# 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 $b_1$..." May 19, 2012 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, 2012 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 $p_2$. May 19, 2012 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, 2012 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, 2012 at 5:45

No.

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, 2012 at 1:55
• Yes. The method does not rely on the rationality of $p_i$, but only on the $p_i$ finding a root. May 19, 2012 at 2:04
• @Will: It may also serve as an answer to the original question of Joseph O'Rourke. It shows that there are no polynomial conditions defining the set of polynomials with all roots rational if the degree is at least 5 (because that set is Zariski dense).
– user6976
May 19, 2012 at 2:13
• 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. May 19, 2012 at 2:19
• @Will: You may want to add this answer to Joseph's question.
– user6976
May 19, 2012 at 5:27