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16th November 12:19
External User
Posts: 1

normal forms for algebraic numbers
>OK, it's easy to see that this is unique. (aren't algebraic integers
>such things with the extra restriction that the leading coeff is 1?) >However, I have trouble figuring out how to compute with this, that is: >1) given an algebraic number, say > sqrt(1+sqrt(17+cuberoot(19))) cuberoot(5) >Oh. OK, that's not necessarily algebraic according to my definition. >So...uh... how about I change the set on you but keep the same question. >For forms involving arbitrary roots, etc, no vars, is there a unique >normal form? I simplemindedly can't see an obvious way to convert the >above form to a poly of which it is a root. (for some forms yes, but not >all (e.g. like the one above)). >Likewise is there a way (algorithm) to get a such a form from a polynomial? Yes, there is. In Maple: P:=sort(evala(Norm(convert((qx),RootOf)))); 36 34 33 31 30 29 P := x  18 x + 60 x  720 x + 3282 x + 9720 x 28 27 26 25  16146 x  78820 x + 295956 x  287820 x 24 23 22 21  1792506 x + 13743360 x + 40209912 x  14096280 x 20 19 18  279700236 x + 387516960 x  523775078 x 17 16 15 + 730492920 x + 11686352778 x + 85080734460 x 14 13 12 + 9488713392 x  254456753220 x  616299273531 x 11 10 9 + 857922985800 x + 1211443482834 x + 277908076980 x 8 7 6  5266258028172 x + 3345011588160 x + 1484108332884 x 5 4 3 + 4356940483080 x  14804808163020 x + 15364752902640 x 2  7922401473600 x + 2087405575200 x  225309308400 Of course, then you have to figure out which of the 36 roots of P this is, which might not be so simple (requiring highprecision numerical calculation). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 
4
16th November 12:19
External User
Posts: 1

normal forms for algebraic numbers
It's not undecidable  it's impossible.
 Timothy Murphy email: tim@birdsnest.maths.tcd.ie tel: +35386233 6090 smail: School of Mathematics, Trinity College, Dublin 2, Ireland 
6
17th November 04:42
External User
Posts: 1

normal forms for algebraic numbers
It is decidable.
There is an algorithm for determining the galois group of a polynomial (ie of the splitting field of the polynomial). IIRC, such an algorithm is given in van der Waerden, Modern Algebra. (I'm not certain of that, as both my volumes have been stolen. However, I am certain there is such an algorithm.) Since the galois group is finite, any question about it is decidable  eg to determine if it is simple one goes through all subsets, and decides if each one is a normal subgroup. [I'm not suggesting that is a sensible thing to do; just that it is a possible procedure.] I don't know the complexity of the algorithm, but I would be slightly surprised if it were not polynomial time in the degree of the polynomial. Yes, the other roots (ie the conjugates of the given root) are obtained by replacing each nth root by the other nth roots, eg replacing a^{1/n} by wa^{1/n} where w is an nth root of 1.  Timothy Murphy email: tim@birdsnest.maths.tcd.ie tel: +35386233 6090 smail: School of Mathematics, Trinity College, Dublin 2, Ireland 