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About LinkBacksIf a,b,c and p are rational numbers,
and a+bp^1/3+cp^2/3=0
and p is not a perfect cube.
prove that a=b=c=0
Originally Posted by jashansinghal
First proof (requires knowledge in fields extensions): the cubic rational extensionis of degree 3 since
is the minimal polynomial of
(why?), and
a basis for it is preciselythese three elements are linearly independent over
, and this is precisely what had
to be proved.
Second proof:![a+b\sqrt[3]{p}+c\sqrt[3]{p^2}=0\iff (a+b\sqrt[3]{p})^3=](http://latex.codecogs.com/png.latex?a+b\sqrt[3]{p}+c\sqrt[3]{p^2}=0\iff (a+b\sqrt[3]{p})^3=)
![-c^3p^2\iff a^3+3a^2b\sqrt[3]{p}+3ab^2\sqrt[3]{p^2}+b^3p=-c^3p^2](http://latex.codecogs.com/png.latex?-c^3p^2\iff a^3+3a^2b\sqrt[3]{p}+3ab^2\sqrt[3]{p^2}+b^3p=-c^3p^2)
.
Note that(I assume
; if some of these two numbers is zero I leave to you to do the little changes necessary to fix the proof).
Raise to the 3rd power the expression, and now make some order here and deduce a contradiction to p not being a
perfect (rational, of course) cube.
Tonio
When p is not a perfect square, one can use Eisenstein's criterion to prove x^3-p is irreducible. But it's nontrivial (at least I think) to show x^3-p is irreducible in general case, i.e. merely if p is not a perfect cube of some element in the ground field (
And a warning if anyone want to generalize it: it's not true that if q is not a perfect 4-th power, then x^4-q is irreducible. e.g. if q=-4, then one can factor x^4+4=(x^2-2x-2)(x^2+2x-2)
EDIT: OK I forgot that a cubic polynomial is irreducible iff it doesn't have a root, so in this case it's trivial to show x^3-p is irreducible.
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confused first but clear now