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Let $p(x)=\sum_{i=1}^n a_ix^i$ with $a_i$ an integer for all $i$ and $a_n=1$ such that $p(x)$ has only real roots, and let $\lambda_1,\ldots,\lambda_n$ be the $n$ roots of this polynomial. Then the Vandermonde determinant

$\begin{vmatrix} 1 & \lambda_1 & \cdots & \lambda_1^{n-1} \\ 1 & \lambda_2 & \cdots & \lambda_2^{n-1} \\ \vdots & \vdots & & \vdots \\ 1 & \lambda_n & \cdots & \lambda_n^{n-1}\end{vmatrix}$

is an integer. How do I prove this? Thanks.

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3 Answers 3

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In fact, it is not an integer. The formula for the determinant, compared to that for the discriminant $D$ of a polynomial, shows that $\det =\sqrt D$, up to signs. As an example, if your polynomial is $x^2+x+1$, the Vandermonde equals $\pm i\sqrt{3}$.

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  • $\begingroup$ Thanks. What about if all the roots of the polynomial are real? I'm sorry but I didn't mention this in the question... maybe I should. $\endgroup$ Commented Mar 22, 2015 at 15:56
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    $\begingroup$ Same. For $p(x)=x^2-2$ you get $\pm 2\sqrt{2}$. BTW, this is not a research question. $\endgroup$
    – abx
    Commented Mar 22, 2015 at 16:07
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To show that the square of the Vandermonde is an integer (and to compute it explicitly), note that the square is the resultant of the polynomial and its derivative, which is given by the Sylvester formula (which obviously gives you an integer).

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    $\begingroup$ the square of the Vandermonde is a symmetric function, and thus Per's argument applies. $\endgroup$ Commented Mar 22, 2015 at 16:13
  • $\begingroup$ @DimaPasechnik Yes, but I, personally, prefer to know what the value actually is. $\endgroup$
    – Igor Rivin
    Commented Mar 22, 2015 at 16:14
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It is symmetric in the $\lambda_i$, so can therefore be expressed in the elementary symmetric polynomials in $\lambda_i$. The latter are exactly the coefficients of the original polynomials, so these are integers.

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    $\begingroup$ But isn't the Vandermonde determinant an alternating polynomial, rather than a symmetric one? $\endgroup$ Commented Mar 22, 2015 at 15:47
  • $\begingroup$ Ah, right, that's true... was too quick there, haha. $\endgroup$ Commented Mar 22, 2015 at 15:50

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