It is well known that one cannot solve every polynomial equation over $\Bbb Q$ using just radicals. In other words, let $A_n = \{x^n - a\mid a \in \Bbb Q\}$, $A = \cup_n A_n$ and $\bar{\Bbb Q}$ the algebraic closure of $\Bbb Q$.

Then we know that the splitting field of $A$ over $\Bbb Q$ is not all of $\bar{\Bbb Q}$. Given this, is there a proper subset $S$ of all irreducible polynomials over $\Bbb Q$ whose splitting field is indeed all of $\bar{\Bbb Q}$?

Can we say anything non trivial about such sets(like finding a minimum such one if it exists)? Can we define this set by induction on the degree of the polynomial? For instance, square roots are enough to solve quadratic equations and so on till fifth degree where radicals no longer suffice.

There are a lot of questions here but I am only interested in understanding these sets in general and only suggest the questions as possibly interesting.

  • $\begingroup$ You're welcome, cheers. $\endgroup$ – Travis May 6 '15 at 13:52
  • $\begingroup$ A trivial answer is: yes, there is such a set and you can define it as the set S of all irreducible polynomials $f$ over $\mathbb Q$ such that $\mathbb Q[x]/(f)$ is Galois over $\mathbb Q$. Note that morally this set is "much smaller" than the set of all irreducible polynomials, since the generic condition for a polynomial of degree $n$ is to have Galois group $S_n$. $\endgroup$ – Ferra May 6 '15 at 14:09
  • $\begingroup$ @Ferra Is that easy to see? $\endgroup$ – Asvin May 6 '15 at 15:42
  • $\begingroup$ Yes it is, because the roots of every irreducible $f$ lie in the normal closure of $\mathbb Q[x]/(f)$, which is a finite Galois extension of $\mathbb Q$! $\endgroup$ – Ferra May 7 '15 at 7:55

Even for polynomials up to degree four, note that splitting the polynomials $x^2-a$, $x^3-a$, $x^4-a$ for $a \in \mathbb Q$ is not sufficient - for Cardano's formulas for third degree for example, we may have to take the third root of an element already living in some quadratic extension of $\mathbb Q$. What happens for the soluble-by-radical equations is that
1. their Galois groups (by definition) only contain cyclic factors in their decomposition series,
2. cyclic groups correspond (via Kummer extensions) to equations soluble by radicals.

So, to find a “complete” set of polynomials, you have to transfer both 1. and 2. above to the full set of polynomials. Even question 1. above is hard (it is known as inverse Galois theory: which finite groups appear as Galois groups over the rationals? I don't believe the answer is known in full generality, or even when restricted to simple finite groups - for example, is the Monster group a Galois group over the rationals ?). Question 2. is even less well-defined. One could think of point division on elliptic curves to generate extensions with Galois group $\mathrm{PGL}_2 (\mathbb Z/n\mathbb Z)$.

  • $\begingroup$ So, an element of the third degree can also be of the second degree. I don't believe it. $\endgroup$ – Piquito May 6 '15 at 18:21
  • $\begingroup$ What I mean is that for solving, e.g., $x^3-12x-24$, taking cube roots in $\mathbb Q$ is not enough since the real root is $\sqrt[3]{12+4\sqrt{5}}+\sqrt[3]{12-4\sqrt{5}}$. $\endgroup$ – Circonflexe May 7 '15 at 7:28

What you want implies that some irreducible f not in S (and whose roots are obviously in $\bar{\Bbb Q}$) is such that it has a root belonging to the union of some splitting fields of irreducible $g_1$, $g_2$,….$g_n$ each belongs to S – {f} . This is not possible, because two distinct irreducible polynomial over Z have its roots Q-linearly independent each other. (It is very hard for me to write in English; sorry if not well explained). My answer is Not possible, you need all of S.

  • $\begingroup$ The roots of $x^2 + x + 1$ and $x^2 + 3$ are not linearly independent over $\Bbb Q$, are they? Sorry if I misunderstood you! $\endgroup$ – Asvin May 6 '15 at 19:04
  • $\begingroup$ You are right.....and I was wrong. But this simple example, is it not enough to answer YES to your question?. I should have added "f not generating the same field of the other" but if this is possible your YES is in hand. A little almost objection however, the roots of $x^2$ + x +1 are units of the corresponding field of $x^2$ + 3. Regards. $\endgroup$ – Piquito May 6 '15 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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