Trisecting the Angle
 |
| Credit: Teodomiro, wikipedia.org |
It turns out that trisecting an angle is in general impossible, but the proof that this is the case had to wait for some mathematics developed by Galois. In this post I'll give a proof using polynomials, fields, and vector spaces.
Most impossibility proofs work the same way. First you identify some property which remains invariant with each step, and then you show that the property would need to change to get to the final state. In this case the property is very abstract....
The Invariant Property
Let $\mathbb{F}_0$ be the minimal subfield of $\mathbb{R}$ to contain all the coordinates $a_i, b_i$ where $(a_1,b_1),..., (a_m,b_m)$ are the initial points. Suppose we construct points $(x_1, y_1),...,(x_n,y_n)$ from these. This allows us to define $\mathbb{F}_k$ to be smallest subfield of $\mathbb{R}$ to contain $\mathbb{F}_0 \cup \{x_1,y_1,...,x_k,y_k\}$. The property we are interested is the dimension of $\mathbb{F}_k$ considered as a vector space over $\mathbb{F}_0$. The invariant property is that this is always a power of 2.Minimal Fields
Suppose $\mathbb{F}$ is a subfield of $\mathbb{R}$ and $S \subseteq \mathbb{R}$. It's not immediately obvious that the minimal field $\mathbb{F}(S)$ which includes both $\mathbb{F}$ and $S$ is a well defined concept. However, it's easy to show that the intersection of subfields is always a subfield, and this means we can unambiguously define $\mathbb{F}(S)$ as simply the intersection of all subfields of $\mathbb{R}$ that contain $\mathbb{F}\cup S$.Constructed Points and Quadratic Roots
Each constructed point is either a place where two lines meet, or two circles meet, or a place where a line and a circle meet.Suppose we are constructing point $(x_k, y_k)$.
- If it's on a line then $(y_k-\beta_1)(\alpha_2-\alpha_1) =$ $(x_k - \alpha_1)(\beta_2-\beta_1)$
- If it's on a circle then $(x_k - \alpha_1)^2 + (y_k - \beta_1)^2 =$ $(\alpha_2 - \alpha_1)^2 + (\beta_2 - \beta_1)^2$
Irreducible Polynomials
Irreducible polynomials are to polynomials what primes are to natural numbers. The space of polynomials over a field $\mathbb{F}$ is denoted $\mathbb{F}[X]$. Some of these can be factored and some cannot. For example the polynomial $X^2-1 \in \mathbb{R}[X]$ can be factored:$$
X^2 - 1 = (X+1)(X-1)
$$
Whereas the polynomial $X^2+1$ cannot. (Although it can be factored in $\mathbb{C}[X]$) Now I showed earlier that $x_k, y_k$ are roots of degree 1 or 2 polynomials over $\mathbb{F}_{k-1}$. If the degree is 1 then it is clear that they must be in $\mathbb{F}_{k-1}$ too. But if the degree is 2 then it depends on whether the polynomial is irreducible over $\mathbb{F}_{k-1}$. If the polynomial is not irreducible then the roots will be in $\mathbb{F}_{k-1}$. This means the only non trivial case is when one or both of the new coordinates are the solution to an irreducible quadratic polynomial in $\mathbb{F}_{k-1}[X]$.
The Field Extension $\mathbb{F}(\{x\})$
Let's suppose we have fields $\mathbb{F} \subset \mathbb{G}$ and $x \in \mathbb{G}$ is a root of a degree $n$ irreducible polynomial $a \in \mathbb{F}[X]$. What does the extension $\mathbb{F}(\{x\})$ look like?Lemma
$$\mathbb{F}(\{x\}) = \{ p(x) | p \in \mathbb{F}[X], deg(p) < n \}
$$
Proof
Clearly the RHS is a subset of the LHS so all we need to do is show it's a field. This is equivalent to showing that is is closed under addition, multiplication, and the taking of inverses. If you add two polynomials the degree of the sum is at most that of the largest summand, so the set is definitely closed under addition. All that remains is to show it is closed under multiplication, and the taking of inverses.Suppose $p, q$ are both at most degree $n-1$ then we can use long division to divide $pq$ by $a$, getting:
$$
pq = ab + r
$$
where $deg(r) < n$. Substituting in $x$ gives $p(x)q(x) = r(x)$ since $x$ is a root of $a$. This shows that the set is closed under multiplication.
We know that $a$ and $p$ must be coprime, since anything that divides $a$ must be degree zero or degree $n$, and because $deg(p) <n$. This means that Euclid's algorithm to find the greatest common divisor must produce a degree zero polynomial. Thus for some $q$ and $b$
$$
pq + ab = 1
$$
And substituting in $x$ gives $p(x)q(x) = 1$. This shows that the set is closed under the taking of inverses and this completes the proof$^{(\dagger)}$.
