factors into irreducible polynomials over . Your expression is not a root in the first one, so...
Here's an easy question: What is the minimal polynomial of (1/3)+(1/6)(1+sqrt(-3))((1/2)(25-3sqrt(69)))^(1/3) + (1-sqrt(-3))/(3*2^(2/3)(25-3sqrt(69))^(1/3)
(insert this into Mathematica or your favorite program to make sense of it. If there's something missing, it's a root of x^5+x+1)
The minimal polynomial is, according to Mathematica, x^3-x^2+1, question is how to come to that conclusion? I've only seen this done with simple examples.
True, it's a root of the second factor, but I don't see how either of that is relevant to the question. (here's where I'm not sure how to relay that I'm really asking...)
Perhaps I phrased my question poorly: Given the root above, how do I manually find its minimal polynomial? What I mean is a sequence of steps like this one:
Minimal polynomial for a:=sqrt(2)+sqrt(3):
x=sqrt(2)+sqrt(3)
(x-sqrt(2))^2=sqrt(3)^2
x^2-2xsqrt(2)+2=3
x^2-1=2x sqrt(2)
(x^2-1)^2 = (2x sqrt(2))^2
x^4-2x+1=8x^2
So minpoly(a) = x^4-10x+1
I'll work on that and update this post later. May I ask how you came by such an equation? The method I showed you was a good reason why Galois theory is invented...
EDIT: Sorry, but I have no wish to work this out by hand. If you're looking for a purely mechanical way to do this with no intuition whatsoever, the computer is the way to go. However if you know the degree (or approximate range) of the minimal polynomial , say degree , you can calculate the explicit powers . Then write and solve for the coefficients using a matrix and an appropriate basis depending on what is.
Alright thanks. I've been reading up on GT for a way overdue bachelor thesis, hence been doodling with polynomials quite a bit, I guess the question just popped up in my head how one would find the minimal polynomial for a more difficult radical expression than the ones featured in standard textbooks
I made some "conceptual" progress on my own: take the easier example of finding the minimal polynomial of a:=2^(1/3)+2^(2/3). To see if we're heading in the right direction, we use Mathematica to see that its minimal polynomial is x^3-6x-6. What next? We can subtract one or both terms from RHS, square or cube, or leave as is and square or cube. We'll cube as is. Put:
x = 2^(1/3)+2^(2/3) (=a)
x^3 = 6+6a = 6+6x.
Hence the minimal polynomial of a is x^3-6x-6, like we wanted. I was a bit puzzled by this result, didn't it just produce a polynomial not in Q[x], I thought. But, the ansatz was explicitly that x=a, or basically to produce a polynomial "in a" over Q, of smallest degree >1, which we did. This particular example isn't very enlightening really, but the reasoning is the same for, say, b=2^(1/3)+3^(1/3); we'll start raising powers, and not until we hit degree 9 will be able to factorize the resultant polynomials so that it's a polynomial "in b".
edit: Didn't see your edit there above. I'll check out your approach, thanks!