Re: a question about algebraic numbers

On Mon, 3 Apr 2006 18:22:17 -0500, ol3@xxxxxxxxx (Oscar Lanzi III)
wrote:

If y is a complex cube root of 1/3, its own algebraic degree is 6, not
3. You have an algebraic degree of 3 for the cube root of 1/3 itself,
and multiplying by a complex cube root of unity (degree = 2) perforce
doubles that degree of 3.


That's not right.

If y is a cube root of 1/3, then y^3=1/3 which is an irreducible
degree 3 polynomial over the rationals. Hence all cube roots of 1/3
have degree exactly 3.

You somehow missed the key point of this thread.

deg(ab) is not necessarily equal to deg(a)*deg(b) -- it can be less,
even when the deg(a) and deg(b) are relatively prime.

On the other hand, as I've pointed out in a prior reply, if deg(a) and
deg(b) are relatively prime then deg(a+b) = deg(a)*deg(b).

quasi
.

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