It is a famous theorem that there is no "quintic formula", i.e. there is no formula which expresses the roots of a quintic polynomial $x^5+a_4x^4+\cdots+a_0$ in terms of $a_4,\ldots,a_0$ and rational constants using the field operations and taking $n$th roots. But the standard proof of this actually proves something which on its face seems stronger: that there are particular quintics with roots that cannot be expressed in terms of rational numbers using these operations. The non-existence of a quintic formula merely requires that there be no "uniform" way of doing so. I'm wondering if this is actually a stronger statement.

To make this rigorous: Given $f(a_0,\ldots,a_k,x)\in \mathbb Q(a_0,\ldots,a_k)[x]$, we can define the family of polynomials associated to $f$ to be the set of polynomials $f(t_0,\ldots,t_k,x)\in \mathbb Q[x]$ where $t_0,\ldots,t_k\in \mathbb Q$ are such that this expression is defined. For example, the family of polynomials associated to $x^2+a_1x+a_0$ is the set of monic quadratic polynomials. If there an $f\in \mathbb Q(a_0,\ldots,a_k)[x]$ which does not admit a formula for its roots (in the variable $t$) in terms of elements of $\mathbb Q(a_0,\ldots,a_k)$, field operations and taking $n$th roots, yet every polynomial in the family of polynomials associated to $f$ does?

From the point of view of Galois theory, the question becomes whether there exist polynomials $f\in \mathbb Q(a_0,\ldots,a_k)[x]$ which are not solvable over $\mathbb Q(a_0,\ldots,a_k)$, but evaluating them at any point $(t_0,\ldots,t_k)\in\mathbb Q$ (for which the result is defined) gives a polynomial solvable over $\mathbb Q$?

  • $\begingroup$ Very interesting! Perhaps it would be better to consider the question in the setting of commutative algebra: assume $f$ is monic, irreducible, and separable; let $A$ be the subring of the coefficient field generated by $\mathbb{Q}[a_0, \ldots, a_k]$ and the coefficients of $f$, and consider the $A$-algebra $A [x] / (f)$. Then the automorphism group of $A [x] / (f)$ (as an $A$-algebra) will also act on each of the specialisations $A [x] / (f) \otimes_A A / \mathfrak{m}$, for each maximal ideal $\mathfrak{m}$ of $A$, and the "generic" situation is obtained by considering $\mathfrak{p} = (0)$. $\endgroup$ – Zhen Lin May 18 '13 at 16:02
  • $\begingroup$ @ZhenLin I'm not familiar with this approach. Can you point me towards some references? $\endgroup$ – Alex Becker May 18 '13 at 17:50
  • $\begingroup$ It's basically a translation of your problem into algebraic geometry. Rather than thinking of $f(a_1, \ldots, a_n; x)$ as a family of one-variable polynomials, think of it as just one; then it defines a hypersurface $V$ in $U \times \mathbb{A}^1$ where $U$ is an open subset of $\mathbb{A}^n$, and what you want to do is to study the properties of the projection $V \to U$ in terms of the generic fibre. $\endgroup$ – Zhen Lin May 18 '13 at 18:07

The answer is no. The following is a theorem of Hilbert:

Let $f(x_1, x_2, \ldots, x_n, t)$ be a polynomial with rational coefficients. Let $K_{gen}$ be the splitting field of $f$ over $\mathbb{Q}(x_1, \ldots, x_n)$, thought of as a polynomial in $t$. For $a_1$, ..., $a_n$ in $\mathbb{Q}$, let $K(a_1, \ldots, a_n)$ be the splitting field of $f(a_1, \ldots, a_n, t)$ over $\mathbb{Q}$. There are infinitely many $(a_1,\ldots, a_n)$ for which $\mathrm{Gal}(K_{gen}/\mathbb{Q}(x_1, \ldots, x_n)) \cong \mathrm{Gal}(K(a_1, \ldots, a_n)/\mathbb{Q})$.

Infinitely many is a very weak statement; the right statement is that this is true for all but a "thin set" of $n$-tuples of rational numbers. There are many theorems saying that thin sets are small. See this question for some more discussion in the case that $G$ is $S_n$.

The theorem which gets the famous name is Hilbert's Irreducibility Theorem: If $f(x_1, \ldots, x_n, t)$ is irreducible as a polynomial in $\mathbb{Q}(x_1, \ldots, x_n)[t]$ then, for all most all $(a_1, \ldots, a_n) \in \mathbb{Q}^n$, the polynomial $f(a_1, \ldots, a_n)(t)$ is irreducible as a polynomial in $\mathbb{Q}[t]$. The deduction of the result I stated from this one is pretty straightforward (it is the first application of Hilbert's irreducibility theorem mentioned in Wikipedia) but I had trouble finding you a reference that writes it out carefully.

If you are happy with the language of modern algebraic geometry, I recommend the first three chapters of these notes by Serre. If you are willing to read something lengthy, this Master's thesis looks like a beautiful winding tour through Hilbert's Irreducibility Theorem and its many applications

  • $\begingroup$ I can see how to deduce the desired result from Hilbert's Irreducibility Theorem now that it's pointed out to me. Thanks. $\endgroup$ – Alex Becker Aug 24 '13 at 21:00

It appears to me the answer is essentially no, but there are some details to be checked. The main observation is this: under reasonable conditions, the Galois group of the generic fibre is canonically isomorphic to the Galois group of almost all fibres; thus, the generic fibre is solvable if and only if almost all fibres are solvable.

