5
$\begingroup$

A problem in my Galois theory syllabus in the chapter on constructible numbers is as follows:

Check if the small square if constructible from the big square. [Hint: Choose coordinates $0,1\in\mathbf{C}$ as in the picture, and find $z=a+bi$.]

enter image description here

I found two equations from which I can solve $a$ and $b$. Adding up the area's of the four congruent triangles and the small square, we get $1=4\cdot \frac{1}{2}b+(1-a)^2$. Using similar triangles, I can extract a second relation, which is $\frac{1}{\sqrt{b^2+(1-a)^2}}=\frac{\sqrt{b^2+(1-a)^2}}{1-a}$ or $b^2=a(1-a)$.

This gives the following system of equations to solve:

$$\begin{cases} 2b+(1-a)^2=1 \\ b^2=a(1-a) \end{cases}$$

I tried finding the real solution of this system of equations, in vain, which after seeing the solution from Mathematica is not such a surprise.

enter image description here

The minimal polynomial of $z=a+bi$ is, according to Mathematica, $X^3 - 4 X^2 + 6 X - 2$. This polynomial is Eisenstein with $p=2$, so irreducible. By Galois theory, we can then conclude that $z$ is not constructible.

My question is: how can I find either a closed expression for $z$ or it's minimal polynomial without a ton of calculations or use of Mathematica? The author of this problem in my syllabus obviously intended an approach by hand.

$\endgroup$
2
+25
$\begingroup$

We can get the polynomial which $a$ satisfies: $4a(1 - a) = 4b^2 = (1 - (1 - a)^2)^2$ and thus $4(1 - a) = a(2 - a)^2$, equivalently: $a^3 - 4a^2 + 8a - 4 = 0$ which is irreducible $\mathrm{mod}$ $3$ (because it got no root $\mathrm{mod}$ $3$ ). So we can't construct $a$. But $z$ is constructible if and only if both $a, b$ are.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! Can you elaborate on your last statement? $\endgroup$ – rae306 May 16 '19 at 13:44
  • $\begingroup$ Suppose $a$ and $b$ are constructible. Then $bi$ is a $90°$ counterclockwise rotation of $b$ and adding that vectorially to $a$ gives $a+bi$. Now suppose $a+bi$ is constructive. Drop a perpendicular from there to the real axis giving $a$ as the intersection. The vector from $a$ to $a+bi$ is $bi$ which is rotated $90°$ clockwise to resolve the real quantity $b$. $\endgroup$ – Oscar Lanzi May 16 '19 at 14:01
  • $\begingroup$ @RybinDmitry your equation is not correct $\endgroup$ – rae306 May 17 '19 at 6:37

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.