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 . 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- and 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 and the complex numbers . Here we only care about fields included in the complex numbers.
- A number is
**algebraic** (in ) if it is the root of a polynomial equation (with coefficients in ).
- The
**minimal/ irreducible polynomial** of an algebraic number in a field is an irreducible monic polynomial with coefficients in that has 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 as root. Any polynomial with as a root must be a multiple of the minimal polynomial– see this by factoring it into irreducible factors and taking the factor with as root.
- We can
**adjoin** an element to a field (such as ) to get a **field extension** , which consists of all combinations of elements of and with +,-,*,/.
- If are fields, the
**degree** is the number of elements of needed to generate over as a vector space, that is, so that any element of can be written as a sum of terms in the form (element of F)(chosen element). If has irreducible polynomial of degree , then as is generated by .

Here’s the proof.

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

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See also:

http://www.thehcmr.org/issue2_1/mfp.pdf

By:

holdenleeon June 5, 2010at 8:45 pm