Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
@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. – Arash Sep 13 '13 at 9:46
@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. – Mark Bennet Sep 13 '13 at 9:51
I don't understand this argument. A polynomial with a $p$-fractional leading coefficient and all other coefficients $p$-integral can still factor completely over $\bf Q$. For example, consider $p^{-1} \prod_{i=1}^n (x-x_i)$ with $p\mid x_i$ for each $i$. And even if all the $a_k$ are $p$-integral, they could eventually all be multiples of $p$, and again there would be no $p$-adic contradiction. – Noam D. Elkies Dec 8 '13 at 3:41
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 $\Pi_{i=1}^{n} (x-a_i)$. ?? – Arbias Hashani Jan 10 '14 at 19:33
@ 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. – DiffeoR Jan 11 '14 at 14:13

1 Answer 1

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


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


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.

share|cite|improve this answer
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. – Norbert Pintye Mar 15 '14 at 20:09
And $D$? What does it mean / denotes? – leo Mar 22 '14 at 5:23
$D(f)$ is the complement of $V(f)$. It is the set of those points where $f$ does not vanish. – Norbert Pintye Mar 22 '14 at 12:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.