2

To be very concrete, I want to show that every element of the form $1/(p+qx+rx^2)$ where $x=\sqrt[3]{2}$ where $p,q,r$ are rationals can be written in the form $a+bx+cx^2$ where again $a,b,c$ are integers. Clearly, it is enough to show that certain 3x3 matrix with coefficients involving $p,q,r$ has nonzero determinant. However, I cannot show why it cannot be zero for rational values of $p,q,r$. Any simple way to do this?

MJD
  • 65,394
  • 39
  • 298
  • 580
a12345
  • 1,323
  • Assume $p+qx+rx^2 \neq 0$ then look to see if $(p+qx+rx^2)(a+bx+cx^2)-1=0$ is soluble. – John Fernley Jul 28 '14 at 13:47
  • That's what I did. By taking the determinant, you get an expression for $p,q,r$. I want to show that it cannot be zero for rational values of $p,q,r$. – a12345 Jul 28 '14 at 13:51
  • Simpler, $,x^3!-!2,$ is irreducible over $,\Bbb Q,,$ so $,\deg f < 3,\Rightarrow, (f,x^3!-!2) = 1,,$ so, by Bezout's identity for the gcd we get $, a(x) f(x) + b(x) (x^3!-!2) = 1,,$ so $, a f = 1\ $ in $,\Bbb Q[x]/(x^3!-!1) \cong \Bbb Q(\sqrt[3] 2)\ \ $ Or, by finite dimensionality, $,w = f(x),$ is algebraic over $,\Bbb Q,$ so its inverse can be read off from its minimal polynomial $, c+w, g(w) = 0,\Rightarrow w (-g(w)/c) = 1.\ \ $ – Bill Dubuque Jul 28 '14 at 13:55
  • 1
    See here for a proof that if $a$ is algebraic, $F(a)=F[a]$. – Meow Jul 28 '14 at 14:04
  • Also you can use that the map $\ x \to \alpha x \ $ is $1$-$1$ so onto (being on a finite dim vector space), so $,\alpha\beta = 1,$ for some $,\beta.,$ No need to calculate $,\det \alpha.,$ Theory trumps calculation! – Bill Dubuque Jul 28 '14 at 14:09
  • You make a $\sqrt[3]2$ by writing \sqrt[3]{2}. Reference guide here. – MJD Jul 28 '14 at 14:17
  • See also: http://math.stackexchange.com/questions/422440/does-a3-2b3-4c3-6abc-have-solutions-in-mathbbq – Martin Sleziak Jan 22 '16 at 16:59

1 Answers1

5

I assume the determinant you want is $$\det\pmatrix{p&2r&2q\cr q&p&2r\cr r&q&p\cr}=p^3+2q^3+4r^3-6pqr\ .$$ The determinant obviously is zero when $p=q=r=0$; suppose that this is not the case and the determinant is still zero, $p,q,r$ being rational. Multiplying by a common denominator and then cancelling any common factor, we may assume that $p,q,r$ are integers with no common factor. If $$p^3+2q^3+4r^3-6pqr=0$$ then $2\mid p^3$ so $2\mid p$, say $p=2s$. Substituting back and cancelling $2$ gives $$4s^3+q^3+2r^3-6qrs=0\ .$$ This shows that $2\mid q$; then by a similar process you get $2\mid r$; this contradicts the fact that $p,q,r$ have no common factor. Therefore the determinant cannot be zero for rational $p,q,r$, not all zero.

I am sure you can see the similarity between this proof and the standard irrationality proof for $\sqrt2\,$.

It is actually easier to do this by using the fact that $x^3-2$ is irreducible, but you asked for a determinant proof so that's what you've got ;-)

David
  • 82,662