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.

Advertisements

Responses

  1. See also:
    http://www.thehcmr.org/issue2_1/mfp.pdf


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Categories

%d bloggers like this: