# Solving polynomial equations in radicals provided all roots are rational

This question is related to this question of Joseph O'Rourke and this question of mine.

Question. Let $f$ be a polynomial with integer coefficients. Suppose that all roots of $f$ are rational. Can one find a rational root by the following procedure (similar to solving equations in radicals):

Step 1. Find a number $p_1$ that is polynomially dependent on the coefficients of $f$ (say, the discriminant of $f$) and check if $p_1$ has a rational root $b_1$ of degree $k_1\le n$. Thus $p_1=g_1(a_0,...,a_n)$ for some polynomial $g_1$, and $b_1^{k_1}=p_1$.

Step 2. Find a number $p_2$ that is polynomially dependent on the coefficients of $f$ and $b_1$ and check if it has a rational root of degree $k_2\le n$. Thus $p_2=g_2(a_0,...,a_n,b_1)$ for some polynomial $g_2$, and $b_2^{k_2}=p_2$.

....

The number $p_m$ ($m$ polynomially depends on $n$) is a rational root of $f$. Of course all polynomials $g_i$ involved in this procedure should not depend on the coefficients of the polynomial $f$, only on its degree (like the discriminant).

Note that existence of such a procedure does not contradict unsolvability of the symmetric group $S_n$, $n\ge 5$, because we assume that the Galois group of $f$ is trivial.

• You probably did not mean what you wrote in steps 1 and 2 as they make no sense as written: "number $p_1$ has a rational root $b_1$..." – Felipe Voloch May 19 '12 at 1:39
• @Felipe: I meant that the root of $p_1$ of degree $k_1$ is rational. Isn't that what is written there? – user6976 May 19 '12 at 1:47
• $p_1$ is a number and numbers don't have roots. Did you mean to write that $p_1$ is a polynomial? Same with $p_2$. – Felipe Voloch May 19 '12 at 2:36
• @Felipe: Why do you think numbers do not have roots? Number $64$, for example, has root of degree $3$, which is $4$. Number $p_1$ is a value of a polynomial $p_1=g_1(a_0,...,a_n)$. Its root of degree $k_1$ is $b_1$. Thus $p_1=b_1^{k_1}$. Is that clearer? – user6976 May 19 '12 at 5:19
• @Felipe: Many things have roots, not only polynomials. Trees, for example, have roots (and I am not talking about rooted trees here), carrots themselves are roots. Afghanistan’s current problems have roots in its political structure: bamdad.af/english/text/story/1606 . – user6976 May 19 '12 at 5:45

Consider a generic polynomial $x^n+a_1x^{n-1}+a_2x^{n-2}+....+a_n$ over $\mathbb Q(a_1,...,a_n)$. Adjoin $b_1$, a root of $p_1$ of degree $k_1$, then adjoin $b_2$, and so on. This is contained in some solvable galois extension of $\mathbb Q(a_1,...,a_n)$.
Let $q$ be the generic polynomial evaluated at $p_m$, and let $r$ be the norm to $\mathbb Q(a_1,...,a_n)$ of $q$. Then $r$ lies is $\mathbb Q(a_1,...,a_n)$, and is zero whenever the polynomial has all rational roots. But this is a Zariski dense subset, so $r$ is identically zero, so this solvable Galois extension always contains a root, which does violate unsolvability of the symmetric group.
• @Will: I think I understand your answer. In fact you show that a much more general procedure, when only $p_m$ is supposed to be rational, is impossible? Thank you! – user6976 May 19 '12 at 1:55
• Yes. The method does not rely on the rationality of $p_i$, but only on the $p_i$ finding a root. – Will Sawin May 19 '12 at 2:04
• I didn't really show that, so much as say that. It's true because the set of polynomials with rational roots is Euclidean-dense in the set of polynomials with all real roots whose tangent space at most points, tensored with $\mathbb C$, is the tanegnt space of the whole space of polynomials. – Will Sawin May 19 '12 at 2:19