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Thread: Finding minimal polynomial

  1. #1
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    Finding minimal polynomial

    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.
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    Re: Finding minimal polynomial

    $\displaystyle x^5+x+1$ factors into irreducible polynomials $\displaystyle (x^2+x+1)(x^3-x^2+1)$ over $\displaystyle \mathbb{Q}$. Your expression is not a root in the first one, so...
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  3. #3
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    Re: Finding minimal polynomial

    Double checked the expression and it's correct except for missing the very last parenthesis, so, yes, it's a root of x^5+x+1.
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    Re: Finding minimal polynomial

    Quote Originally Posted by spudwish View Post
    Double checked the expression and it's correct except for missing the very last parenthesis, so, yes, it's a root of x^5+x+1.
    Sorry, what I meant was that your expression is not a root of the first factor $\displaystyle x^2+x+1$.
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  5. #5
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    Re: Finding minimal polynomial

    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
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  6. #6
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    Re: Finding minimal polynomial

    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 $\displaystyle f(x)$, say degree $\displaystyle n$, you can calculate the explicit powers $\displaystyle \alpha^2,\alpha^3,...\alpha^n$. Then write $\displaystyle f(\alpha)=\displaystyle{\sum_{i=0}^n c_i\alpha_i}=0$ and solve for the coefficients $\displaystyle c_i$ using a matrix and an appropriate basis depending on what $\displaystyle \alpha$ is.
    Last edited by Gusbob; Mar 24th 2013 at 06:59 PM.
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  7. #7
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    Re: Finding minimal polynomial

    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
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  8. #8
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    Re: Finding minimal polynomial

    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!
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