I'm considering a notion of quintic-constructible numbers. This is a follow-up from the cubic case.

Suppose a real number $\alpha$ is contained in a tower of fields

$$\alpha\in\mathbb F_m\supset\mathbb F_{m-1}\supset\cdots\supset\mathbb F_1\supset\mathbb F_0=\mathbb Q$$

where each extension $\mathbb F_{k+1}/\mathbb F_k$ has degree $5$ or less. Is $\alpha$ contained in a tower of real fields

$$\alpha\in(\mathbb R\cap\mathbb G_{m'})=\mathbb G_{m'}\supset\mathbb G_{m'-1}\supset\cdots\supset\mathbb G_1\supset\mathbb G_0=\mathbb Q$$

where each extension $\mathbb G_{k+1}/\mathbb G_k$ has degree $5$ or less?

If a complex number has the form $a+bi$, where $a$ and $b$ are real-quintic-constructible, then its $5$th root also has that form. Proof:

$$(x+yi)^5=a+bi$$ $$x^5-10x^3y^2+5xy^4=a,\quad 5x^4y-10x^2y^3+y^5=b$$

Note that $x=0$ can be a solution only if $a=0$, in which case $y^5=b$, so both $x$ and $y$ are obviously real-quintic-constructible. Hence, assume $x\neq0$, and apply the quadratic formula to the left equation:

$$y^2=\frac{10x^3\pm\sqrt{(10x^3)^2-4(5x)(x^5-a)}}{2(5x)}$$ $$=x^2\pm\frac{\sqrt{20x^6+5ax}}{5x}$$

Now let's look at the right equation and try to eliminate $y$:

$$\big(5x^4-10x^2y^2+(y^2)^2\big)y=b$$ $$\Big(-\frac{16x^4}{5}+\frac{a}{5x}\mp\frac{8x}{5}\sqrt{20x^6+5ax}\Big)y=b$$ $$\Big(-16x^5+a\mp8x^2\sqrt{20x^6+5ax}\Big)y=5bx$$ $$\Big(1536x^{10}+288ax^5+a^2\pm16x^2(16x^5-a)\sqrt{20x^6+5ax}\Big)\Big(x^2\pm\frac{\sqrt{20x^6+5ax}}{5x}\Big)=25b^2x^2$$ $$\Big(1536x^{10}+288ax^5+a^2\pm16x^2(16x^5-a)\sqrt{20x^6+5ax}\Big)\Big(5x^3\pm\sqrt{20x^6+5ax}\Big)=125b^2x^3$$ $$x^3(12800x^{10}+2400ax^5-75a^2)\pm(2816x^{10}+208ax^5+a^2)\sqrt{20x^6+5ax}=125b^2x^3$$ $$x^6(12800x^{10}+2400ax^5-75a^2-125b^2)^2=(2816x^{10}+208ax^5+a^2)^2(20x^6+5ax)$$ $$x^5\big(12800(x^5)^2+2400ax^5-75a^2-125b^2\big)^2=\big(2816(x^5)^2+208ax^5+a^2\big)^2(20x^5+5a)$$

This latter equation is $25$th degree in $x$, but it's $5$th degree in $x^5$, so we can solve for $x$ in two steps; thus $x$ is real-quintic-constructible. It follows that $y$ is also.

Since any quintic can be solved using the Bring radical (along with $n$th roots, for $n\leq5$, and field operations), we only need to show that the Bring radical is real-quintic-constructible:

$$(x+yi)^5+(x+yi)=a+bi$$ $$x^5-10x^3y^2+5xy^4+x=a,\quad 5x^4y-10x^2y^3+y^5+y=b$$ $$y^2=x^2\pm\frac{\sqrt{20x^6-5x^2+5ax}}{5x}$$ $$\big(5x^4-10x^2y^2+(y^2)^2+1\big)y=b$$ $$(-16x^5+4x+a\mp8x^2\sqrt{20x^6-5x^2+5ax})y=5bx$$ $$\big(1536x^{10}-448x^6+288ax^5+16x^2+8ax+a^2\pm16x^2(16x^5-4x-a)\sqrt{20x^6-5x^2+5ax}\big)(5x^3\pm\sqrt{20x^6-5x^2+5ax})=125b^2x^3$$ $$x^3(12800x^{10}-4800x^6+2400ax^5+400x^2-200ax-75a^2)\pm(2816x^{10}-768x^6+208ax^5+16x^2+8ax+a^2)\sqrt{20x^6-5x^2+5ax}=125b^2x^3$$ $$x^5(12800x^{10}-4800x^6+2400ax^5+400x^2-200ax-75a^2-125b^2)^2=(2816x^{10}-768x^6+208ax^5+16x^2+8ax+a^2)^2(20x^5-5x+5a)$$

At last we get another equation which is $25$th degree in $x$, but this time I don't see any way to write it in terms of $x^5$ or $x^5+x$. Is there any quantity $u$ which is quintic in $x$ such that this equation is quintic in $u$?

If that fails, how else can we approach this general problem?

Given any quintic polynomial $p$ (such as the minimal polynomial of something in $\mathbb F_{k+1}$ over $\mathbb F_k$), the splitting field of $p$ can be constructed using extensions with degree $5$ or less. So we may assume that $\mathbb F_m/\mathbb F_0$ is a Galois extension... if that helps.(?)

We may also adjoin to $\mathbb F_m$ any $n$th roots of unity for $n\leq5$, in particular $i=\sqrt{-1}$.

If a complex number $z$ is quintic-constructible (meaning it's contained in a tower of quintic-or-lower extensions), then its conjugate $z^*$ is as well. It follows that the real part $\tfrac{z+z^*}{2}$ and the imaginary part $\tfrac{z-z^*}{2i}$ are quintic-constructible numbers which happen to be real, so they are subjects of the main question.

  • $\begingroup$ I'm not requiring radical extensions. $\endgroup$
    – mr_e_man
    Jan 26, 2022 at 0:05
  • $\begingroup$ So why doesnt intersecting with the reals work ? $\endgroup$ Jan 26, 2022 at 0:12
  • $\begingroup$ See the example in the linked question. If $\mathbb F/\mathbb G$ has degree $n$, it is possible that $(\mathbb R\cap\mathbb F)/(\mathbb R\cap\mathbb G)$ has degree greater than $n$. $\endgroup$
    – mr_e_man
    Jan 26, 2022 at 0:14

1 Answer 1


The answer is no.

Let $\Gamma = A_5 \wr C_2 := (A_5 \times A_5) \ltimes C_2$ where the action of the group $C_2$ of order $2$ swaps the factors (i.e. $c(x,y)c^{-1}=(y,x)$).

Lemma: The only proper subgroup of $\Gamma$ of index less than $10$ is $A_5 \times A_5$.

Proof: Any such subgroup would give rise to a transitive action of $\Gamma$ on less than $10$ points (by conjugation). The kernel of such an action would be non-trivial, because $5^2$ divides $\Gamma$ but not $|S_n|$ for $n < 10$. But the only proper normal subgroup of $\Gamma$ is $(A_5 \times A_5)$, and if this is the kernel of the action then the index is $2$, and the subgroup must $A_5 \times A_5$.

Now let $L/\mathbf{Q}$ be a Galois extension with $\mathrm{Gal}(L/\mathbf{Q}) = \Gamma$. Suppose that $L$ comes with a fixed embedding into $\mathbf{C}$ and that complex conjugation (which is well-defined since we have fixed an embedding) is given by a generator of $C_2$. Now let $K = L^{C_2}$ be the fixed field (of degree $3600$ over $\mathbf{Q}$), so $K$ is real. By the primitive element theorem, $K = \mathbf{Q}(\alpha)$ for some $\alpha \in \mathbf{R}$.

I claim that $\alpha$ does not live inside a tower of real fields each of which has degree at most $5$ over the smaller field. In fact, there has to be an extension of degree at least $10$. On the other hand, it certainly does live in a tower of extensions of degree at most $5$ because $L$ has that property.

