# implication of the Abel–Ruffini theorem

I am taking a course in abstract algebra, and we proved the following theorem: I want to prove something more specific. Let's look at polynomials of degree 5 over C. Someone is claiming he has a magic formula, which receives the coefficients of a polynomial of degree 5, and returns its roots using only basic operations and radicals. I want to understand, how I can prove this person wrong using the theorem above. In this case, the theorem talks about the field of rational functions with 5 variables over C. It shows that I can't express $$t_1, ..., t_5$$ (the roots of f) in terms of $$s_1,...,s_5$$, in this abstract field. I understand the proof in this context, but I want to understand how I can use it concretely in order to prove this person wrong. In the sources that I have seen, they say that Abel–Ruffini theorem implies what I want to prove, but they don't show how. Can someone help me understand how you can show this? I am adding the proof we saw in the course: • It seems clear: the corollary you quote asserts there exists a general formula if and only if the degree is no more than $4$. Jul 12, 2020 at 15:39
• @Bernard I am trying to understand how the formal corollary, which talks about the field of rational functions with d variables over C, implies that there is no formula for polynomials over C Jul 12, 2020 at 15:43
• I see – I misunderstood what you were asking. I think it's a matter of Galois group (there exist quintic equations which are solvable by radicals, and there is a Maple module which solves them when they are. If I remember well, it's a matter of the Galois group of the polynomial being metacyclic). Jul 12, 2020 at 15:49
• Switch to Galois. Jul 12, 2020 at 16:51

Nice question. I was annoyed by the same thing after learning Galois theory. Before learning it I thought that I will see a proof that there is no formula which works specifically for polynomials over $$\mathbb{C}$$, but instead of that I only saw a proof that there is no formula which works for polynomials over the field $$\mathbb{C}(t_1,...,t_n)$$, which is a much bigger field. I was very disappointed for a while, but lucky for me I managed to think of a proof myself.

Suppose there is a formula which works for polynomials of degree $$n\geq 5$$ over $$\mathbb{C}$$. The formula contains the field operations, taking roots, the coefficients of a polynomial and some complex constants. (for example the quadratic formula $$\frac{-b+\sqrt{b^2-4ac}}{2a}$$ uses the constants $$2,4,...$$). The main thing we should note is that the formula can contain only finitely many constants. Let's call them $$z_1,...,z_k$$. So our formula is actually a formula over the field $$K:=\mathbb{Q}(z_1,...,z_k)$$. Note that this field is countable, since it is finitely generated over $$\mathbb{Q}$$. Since $$\mathbb{C}$$ is uncountable there must be an element $$t_1\in\mathbb{C}$$ which is transcendental over $$K$$. Again, $$K(t_1)$$ is countable, so there is $$t_2\in\mathbb{C}$$ which is transcendental over $$K(t_1)$$. We continue this way, and finally get a field $$K(t_1,...,t_n)$$ where $$t_i$$ is transcendental over $$K(t_1,...,t_{i-1})$$ for all $$i$$. Now define the polynomial $$f=(x-t_1)...(x-t_n)$$ and call its symmetric functions $$s_1,...,s_n$$. (the coefficient of $$x^n$$ is $$1$$). Finally, let $$L=K(s_1,..,s_n)$$, this is a subfield of $$K(t_1,...,t_n)$$.

Now, what can we say about $$L$$? I claim that every polynomial of degree $$n$$ in $$L[x]$$ is solvable by radicals. Why? Well, take a polynomial $$g\in L[x]$$ of degree $$n$$ and put its coefficients in the formula we have. The coefficients of $$g$$ are obviously in $$L$$, and remember that the constants in the formula belong to $$K\subseteq L$$, so they are in $$L$$ as well! So this shows $$g$$ is solvable by radicals over $$L$$.

But now let's go back to the polynomial $$f=(x-t_1)(x-t_2)...(x-t_n)\in L[x]$$. As we showed above it must be solvable by radicals over $$L$$, so its Galois group over $$L$$ is solvable. On the other hand, using the fact that $$t_1,...,t_n$$ are algebraically independent over $$K$$ (which means $$t_i$$ is transcendental over $$K(t_1,...,t_{i-1})$$ for all $$i$$) we can conclude that $$Gal(K(t_1,...,t_n)/K(s_1,...,s_n))\cong S_n$$, this is exactly the same proof as the proof that the Galois group of the general polynomial in the field of rational functions is $$S_n$$. This means the Galois group of $$f$$ over $$L$$ is $$S_n$$, a contradiction.

The difference is that in my proof $$t_1,...,t_n$$ are all complex numbers, and not just formal variables like in the field of rational functions. So here $$f=(x-t_1)...(x-t_n)$$ is a specific polynomial over $$\mathbb{C}$$ for which the formula we took fails.

• hey, thank you for the answer. I think you need to define the polynomial as $f\left(t\right)=t^{n}-t_{1}\cdot t^{n-1}+...+\left(-1\right)^{n}\cdot t_{n}$ , since you don't want the roots to be in the field, just the coefficients Jul 12, 2020 at 16:32
• Yeah, you are right. I already edited my answer. The important field is indeed $L=K(s_1,...,s_n)$. The field $K(t_1,...,t_n)$ is the splitting field.
– Mark
Jul 12, 2020 at 16:33
• maybe it's just not updating for me, but I think there are still places you need to switch between s and t. Either way, I understood what you meant, thank you Jul 12, 2020 at 16:41
• I think as it is written now it is fine. The roots $t_1,...,t_n$ are algebraically independent, and this can be used to prove that $Gal(K(t_1,...,t_n)/K(s_1,...,s_n))\cong S_n$. The algebraic independence implies that any permutation of the roots can be extended to an automorphism.
– Mark
Jul 12, 2020 at 16:47
• we actually get that the field K is algebraically closed and therefore contains the algebraic closure of Q. If we show that the algebraic closure of Q can't be generated by a finite amount of elements (which is a bit tricky because the elements here aren't necessarily algebraic) we can get an alternative proof Jul 12, 2020 at 17:41