials and get a smaller one with $\alpha$ as root. Any polynomial with $\alpha$ as a root must be a multiple of the minimal polynomial– see this by factoring it into irreducible factors and taking the factor with $\alpha$ as root.
• We can adjoin an element $\alpha$ to a field $F$ (such as $\mathbb{Q}$) to get a field extension $F(\alpha)$, which consists of all combinations of elements of $F$ and $\alpha$ with +,-,*,/.
• If $F\subseteq K$ are fields, the degree $[K:F]$ is the number of elements of $K$ needed to generate $K$ over $F$ as a vector space, that is, so that any element of $K$ can be written as a sum of terms in the form (element of F)$\times$(chosen element). If $\alpha$ has irreducible polynomial of degree $n$, then $[F(\alpha):F]=n$ as $F(\alpha)$ is generated by $\{1,\alpha,\ldots ,\alpha^{n-1}\}$.

Here’s the proof.

When I learn the “advanced” abstract algebra proof of this (which I hope to soon) I’ll post it too.