The key point is that if $F/E$ are finite extensions over $\mathbf{Q}$ and if $L/\mathbf{Q}$ is Galois over $\mathbf{Q}$, then $(L \cap F)/(L \cap E)$ has degree at most $[F:E]$. To prove this, let $M$ be some Galois extension containing $L$, $E$, and $F$, with $A = \mathrm{Gal}(M/E)$ and $B = \mathrm{Gal}(M/F) \subset A$ and $H = \mathrm{Gal}(M/L)$ so $H$ is normal in $G = \mathrm{Gal}(M/\mathbf{Q})$. By Galois theory, the statement to be proved is that

$$[A:B] \ge [AH:BH],$$

where $AH$ (respectively $BH$) is the smallest group containing $A$ and $H$ (resp $B$ and $H$). Since $H$ is normal, the group $AH$ just consists of elements of the form $ah$ with $a \in A$ and $h \in H$ since (by normality) one can move all the elements of $H$ to the right (e.g. $ha=a a^{-1}ha = ah'$ since $H$ is normal). There is a well-defined map of cosets $A/B \rightarrow AH/BH$ given by sending $aB$ to $aBH$. But now using that $H$ is normal we see that any coset of $BH$ is of the form $ahBH = aBH$ (moving the $h$ to the right by normality) so this map of cosets is surjective. Thus the claim.

It follows that if $\alpha$ lies in such a tower starting at the top with $M$, it remains in such a tower after intersecting the fields with $L$. But then the first field $M \cap L$ is totally real, is contained in $L$, and contains $K$ since it contains $\alpha$ and $K = \mathbf{Q}(\alpha)$. It follows that this intersection is $K$. Hence it suffices to show that $K$ does not admit such a tower of fields over $\mathbf{Q}$. But consider the first extension in this tower over $\mathbf{Q}$ of degree $n \le 5$. By Galois theory, it has to correspond to a subgroup of $\Gamma$ of index $n$ which contains $C_2 = \mathrm{Gal}(L/K)$, and by the lemma above no such subgroup exists.

Just to see that $\Gamma$ really does exist as a Galois group with the corresponding complex conjugation, take a random $A_5$ extension of $\mathbf{Q}(i)$ and take the Galois closure over $\mathbf{Q}$. For example,

$$x^5 + (50 i - 5) x + 40 i - 4,$$

which is also a root of

$$x^{10} - 10 x^6 - 8 x^5 + 2525 x^2 + 4040 x +1616,$$

and this does turn out to have Galois group $\Gamma$ and the correct complex conjugation since $\mathbf{Q}(i)$ is complex.

  • $\begingroup$ I haven't studied enough group theory or Galois theory. Could you add more details to the beginning (the lemma proof) and the end (the existence proof)? I understand the rest... including "the key point", which answers a related question I was thinking about: We can tell whether $\alpha\in\mathbb C$ is quintic-constructible just by looking at the Galois closure of $\mathbb Q(\alpha)$; we don't need to consider infinitely many fields. $\endgroup$
    – mr_e_man
    Feb 8, 2022 at 6:05
  • $\begingroup$ In the middle of the answer you define a group $K$ (conflicting with the field $K$) but then refer to the group as $H$. $\endgroup$
    – mr_e_man
    Feb 8, 2022 at 6:06
  • $\begingroup$ Typo fixed. If you have more precise clarifications you can ask (here or elsewhere) but there's plenty of detail. For the last example, if you take any Galois extension $F/E$ with Galois group $G$ such that $E/\mathbf{Q}$ is quadratic, the generic behavior of the Galois closure of $F$ over $\mathbf{Q}$ is that it will have Galois group $G \wr C_2 = (G \times G) \rtimes C_2$. Generic is in a similar sense to saying that a generic degree $n$ polynomial has Galois group $S_n$. Doing the computation in any particular case can be annoying though. $\endgroup$
    – user994373
    Feb 8, 2022 at 14:45

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