Consider (irreducible) monic polynomials with integer coefficients which satisfy some fixed linear condition $L$ (for example, the coefficient of $x$ is $1,$ or something more complicated). The question is whether the set of real roots of such polynomials is dense in $\mathbb{R}.$ My previous question is for a particular such $L.$ For simple conditions like $a_0 = 1,$ or, indeed, any fixed subset subset of the coefficients fixed, it is known (by the work of S. D. Cohen) that the Galois group of such a polynomial is the full symmetric group, but this is almost certainly completely irrelevant.


2 Answers 2


In general, the answer is yes (unless the linear condition is like $1=0$). We want to a find an irreducible polynomial $f(x)=\sum_{k=0}^N a_kx^k$ with a root in the interval $(\alpha,\beta)$. Without loss of generality $\alpha,\beta$ are algebraically independent over $\mathbb{Q}$ (else change them a bit). We try to choose $f(x)$ so that $f(\alpha)<0$, $f(\beta)>0$. Note that $f(\alpha)$, $f(\beta)$ are linear forms in coefficients of $a_0,\dots,a_N$ with linearly independent coefficients. For such forms the pairs $(f(\alpha),f(\beta))$ are dense on $\mathbb{R}^2$. We should also care that $f$ is irreducible. This should be doable by a priori fixing all coefficients modulo some $p^2$ ($p$ is prime) and using Eisenstein criterion for irreducibility.

Well, let me now be more specific.

Claim. Assume that numbers $a,b,c,A,B,C,1,aB,bA,aC,\dots$ are linearly independent over $\mathbb{Q}$. Then the subgroup $G:=\{(ax+by+cz,Ax+By+Cz),x,y,z,\in \mathbb{Z}\}$ is dense in $\mathbb{R}^2$.

Proof. Note that all points $P(x,y,z)=(ax+by+cz,Ax+By+Cz)$ are mutually different. Choose large integer $R>0$. There are $R^3$ such points for $0<x,y,z\leq R$, and they lie inside a circle of radius, say, $10R$. Thus there exist two points which we are $O(R^{-1/2})$-close. So, our subgroup $G$ contains arbitrarily small vectors $v_1,v_2,\dots$, $\lim \|v_i\|=0$. Next, we may assume that directions of $v_i$ converge to direction of some unit vector $u=(\alpha,\beta)$. It follows that the closure of $G$ contains the whole line $\mathbb{R}\cdot u$. Now consider $G$ modulo this line, that is, consider the set $M:=\{\beta(ax+by+cz)-\alpha(Ax+By+Cz))\}\subset \mathbb{R}$. It suffices to prove that it is dense in $\mathbb{R}$. Indeed, this would just mean that the closure of $G$ contains a dense set of lines parallel to $u$. $M$ is not dense in $\mathbb{R}$ if and only if numbers $\beta a-\alpha A$, $\beta b-\alpha B$, $\beta c-\alpha C$ have rational mutual ratios. In other words, $\beta a-\alpha A=\lambda q_1$, $\beta b-\alpha B=\lambda q_2$, $\beta c-\alpha C=\lambda q_3$ for rationals $q_1,q_2,q_3$ (not all of $q_1,q_2,q_3$ are zero) and some real $\lambda$. Regarding this as a linear system on $\alpha,\beta,\gamma$ we see that a determinant $\det \pmatrix{a&-A&-q_1\\b&-B&-q_2\\c&-C&-q_3}$ does vanish. This contradicts to our assumption on linear independence.

  • $\begingroup$ Why is the statement about pairs of linear forms being dense in $\mathbb{R}^2$ true? $\endgroup$
    – Igor Rivin
    May 9, 2015 at 0:09
  • 1
    $\begingroup$ I think you're assuming that the class of polynomials has at least two degrees of freedom. The values aren't always dense, but do reach each quadrant because they are a subgroup of $\mathbb R^2$ not contained in any line. If there are three degrees of freedom it will be dense. $\endgroup$
    – Will Sawin
    May 9, 2015 at 1:00
  • $\begingroup$ But notice that requiring Eisenstein mod a FIXED prime kills another degree of freedom. Presumably, being able to vary the prime gets around this, but this is not 100% obvious. $\endgroup$
    – Igor Rivin
    May 10, 2015 at 2:38
  • $\begingroup$ I think, it does not kill degree of freedom, because you fix only remainders, while partial quotients are free integer variables. In any case, if I understand the question correctly, we have as many degrees of freedom as we need. $\endgroup$ May 10, 2015 at 8:00

If being monic is not required, then the real roots of cubic polynomials with coefficient of $x^2$ zero will be a dense set in real numbers.

Just take one irreducible cubic polynomial $f(x)=x^3+px+q$ with one real root $b$, and two (possibly complex) other roots $\alpha,\beta$. The the set of polynomials with roots $rb,r\alpha, r\beta$ with $r$ running over all the rational numbers will have minimal polynomial of the type $f_r(x)=ux^3+vx+w$, for some integers $u,v,w$. As the set of rational multiples of a fixed non-zero number is a dense set we are done.


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