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Can one construct a sequence $(a_k)_{k\geqslant 0}$ of rational numbers such that, for every positive integer $n$ the polynomial $a_nX^n+a_{n-1}X^{n-1}+\cdots +a_0$ has exactly $n$ distinct rational roots ?

If we cannot construct it explicitly, can we show that such a sequence exists?

PS: One can show (not easily) that the polynomial $\displaystyle \sum_{k=0}^n\frac{1}{3^{k^2}}X^k$ has $n$ distinct real roots.

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    $\begingroup$ @DanielR, I just wanted to say that having rational coefficients is neither sufficient nor necessary. By the way the suggested answer for $n=2$ does not seem to have rational roots. $\endgroup$
    – Arash
    Sep 13, 2013 at 9:46
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    $\begingroup$ @DanielR The problem is that having chosen coefficients which work in degree $n$ you have to find one additional coefficient (keeping the others the same) which gives all rational roots in degree $n+1$ - but you can't keep the $n$ roots you had before. $\endgroup$ Sep 13, 2013 at 9:51
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    $\begingroup$ What if we we had $(x-a_1)(x-a_2)...$ with $a_i$ to be distinct and rational for all $i=1,2,...$. then for any natural number $n$, pick $n$ distinct numbers $\lbrace a_1, \ldots, a_n \rbrace$ and just consider the polynomial $\Pi_{i=1}^{n} (x-a_i)$. ?? $\endgroup$
    – AXH
    Jan 10, 2014 at 19:33
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    $\begingroup$ @ Arbias : But the values of the coefficients changes as you raise the polynomial. Note the constant term of the polynomial is (-1)^n product of roots in every polynomial, so what is $a_0$ for n - deg polynomial in your construction is different for the (n+1) - deg polynomial. $\endgroup$
    – DiffeoR
    Jan 11, 2014 at 14:13
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    $\begingroup$ One small thing that I don't think anyone has noted explicitly yet: the $a_n$ must have unbounded height (where I'm using $\max(|p_n|, |q_n|)$ as my specific definition of height for $a_n=\frac{p_n}{q_n}$). Otherwise, by the rational root theorem there would be a bounded number of possible roots of the polynomial, and so once the degree gets high enough some root would have to be multiple. $\endgroup$ Mar 9, 2014 at 20:29

1 Answer 1

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This does not answer the OP's question, rather gives a partial result. Anyhow, it is too long to fit into a comment. From now on, I work with the Zariski topology.

Let $p(x)=a_0+a_1x+\cdots+a_nx^n=(x-x_1)\cdots(x-x_n)\in\mathbb{Q}[x]$.

First, let us describe the set of coefficients $(a_0,\ldots,a_n)\in\mathbb{A}^{n+1}_{\mathbb{Q}}$ for which $p$ has distinct rational roots. Relations between the roots and coefficients (Vieta's formulae) tell us that

$$a_i=(-1)^{(n-i)}a_ns_{n-i}(x_1,\ldots,x_n)$$

for $i=0,\ldots,n-1$, where $s_j$ denotes the $j$-th elementary symmetric polynomial. Now, let $X=Y=\mathbb{A}^{n+1}_{\mathbb{Q}}$, and consider the regular map $\varphi:X\longrightarrow Y$ defined by

$$\varphi(x_1,\ldots,x_n,a_n)=(a_0,\ldots,a_n).$$

I claim that the set of coefficients we are looking for is precisely $\varphi(X)\cap D(\Delta_p)\cap D(a_n)$, where $\Delta_p$ denotes the discriminant of $p$. Indeed, while $\varphi(X)$ guarantees rational roots, $D(\Delta_p)$ that they are distinct, $D(a_n)$ ensures that the leading coefficient is not zero.

Lemma. The map $\varphi:X\longrightarrow Y$ is dominant, that is, $\overline{\varphi(X)}=Y$.

Proof. We argue by contradiction. Suppose that $Z\subset Y$ is a proper closed subset containing $\varphi(X)$. Without loss of generality, we may assume that $Z=V(f)$ for some non-zero polynomial $f\in\mathbb{Q}[y_0,\ldots,y_n]$. Define $f_{a_n}=f(y_0,\ldots,y_{n-1},a_n)$. Now, by assumption, $f\circ\varphi$ vanishes identically on $X$, but then $f_{a_n}=0$ for all $0\neq a_n\in\mathbb{Q}$ since (and this is the key point) the elementary symmetric polynomials are algebraically independent over $\mathbb{Q}$. This implies that $f=0$, a contradiction.

As a dominant map, $\varphi$ contains a non-empty open set $U\subset Y$ in its image (the proof uses Noether normalisation). Therefore, $U\cap D(\Delta_p)\cap D(a_n)$ is a non-empty open set of coefficients for which $p$ has distinct rational roots. Let us denote this set by $D_n$, emphasising the degree of the polynomial. Now, if $(a_0,\ldots,a_n)$ is in the non-empty open set

$$V_n=\bigcap_{k=0}^n D_k\times\mathbb{A}^{n-k}_{\mathbb{Q}}\subset\mathbb{A}^{n+1}_{\mathbb{Q}},$$

then $p_k(x)=a_0+a_1x+\cdots+a_kx^k$ has distinct rational roots for all $k=0,\ldots,n$.

Therefore, in degree $n$, we have very many solutions.

Unfortunately, it is not clear how we can construct an infinite sequence in this way since $V_n$ might have empty intersection with the closed set $\{a_0\}\times\cdots\times\{a_{n-1}\}\times\mathbb{A}^1_{\mathbb{Q}}$ for a given $(a_0,\ldots,a_{n-1})\in V_{n-1}$. If we could bound the dimension of the complement of an open subset of $\varphi(X)\cap D(\Delta_p)\cap D(a_n)$ independently of $n$, then one could argue inductively. However, all we know for sure is that $\dim(U^c)\leq n-1$.

EDIT. I have just realised that, though it is dominant, $\varphi$ might not have any non-empty open set in its image, for $\mathbb{Q}$ is not algebraically closed. Unfortunately, this means that the conclusion I drew is wrong. However, the first part still makes sense.

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  • $\begingroup$ I managed to prove that for $n=2$, there is no non-empty open subset of $Y$ contained in $\varphi(X)$. Therefore, the method above does not work. $\endgroup$
    – user37391
    Mar 15, 2014 at 20:09
  • $\begingroup$ And $D$? What does it mean / denotes? $\endgroup$
    – leo
    Mar 22, 2014 at 5:23
  • $\begingroup$ $D(f)$ is the complement of $V(f)$. It is the set of those points where $f$ does not vanish. $\endgroup$
    – user37391
    Mar 22, 2014 at 12:12

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