4
$\begingroup$

I am taking a course in abstract algebra, and we proved the following theorem: enter image description here

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:

enter image description here

$\endgroup$
4
  • $\begingroup$ It seems clear: the corollary you quote asserts there exists a general formula if and only if the degree is no more than $4$. $\endgroup$
    – Bernard
    Jul 12, 2020 at 15:39
  • $\begingroup$ @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 $\endgroup$
    – BinyaminR
    Jul 12, 2020 at 15:43
  • $\begingroup$ 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). $\endgroup$
    – Bernard
    Jul 12, 2020 at 15:49
  • $\begingroup$ Switch to Galois. $\endgroup$
    – markvs
    Jul 12, 2020 at 16:51

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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 $\endgroup$
    – BinyaminR
    Jul 12, 2020 at 16:32
  • $\begingroup$ 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. $\endgroup$
    – Mark
    Jul 12, 2020 at 16:33
  • $\begingroup$ 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 $\endgroup$
    – BinyaminR
    Jul 12, 2020 at 16:41
  • 1
    $\begingroup$ 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. $\endgroup$
    – Mark
    Jul 12, 2020 at 16:47
  • $\begingroup$ 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 $\endgroup$
    – BinyaminR
    Jul 12, 2020 at 17:41

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.