A problem in Artin's Algebra, 2nd ed. (16.9.17) reads:

Determine the real numbers $\alpha$ of degree 4 over $\mathbb{Q}$ that can be constructed with ruler and compass, in terms of the Galois groups of their irreducible polynomials.

At first I thought this was a relatively simple exercise. In the case where the Galois group is $C_2$, $V_4$, or $D_4$, it has subgroups of order $2^k$ for each $2^k$ dividing its order, so the full splitting field can be constructed by successive quadratic extensions starting from $\mathbb{Q}$. However, when attempting to prove that the roots are not constructible when the Galois group is $A_4$ or $S_4$, I ran into some difficulties.

Theorem 15.5.6 in the book states,

Let $p$ be a constructible point. For some integer $n$, there is a chain of fields $$ \mathbb{Q} = F_0 \subset F_1 \subset F_2 \subset \ldots \subset F_n = K,$$ such that

  • $K$ is a subfield of the field of real numbers;
  • the coordinates of $p$ are in $K$;
  • for each $i = 0, \ldots, n - 1$, the degree $[F_{i+1} : F_i]$ is equal to 2.

Therefore the degree $[K : \mathbb{Q}]$ is a power of 2.

This theorem does not seem to be powerful enough to establish the desired nonconstructibility result for, say, a root with Galois group $A_4$. Yes, we know that $\mathbb{Q}(\alpha)$ in that case has no subfield that is a quadratic extension of $\mathbb{Q}$. However, this together with Theorem 15.5.6 is not enough, because the theorem only asserts that there is some field $K$ at the top of the tower, and $K$ is not necessarily $\mathbb{Q}(\alpha)$. In that sense, the theorem leaves open the possibility that there exists a constructible field $K$ that is some proper extension of $\mathbb{Q}(\alpha)$, and obviously $K$ contains some quadratic extension $F_1$ of $\mathbb{Q}$, but $F_1$ doesn't need to be a subfield of $\mathbb{Q}(\alpha)$.

If we had a stronger version of Theorem 15.5.6, that might read as follows:

Let $\alpha$ be a constructible number, and suppose $\alpha \in K \subset \mathbb{R}$ where $K$ is a field. Then there exists an integer $n$ and a tower of quadratic extensions $$ \mathbb{Q} = F_0 \subset F_1 \subset \ldots \subset F_n$$ such that $p \in F_n \subset K$.

then it seems that Exercise 16.9.17 would be easy to solve, but this stronger result does not seem trivial to prove.


3 Answers 3


What you have to check is that if $\alpha$ is constructible then all its conjugates are constructible. From here, using composite fields, you should prove that $\alpha$ lies in a finite Galois extension of degree a power of $2$.

Conversely, if a number $\alpha$ lies in a Galois extension of degree a power of $2$, it is constructible.

Therefore the constructible numbers are those for which the Galois group of their minimal polynomial is of order a power of $2$.

Since you know the possiblilities for the Galois group of an irreducible of degree $4$, you should have the answer.

  • $\begingroup$ I don't know how to prove that all conjugates are constructible. It's obvious that if at least 2 roots are constructible than so are the rest, but I can't rule out exactly 1 root being constructible. $\endgroup$
    – Brian Bi
    Apr 13, 2018 at 22:17
  • $\begingroup$ There exists an automorphism of the Galois extension generated by all the roots that takes an element to a given conjugate. That will map the sequence of quadratic extensions to another containing at the end the conjugate. $\endgroup$
    – orangeskid
    Apr 13, 2018 at 22:23
  • $\begingroup$ I think that requires the assumption that the tower of quadratic extensions actually lies in the splitting field $K$. Which is exactly the thing I'm not sure how to prove. $\endgroup$
    – Brian Bi
    Apr 14, 2018 at 7:25
  • $\begingroup$ @Brian Bi:Aha, that was not right. Better, consider a Galois extension that contains all those towers of fields. Now apply Galois transformations to that tower. $\endgroup$
    – orangeskid
    Apr 14, 2018 at 13:54
  • $\begingroup$ It looks to me like the OP may be defining a constructible point to only involve extraction of positive square roots. So, by that definition, $\sqrt{1+\sqrt{2}}$ is constructible, but its conjugate $\sqrt{1-\sqrt{2}}$ isn't. The solution to this is change the definition -- if you define "constructible" as "contained in a tower of quadratic extensions", everything will work as orangeskid says. $\endgroup$ Apr 15, 2018 at 13:47

The answer is those quartics in which the resolvent cubic has a rational root will be Euclidean solvable. We construct the roots of the quartic from this rational root. What happens in Galois theory is that the existence of the rational root in the resolvent cubic denies the presence of a threefold symmetry, therefore the Galois group is reduced to something whose order is divisible into $2^3=8$.