Dimension of $\mathbb{F}(\{x\})$ as a vector space over $\mathbb{F}$
$\mathbb{F}(\{x\})$ is known as an extension of $\mathbb{F}$ and it's dimension as a vector space over $\mathbb{F}$ is known as the degree of the extension. The construction of $\mathbb{F}(\{x\})$ makes it clear that $\{x^0, x^1, x^2,..., x^{n-1}\}$ spans the vector space, where $n=deg(a)$. But it is also the case that they are linearly independent, for otherwise $x$ would be a root of a polynomial $b$ with $deg(b) < n$. If that were true you'd be able to divide $a$ by $b$ to get an even lower degree polynomial $c$ which has $x$ as a root, and keep going to eventually prove that $x$ is in $\mathbb{F}$.Therefore $\{x^0, x^1, x^2,..., x^{n-1}\}$ is a basis of $\mathbb{F}(\{x\})$ which must be of dimension $n$ over $\mathbb{F}$.
What if $y$ is a root of a degree $m$ irreducible polynomial $b \in \mathbb{F}(\{x\})$? What is the dimension of the extension $\mathbb{F}(\{x,y\})$ of $\mathbb{F}$?
It's clear from the construction of $\mathbb{F}(\{x,y\})$ that it is spanned by
$$
\begin{align}
\{x^0y^0,&...,x^0y^{m-1},\\
&...\\
x^{n-1}y^0,&..., x^{n-1}y^{m-1}\}
\end{align}
$$
and we can use exactly the same argument to show that these are linearly independent. Thus the dimension of $\mathbb{F}(\{x,y\})$ is $nm$.
Going back to the ruler and compass construction, this tells us that the degree of the extension $\mathbb{F}_n$ of $\mathbb{F}_0$ must be a power of 2.
The point that completes the proof
All that remains is to show that, were we able to trisect a general angle, we could construct a point whose coordinates are roots of an irreducible polynomial of degree 3.Let's start with the points $\{(0,0),(1,0),(0,1)\}$. The minimal field $\mathbb{F}_0$ containing these coordinates is $\mathbb{Q}$. If we can trisect this right angle we can construct the point $30^\circ$ around the unit circle. If we can trisect that we can construct the point $20^\circ$ around the unit circle. This is $(x,y) = (cos\theta,sin\theta)$ where $\theta = \frac{\pi}{9}$. In which case:
$$
\begin{align}
\frac{1}{2} &= cos 3\theta \\
&= x cos 2 \theta - y sin 2 \theta \\
&= x (x^2-y^2) - y (2xy)\\
&=x^3-3xy^2\\
&=x^3 - 3x(1-x^2)\\
\implies
0 &= 8x^3-6x - 1
\end{align}
$$
We now need to show that $8X^3-6X-1\in \mathbb{Q}[X]$ is irreducible because then $\mathbb{Q}(\{x\})$ will be a degree 3 extension of $\mathbb{Q}$ and so the degree of the extension $\mathbb{F}_n$ of $\mathbb{Q}$ cannot be a power of 2$^{(\dagger^2)}$.
Assume $x = \frac{a}{b}$ is a root of $p$ where $a,b$ are coprime. We'll obtain a contradiction, thereby proving all roots are irrational and $p$ is irreducible. If we multiply by $b^3$ we get
$$
0 = 8a^3 - 6ab^2 - b^3
$$
which tells us that $b$ is even and $a$ is odd. Let $b = 2^nc$ where $c$ is odd. Dividing by 8 we get
$$
0 = a^3 - 3 a 2^{2n-2}c^2 - 2^{3n-3}c^3
$$
But for $a$ to be odd we must have $n = 1$, so
$$
0 = a^3 - 3ac^2 - c^3
$$
where $a$ and $c$ are odd. This is impossible since $c^3$ would be odd and even at the same time.
This completes the proof that nobody can trisect an arbitrary angle with a ruler and compass. Except Chuck Norris.
Having said that...
Why play by the rules? |
| Credit: me |
- $\dagger$: In our discussion of the extension $\mathbb{F}(\{x\})$ we have assumed that there's a larger field $\mathbb{G}$ in which the root $x$ of $a$ exists. We can always do this even if we don't have the larger field to hand since if $a \in \mathbb{F}[X]$ is irreducible then the ring $\frac{\mathbb{F}[X]}{a}$ of polynomials mod $a$ is a field. This field both contains $\mathbb{F}$ (they're the degree zero polynomials) and a root of $a$ (namely $X$). In fact this is one way of defining the complex number field....
- $\dagger^2$: The following is similar to the proof that $\sqrt{2}$ is irrational.
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