I am reading Paolo Ruffini's response in 1803 (see here) to a letter in 1802 from Pietro Abbati to Ruffini (see here), Abbati's letter suggesting improvements to Ruffini's lengthy 1799 'General Theory of Equations' (see here) in which Ruffini first proposes a proof that there is no formula for solving a general quintic. My question involves a passage in Ruffini's response where he states and gives a proof for the result below, my question being whether Ruffini's result and proof are basically correct or whether there are hidden assumptions which would be needed in a modern proof of the same result.
To give some context, following on from Lagrange's 1770 analysis of solving quadratics, cubics and quartics, Ruffini's result forms part of his analysis of transforming a polynomial $x^n+ax^{n-1}\ldots=0$ by taking a rational function of its roots $f(x_1,x_2,\ldots,x_n)$, finding how many different expressions are formed by this function when the roots are permuted in all possible ways, say $y_1,y_2,\ldots$, and forming the new polynomial $(y-y_1)(y-y_2)\ldots$. For example, if $f(x_1,x_2,x_3)$ creates just two fundamentally different expressions when $x_1,x_2,x_3$ are permuted in all $3!=6$ ways (such as $f(x_1,x_2,x_3)=(x_1+\alpha x_2+\alpha^2 x_3)^3$ where $\alpha$ is an imaginary cube root of 1) then the new polynomial in $y$ will be a quadratic. (Lagrange and Ruffini show this is actually the start of the method of solving a general cubic, and while the calculations may be difficult, it is always possible to find the coefficients of the transformed polynomial as the coefficients are symmetric functions of $y_1,y_2,\ldots$ which are functions of $x_1,x_2,\ldots$ permutated in all of the ways which produce different results, and so the coefficients of the transformed are symmetric function of $x_1,x_2,\ldots$ and so can be expressed in terms of the elementary symmetric functions of $x_1,x_2,\ldots$ which can be read off from the coefficients of the original polynomial in $x$.)
Ruffini's argument for there being no formula for a general quintic in his 1799 'General Theory' rests on analysing the structure of $S_5$, i.e. the 120 permutations of the five roots of a general quintic $x_1,x_2,\ldots,x_5$, showing that there is no set of transformations such as the above which lead to a quadratic, cubic or quartic from which one could extract fifth roots, the latter being needed in order to obtain five roots in general for a quintic. He originally just states that it is no use transforming to any polynomials higher than a quartic as there are no known general solutions to such equations, but a possible objection might be that it could be useful to transform to a sextic, say, specifically designed to be solvable by factorisation. Transforming to a sextic seems possible as there are permutation groups in $S_5$ of order 20, leading to the possibility of creating a rational function of five variables which produces $5!\div 20=6$ different expressions when the variables are permuted. The following result from his 1803 reply to Abatti, rules out this out this sextic being factorisable, however, i.e. showing that this route to solving a general quintic won't work.
Incidentally, I'm aware that Ruffini is assuming that the function $f$ is a rational function, something which needs Abel's later work in order to complete Ruffini's overall argument, but this assumption of the rationality of $f$ is explicitly stated in the following and so I assume isn't an unwarranted assumption in terms of the result alone (I'd welcome any comments on this, though).
So, to get to the result in question: in Section 28 of his reply to Abbati, Ruffini says that if we start with a general polynomial $$\tag{1} x^n+Ax^{n-1}+Bx^{n-2}+\ldots=0$$ and transform by a suitable rational function $f(x_1,x_2,\ldots,x_n)$ as described above to another polynomial $$\tag{2} y^m+My^{m-1}+Ny^{m-2}+\ldots=0$$ then the latter cannot be factorised, i.e. doesn't have a factor of the form $$\tag{3} y^r+Py^{r-1}+\ldots+Ty+U=0$$ where $0<r<m$ and $P,\ldots,T,U$ are rational functions of the coefficients $M,N,\ldots$ (the reason for the slightly different form of $(3)$ is seen next).
He justifies this by the steps:
(a) If $(2)$ has a factor of the form $(3)$ then of the $m$ roots of $(2)$ there are $r$ which are roots of $(3)$, calling the latter $y_1,y_2,\ldots,y_r$ and the remaining roots of $(2)$ $y_{r+1},y_{r+2},\ldots,y_m$. As the quintic is general, all of these roots are assumed different to each other.
(b) As $y_1$ is a root of $(3)$ we have $$\tag{4}(y_1)^r+P(y_1)^{r-1}+\ldots+Ty_1=-U$$
(c) As $y_{r+1}$ is NOT a root of $(3)$, given that all of the $r$ roots of $(3)$ are one of $y_1,y_2,\ldots,y_r$, we have $$\tag{5}(y_{r+1})^r+P(y_{r+1})^{r-1}+\ldots+Ty_{r+1}\ne -U$$
(d) Now $y_1$ and $y_{r+1}$ are both rational functions of $x_1,x_2,\ldots,x_n$, with $y_{r+1}$ obtainable from $y_1$ by a permutation of the $x_1,x_2,\ldots,x_n$ contained in $y_1$. Let this permutation be $\sigma$.
(e) Applying $\sigma$ to the left hand side of $(4)$ produces $(y_{r+1})^r+P(y_{r+1})^{r-1}+\ldots+Ty_{r+1}$ which we know from (c) is not equal to $-U$.
(f) However, applying $\sigma$ to the right hand side of $(4)$ leaves the expression unchanged given that $-U$ is a rational function of the coefficients $M,N,\ldots$ which are symmetric functions of $x_1,x_2,\ldots,x_n$.
(g) This implies that the result of applying $\sigma$ to the left hand side of $(4)$ result in an expression that both does and does not equal $-U$ which is a contradiction, showing that it is not possible to find a factor such as that in $(3)$.
My thoughts are that the main assumption, and the key to the whole argument, seems to be step (e), i.e. the statement that applying the permutation $\sigma$ to the left hand side of $(4)$ leads to $y_1$ being replaced with $y_{r+1}$ (i.e. essentially that $\sigma$ is a $\mathbb{C}$-automorphism). Given that Ruffini is treating the left hand side of $(4)$ as a rational function of $x_1,x_2,\ldots,x_n$ this seems valid to me, but I'd welcome any thoughts.
The other assumption I can see is that a general polynomial of degree $n$ having $n$ roots was not proved rigorously until Argand in 1806, i.e. after Ruffini's reply to Abbati, but it seems that the result was widely accepted by mathematicians at the time of Ruffini's writing.
As I say, any thoughts on the validity of the result and proof (noting again that the result only needs to apply to rational functions of the roots $x_1,x_2,\ldots,x_n$), or ideas of how it translates into modern algebra, would be most welcome.