Let us examine this using a quartic that is already well known for its constructible solutions. That quartic is $x^4+x^3+x^2+x+1=0$. Yes the roots are all complex, but we can render complex quantities by interpreting the plane as an Argand plane. Let us first go through the algebra of setting up the resolvent cubic, for by doing so we can see how the presence of a rational resolvent cubic root renders the quartic Euclidean-soluble.

We first reduce the quartic via $x=y-1/4$. Thereby:


The resolvent cubic root, following Descartes, is set up as a parameter in a quadratic factorization:

$y^4+(5/8)y^2+(5/8)y+(205/256)=(y^2-2\sqrt{s}y+t_1)(y^2+2\sqrt{s}y+t_2)$, Eq 1, where

$t_1+t_2=(5/8)+4s$, Eq 2, matching quadratic terms

$t_1-t_2=5/(16\sqrt{s})$, Eq 3, matching linear terms

Solving for $t_1, t_2$ in terms of $s$ and then matching constant terms gives our resolvent cubic:


We now search for a rational root, which is made easier by putting $u=16s$ to get $u^3+5u^2-45u-25=0$ (fewer candidates to try for $u$ than directly for $s$). Thereby, $u=5$ and $s=5/16$.

Now we see how this renders the quartic equation Euclidean soluble. When we find the rational root $s=5/16$, putting that into Eqs 2 and 3 gives constructible values for $t_1, t_2$. Now the quadratic factorization in Eq 1 has constructible coefficients and may be solved for $y$, then $x=y-1/4$ follows.

This analysis was meant solely to demonstrate how a rational root in the resolvent cubic leads to constructibility. It may not be the most efficient route to actually implement. Here, the roots for $x$, augmented by an additional root at $1+0i$ in the Argand plane, are just the vertices of a regular pentagon.


Galois' theory and all these technical terms are almost unknown to me, all but a few points that I can more or less understand. In particular, I will say a few things about the possibility of constructing the real roots of a quartic equation with ruler and diabetes, and based on these I will generalize the same question for polynomial equations of greater degree. I do not provide any proof here, but I share Richard Feynman's saying: "The single biggest problem we face is that of visualisation".

Geometric interpretation of the quartic equation:

enter image description here

Draw a similar shape, where the radius $R$, the distance $\alpha$ and the angle $\omega$ with $\omega>0°$ have random values.

The consequences are

(1) $x_1 + x_2 + x_3 + x_4 = 0$

Of course, the values ​​of the points $x$ are the roots of a reduced quartic equation. Then

(2) The radius $R$ is calculated by the coefficients of this equation, while it is the only radius for which the rectangularity of the lines of the figure applies.

This is part of the geometric interpretation of the quartic equation that leads to multiple analytical solutions.

Can this work for higher degree equations?

The following result shows whether it would be possible for such an equation (with degree $n$) to have a similar model with parameters $R, \alpha, \omega$ (where the angles formed by the lines at point $0,0$ are equal to $360^ \circ /n$ each).

If $R, \alpha$ and $n$ have constant values with $R>|\alpha| \geq 0$ and $n>2$, then for every angle $\omega$ will be valid

$x_1+x_2+x_3+...+x_n = \delta \cos(n \omega)$

where $\delta$ is a constant. Therefore $\delta$ will be equal to the sum of the roots for $\omega = 0^\circ$. When $n$ is an odd number it will be $\delta=0$ only if $\alpha = 0$, while when $n$ is an even number it will always be $\delta = 0$ due to the symetry that these systems have with respect to the $y$-axis for $\omega = 0^\circ$.

So the indications are encouraging for equations that have an even degree, but the problem remains open.


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