How to state the "unsolvability of quintic by radicals" in 2nd and higher order logic?

Question

I am attempting to express the sentence

"There does not exist a general solution for the 5th degree polynomial equation in radicals"

In 2nd order logic.

So I made the statement clearer by going through a sequence of "specifications" where I take the english statement and make it more rigorous.

There does not exist a general radical expression $Y(a,b,c,d,e,f)$ such that $aY^5 + bY^4 + cY^3 + dY^2 + eY + f = 0$

$\not\exists $ general radical expression $Y(a,b,c,d,e,f)|\forall a,b,c,d,e,f \in \mathbb{Z}, aY^5 + bY^4 + cY^3 + dY^2 + eY + f = 0$

And here is where it gets trick, how do I formalize the notion of a general radical expression? One idea is to create a set $S$ which basically carries the property that $(k \in \mathbb{N}\wedge \mu^k \in S \rightarrow \mu \in S) \wedge (\mathbb{Q} \subseteq S \wedge a,b,c,d,e,f \in S) $ That is any "radical" expression of rationals and the symbols $a,b,c,d,e,f$ is in $S$ giving me

$\not\exists \ Y \in S | (k \in \mathbb{N}\wedge \mu^k \in S \rightarrow \mu \in S) \wedge (\mathbb{Q} \subseteq S) \wedge (a,b,c,d,e,f \in S) |\forall a,b,c,d,e,f \in \mathbb{Q}, aY^5 + bY^4 + cY^3 + dY^2 + eY + f = 0$

But this now leads to a couple issues. How do I make it clear that $Y$ is a function that accepts $a,b,c,d,e,f$ as arguments, and that the initial $a,b,c,d,e,f$ are symbols. And furthermore, is this really the most succint/intuitive way to encode this statement in 2nd order logic? Is there a better way to express the sentence in say 3rd order logic or higher?

Answer 1

What you're looking for is simply to treat the coefficients as symbols (see this post for much more detail), in the same way the indeterminate in a polynomials should be treated as a symbol for the parameter rather than some actual object or number. Just as we use the notation "$R[x]$" for the ring of univariate polynomials over a ring $R$, we use the notation "$F(x)$" for the field of univariate rational functions over a field $F$, which is simply defined as the fraction field of $F[x]$. Similarly one can define $F(x,y)$ as the fraction field of $F[x,y]$, and likewise for any number of indeterminates. (All this is when $x$ is an indeterminate, though $F(x)$ is in fact isomorphic to $F(r)$ for any $r$ that is transcendental over $F$ in some larger field.)

Then define that $K$ is a simple radical extension of a field $F$ iff $K = F(r)$ for some $r \in K$ such that $r^n \in F$. And define a radical extension of $F$ to be a field that is the result of finitely many simple radical extensions.

This is exactly what you are looking for. A radical expression in $a,b,c,d,e,f$ over the rationals is precisely a member of some radical extension of $\mathbb{Q}(a,b,c,d,e,f)$. The unsolvability of the generic quintic (as closest to that in your question) therefore states:

1. No $y$ in any radical extension of $\mathbb{Q}(a,b,c,d,e,f)$ is such that "$ay^5+by^4+cy^3+dy^2+ey+f$" has value $0$ for every $a,b,c,d,e,f \in \mathbb{Q}$. (The value of such expressions has to be recursively defined. In particular, we would have to define evaluation inside-out and the value of "$\sqrt[n]{a}$" to be the primitive $n$-th root of $a$.)

The above definition is in-between two common presentations of the unsolvability of the quintic:

2. No radical extension of $\mathbb{Q}(a,b,c,d,e,f)$ contains the field generated by adjoining a root of "$ax^5+bx^4+cx^3+dx^2+ex+f$" to $\mathbb{Q}(a,b,c,d,e,f)$, where $a,b,c,d,e,f$ are indeterminates.

3. No radical extension of $\mathbb{Q}$ contains the field generated by adjoining to $\mathbb{Q}$ some complex root of some rational quintic.

We can easily prove that (3) implies (1) and that (1) implies (2), but (2) does not easily imply (3), because it is conceivable that every single rational quintic can be solved by radical expressions but there is no single formula that works for all quintics. Of course, (3) is true and arguably easiest to prove, since we just have to give an explicit polynomial such as "$x^5-4x+2$" and prove that the Galois group of the corresponding extension is not solvable.

For completeness, here is a sketch for the above polynomial. Firstly, it is irreducible by Eisenstein's criterion. Also it has exactly $2$ complex roots, and hence the conjugation automorphism is a $2$-cycle on its roots. The Galois group $G$ has a $5$-cycle that maps one of them to the other, and hence must be $S_5$, which is not solvable.

Answer 2

Yes, you have to think about this. Radical expressions include $$ \left(\;1+5\sqrt{2}\;\right)^{1/3} $$ so you have to allow multiple steps, each one involving radicals of the previous one.