Direct proof that $\pi$ is not constructible

Is there a direct proof that $\pi$ is not constructible, that is, that squaring the circle cannot be done by rule and compass?

Of course, $\pi$ is not constructible because it is transcendental and so is not a root of any polynomial with rational coefficients. But is there a simple direct proof that $\pi$ is not a root of polynomial of degree $2^n$ with rational coefficients?

The kind of proof I seek is one by induction on the height of a tower of quadratic extensions, one that ultimately relies on a proof that $\pi$ is not rational. Does any one know of a proof along these lines or any other direct proof?

I just want a direct proof that $\pi$ is not constructible without appealing to transcendence.

• So far as I know, there is no proof that $\pi$ doesn't have degree $2^n$ that doesn't also prove $\pi$ is transcendental. I leave this as a comment, rather than an answer, because I don't know how to substantiate it. Jan 30, 2012 at 0:20
• @lhf: I don't see how knowing that $\pi$ is irrational helps. There are lots of irrational algebraic numbers. Jan 30, 2012 at 0:30
• @Zhen Lin: What lhf means is, I think, consider a tower of quadratic extensions, starting from Q. Note that $\pi\not\in \mathbb Q$ as an induction basis and then, by some magic induction argument, if $\pi\not\in F\implies \pi\not\in F(u)$, where $u^2\in F$. Jan 30, 2012 at 0:49
• @Marc: that's still not the question being asked in the first sentence. Most irreducible polynomials of degree $2^n$ do not have constructible numbers as their roots; in fact, generically they have Galois group $S_{2^n}$. Jan 30, 2012 at 22:52
• Ha. Dave Renfro's sci.math post was a response to one of mine. I guess when the conversation turns to $\pi$, things naturally keep coming 'round.... Jul 4, 2012 at 4:00

I've just read in the book The Number $\pi$ by Eymard and Lafon that no such proof is known.
“The proof that it is impossible to square the circle does not involve direct demonstration of the non-constructibility of the number $\pi$. As far as we are aware, it is not known how to do this! One proves that it is not algebraic, which is much more restrictive, then one uses the fact that a constructible number is algebraic.” [§4.2, p. 134]
• I see that the book was written $8$ years ago, maybe the information is not up to date, though I doubt that it's not [+1] Oct 9, 2012 at 11:09