# If a real number can be expressed in terms of complex solutions of quintics, can it be expressed in terms of real solutions of quintics?

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.

• I'm not requiring radical extensions. Jan 26, 2022 at 0:05
• So why doesnt intersecting with the reals work ? Jan 26, 2022 at 0:12
• 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 $n$. Jan 26, 2022 at 0:14

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.

• 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. Feb 8, 2022 at 6:05
• In the middle of the answer you define a group $K$ (conflicting with the field $K$) but then refer to the group as $H$. Feb 8, 2022 at 6:06
• 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. Feb 8, 2022 at 14:45