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I'm trying to understand Wantzel's original proof of the necessary condition for constructibility with a straightedge and compass. It's expressed in terms of polynomials rather than field extensions. It's outlined on page 7 of this PDF.

There's a point that I don't understand (I've read the original 1837 paper and it's no clearer there). Wantzel has a finite sequence of quadratic polynomials such that the coefficients of each polynomial are rational functions of the roots of all of the previous equations. He now wants to basically "fold up" the sequence into a single polynomial of degree $2^n$.

I'm going to introduce a little bit of notation here and use $R$ to indicate an arbitrary rational function, just like how $o(x)$ means an arbitrary function that's infinitesimal compared to $x$. In particular, $R$ may represent two distinct rational functions even within the same formula.

If the last two equations are:

$$\begin{align} R(x_1...x_{n-2})x_{n-1}^2+R(x_1...x_{n-2})x_{n-1}+R(x_1...x_{n-2})&=0 \\ R(x_1...x_{n-1})x_n^2+R(x_1...x_{n-1})x_n+R(x_1...x_{n-1})&=0 \end{align}$$

Then what Wantzel does is take the two roots of the first equation, and consider the two possible values that the LHS of the second equation might take on when these two values are substituted for $x_{n-1}$. He then multiplies them together, and claims that this eliminates the $x_{n-1}$, in other words that he now has a polynomial of degree $4$ where each coefficient is $R(x_1...x_{n-2})$. Note that Wantzel previously claimed that all of the rational functions in the problem can be assumed to be linear functions.

Now of course, if you multiply together the two roots to a quadratic equation with coefficients in a given field $F$, you'll obtain a value in $F$. But here the two roots are wrapped up inside rational functions, which are then used as coefficients in quadratic polynomials, and yet somehow this still allegedly works. Explicit calculation seems to indicate that it's false in general, which makes me think I might have misunderstood the proof.

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    $\begingroup$ The answer to the question in the title is clearly no: take $x=\pi$, $y={1\over \pi}$, $R(a)=a$, $P(a)=a^2$. $\endgroup$ – Noah Schweber Oct 22 '13 at 20:35
  • $\begingroup$ @user28111 True. I was struggling to come up with a way to succinctly express the problem. $\endgroup$ – Jack M Oct 22 '13 at 20:41
  • $\begingroup$ Could you make clearer in the body what your question is? Also, I think a broader name (e.g., "Question about Wantzel's proof of the necessary condition for compass/straightedge constructibility") would be better than a specific-but-misleading name. $\endgroup$ – Noah Schweber Oct 22 '13 at 20:43
  • $\begingroup$ @user28111 I'm not sure I can make it clearer because it really boils down to "what's going on in this relatively tricky proof?". If I were attempt to phrase it as a specific technical question, I'd probably get it wrong, as my blunder with the earlier question title demonstrates. $\endgroup$ – Jack M Oct 22 '13 at 21:01
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    $\begingroup$ Rotman does it in his Galois Theory book. It may or may not be identical with Wantzel, but useful. $\endgroup$ – Will Jagy Oct 22 '13 at 21:13
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Let $A(x), B(x), C(x)$ be rational functions in a field $K$. Now let $a, a'$ be the roots of an irreducible quadratic equation over $K$. We'll just write $A, B, C$ for $A(a), B(a)$ and $C(a)$, and $A', B', C'$ for $A(a'), B(a'), C(a')$.

Now consider the equation

$$Ax^2+Bx+C=0$$

And modify it to get this equation:

$$(Ax^2+Bx+C)(A'x^2+B'x+C')=0 \tag 1$$

The left hand side can be written:

$$AA'x^4+(AB'+BA')x^3+(AC'+BB'+CA')x^2+(BC'+CB')x+CC'$$

The coefficients of this polynomial are clearly symmetric in $a, a'$. We could have predicted this without actually distributing out (1), since to get the coefficients after swapping $a$ and $a'$, we could simply swap $a$ and $a'$ in (1) and then distribute it out, but that will clearly give the same coefficients as just distributing out (1), since (1) is symmetric in $a$ and $a'$.

Since the coefficients are symmetric in $a$ and $a'$, they lie in $K$, so that (1) gives us an equation (of degree 4) for $x$ with coefficients in $K$.

This is precisely what Wantzel is doing. The last equation in his system is $Ax_n^2+Bx_n+C=0$, where $A, B$ and $C$ are rational in $x_{n-1}, ... x_1$, and $x_{n_1}$ is quadratic over the field $K$ generated by $x_{n-2}, ... x_1$. So he swaps $x_{n-1}$ in $A, B, C$ with its conjugate to get $A', B', C'$ and does exactly what we did above. Iterating, he obtains a rational equation for $x_n$ of degree $2^n$.

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