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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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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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  • $\begingroup$ Sorry, but we are saying real linear forms... $\endgroup$
    – lonekite
    Commented Aug 31, 2010 at 16:18
  • $\begingroup$ Just take a totally real cubic field instead. Davenport studied such products a lot. $\endgroup$
    – Franz Lemmermeyer
    Commented Aug 31, 2010 at 16:27
  • $\begingroup$ Could you give an explicit example or a reference? Thanks. $\endgroup$
    – lonekite
    Commented Aug 31, 2010 at 16:33
  • $\begingroup$ 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 <a href="cems.uvm.edu/~voight/nf-tables/3-25.txt">here</a>. $\endgroup$
    – felix
    Commented Aug 31, 2010 at 16:38
  • $\begingroup$ @lonekite, Robin's forms are linear. The cosine terms are numerical coefficients of the indeterminates $y$ and $z$. $\endgroup$
    – Gerry Myerson
    Commented Sep 1, 2010 at 0:32

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