In more detail: Let $L$ be a finite Galois extension of $K$. By the primitive element theorem, $L$ is generated by a single element over $K$, say $x$. Let $f$ be the (monic) minimal polynomial of $x$ over $K$ and choose a subring $A \subseteq K$ satisfying these conditions:

  • $A$ is a noetherian integral domain.
  • $\operatorname{Frac} A = K$.
  • $f \in A [x]$.
  • The ring $B = A [x] / (f)$, considered as a subring of $L$, is closed under the action of $\mathrm{Gal}(L \mid K)$.
  • $B$ is a projective $A$-module of rank $d = \deg f$.

This is certainly achievable in the case $K = \mathbb{Q} (t_1, \ldots, t_n)$ that you are considering: take $A$ to be the subring generated by $\mathbb{Q} [t_1, \ldots, t_n]$, the coefficients of $f$, and the coefficients of all the Galois conjugates of $x$; then $B$ will even be a free $A$-module of rank $d$.

Lemma 1. $\mathrm{Aut}(B \mid A)$, the automorphism group of $B$ as an $A$-algebra, is canonically isomorphic to $\mathrm{Gal}(L \mid K)$.

Proof. The hypothesis implies that each automorphism of $L \mid K$ restricts to an automorphism of $B \mid A$; since $B$ contains the generator of $L$, the restriction homomorphism $\mathrm{Gal}(L \mid K) \to \mathrm{Aut}(B \mid A)$ is injective. On the other hand, $\operatorname{Frac}$ is functorial, so any automorphism of $B \mid A$ extends to an automorphism of $L \mid K$, so the restriction homomorphism is in fact an isomorphism. ◼

Lemma 2. Let $\mathfrak{m}$ be a maximal ideal of $A$ and let $\kappa (\mathfrak{m}) = A / \mathfrak{m}$ be the residue field.

  1. $B (\mathfrak{m}) = B \otimes_A \kappa (\mathfrak{m})$ is a $\kappa (\mathfrak{m})$-algebra of dimension $d$.
  2. $B (\mathfrak{m})$ is a field if and only if $f$ is irreducible over $\kappa (\mathfrak{m})$.
  3. There is a canonical homomorphism $\mathrm{Aut}(B \mid A) \to \mathrm{Aut}(B (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$, and it is injective if and only if distinct Galois conjugates of $x$ remain distinct in $B (\mathfrak{m})$.

Proof. All straightforward. ◼

Lemma 3. There exists a dense open subset $U \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ that are in $U$, the Galois conjugates of $x$ remain distinct in $B (\mathfrak{m})$.

Proof. Two Galois conjugates of $x$, say $x'$ and $x''$, become equal in $B (\mathfrak{m})$ precisely if a certain finite set of equations holds in $B (\mathfrak{m})$; more precisely, there exist elements $h_1, \ldots, h_d$ in $B$ such that $x' \equiv x'' \pmod{\mathfrak{m}}$ if and only if $h_1 \equiv \cdots \equiv h_d \equiv 0 \pmod{\mathfrak{m}}$. This is because the $A_{\mathfrak{m}}$-module $B_{\mathfrak{m}}$ is free of rank $d$ (since $B$ is a projective $A$-module of rank $d$). Thus $x' \equiv x'' \pmod{\mathfrak{m}}$ if and only if $\mathfrak{m}$ is in a certain closed subset of $\operatorname{Spec} A$ of codimension $\ge 1$ (by Krull's Hauptidealsatz). The complement of such a closed subset is a dense open subset of $\operatorname{Spec} A$ (because $A$ is an integral domain), and the intersection of finitely many dense open subsets in $\operatorname{Spec} A$ is again a dense open subset, so we are done by the finiteness of $\mathrm{Gal}(K \mid L)$. ◼

Theorem. If there exists a dense open subset $U' \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ in $U'$, $f$ is irreducible over $\kappa (\mathfrak{m})$, then there exists a dense open subset $U'' \subseteq \operatorname{Spec} A$ such that, for all maximal ideals $\mathfrak{m}$ of $A$ in $U''$, $B \otimes_A \kappa (\mathfrak{m})$ is a Galois extension of $\kappa (\mathfrak{m})$ and the canonical homomorphism $$\mathrm{Gal}(K \mid L) \cong \mathrm{Aut}(B \mid A) \to \mathrm{Gal}(B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$$ is an isomorphism.

Proof. Let $U$ be as in lemma 3, and take $U'' = U \cap U'$. This is a dense open subset, and by lemma 2, for any $\mathfrak{m}$ in $U''$, $B \otimes_A \kappa (\mathfrak{m})$ is a field and the canonical homomorphism is injective. However, Dedekind's lemma says that the size of $\mathrm{Aut}(B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m}))$ is bounded above by the degree of $B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m})$, so in fact the canonical homomorphism is a bijection, and moreover $B \otimes_A \kappa (\mathfrak{m}) \mid \kappa (\mathfrak{m})$ is a Galois extension. ◼

Remark. It's not clear to me when the main hypothesis of the above theorem holds. There is definitely something non-trivial to be checked: after all, if we take $K = \mathbb{Q}$ and $A = \mathbb{Z}$, then we have to contend with the fact that there exist polynomials that are reducible mod $p$ for every prime $p$ but yet irreducible over $\mathbb{Q}$. But perhaps in the case you are considering this cannot happen.

  • $\begingroup$ Thanks, this looks very interesting. It will take me some time to digest it however; I've never seen this approach to Galois theory before. $\endgroup$ – Alex Becker May 18 '13 at 20:24
  • $\begingroup$ +1 I think I have an intuitive handle on what is going on here. I'll have to think about whether I can find such a set $U'$ in my case. $\endgroup$ – Alex Becker May 18 '13 at 20:37

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