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We say that a linear form $f=ax+by+cz$ of real coefficients is "irrational" if $a,b,c$ are linearly independent over the rationals. My question is: Are there 3 such "irrational" linear forms $f_1,f_2,f_3$ such that the product $f_1f_2f_3$ is of integral coefficients?

Note that for the "2-dimensional" analogue, we have $(x+\sqrt{2}y)(x-\sqrt{2}y)=x^2-2y^2$.

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2 Answers

This should be a comment to Robin's answer.

Take any irreducible polynomial $f \in \mathbb{Q}[x]$ of degree 3 with real roots, say $\alpha, \beta, \gamma$. Set $f_1 = x + \alpha y + \alpha^2 z$, $f_2 = x + \beta y + \beta^2 z$, $f_3 = x + \gamma y + \gamma^2 z$.

You can find plenty of polynomials here.

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How about $x+2\cos(2\pi/7)y+4\cos^2(2\pi/7)z$, $x+2\cos(4\pi/7)y+4\cos^2(4\pi/7)z$, $x+2\cos(6\pi/7)y+4\cos^2(6\pi/7)z$ ?

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Sorry, but we are saying real linear forms... –  lonekite Aug 31 '10 at 16:18
    
Just take a totally real cubic field instead. Davenport studied such products a lot. –  Franz Lemmermeyer Aug 31 '10 at 16:27
    
Could you give an explicit example or a reference? Thanks. –  lonekite Aug 31 '10 at 16:33
    
@lonekite, Robin's forms are linear. The cosine terms are numerical coefficients of the indeterminates $y$ and $z$. –  Gerry Myerson Sep 1 '10 at 0:32
    
@Gerry: Yes, I understand. Robin first gave an example of complex linear forms with the same property, then revised it using the cosine function. Anyway, both Robin's example and felix' are very helpful to me. Thank you all. –  lonekite Sep 1 '10 at 4:22
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