Rewriting a quaternary cubic as sums of $5$ cubes of linear forms

Question

I am currently working on removing 4 terms from the 9th degree equation by this paper by R. Garver. Although everything seems fine, there was a detail that was unclear for me:

A well known theorem of Sylvester says that the general homogeneous cubic polynomial in four variables may, with the aid of an equation of the fifth degree, be reduced to the canonical form $M_1^3+M_2^3+M_3^3+M_4^3+M_5^3$, where the $M_i^3$ are the roots of a quintic whose coefficients are rational in those of the given cubic polynomial, the $M_i$ are linear homogeneous functions of the four variables

After a bit of research, it was contained in here. However, I don't have access to it.

From what I've understand, the quaternary cubic: $$F(a,b,c,d)=\alpha_{3000}a^3+\alpha_{2100}a^2b+\dots+\alpha_{0003}d^3$$ containing terms with total degree $3$, can be rewritten as: $$F(a,b,c,d)=M_1^3+M_2^3+M_3^3+M_4^3+M_5^3$$ Where $M_i=\beta_{i1}a+\beta_{i2}b+\beta_{i3}c+\beta_{i4}d$. And there seems to be no nice way to solve for the coefficient if we substitute it in directly.

According to this post, if we let $$A=\left(\frac{\partial^2F}{\partial x_i\partial x_j}\right)_{i,j=0,1,2,3}\\ H=\det(A)\\ B=\text{adj}(A)\\ R=\sum_{i,j=0}^{3}B_{i,j}\frac{\partial H}{\partial x_i}\frac{\partial H}{\partial x_j}\\ T=\sum_{i,j=0}^{3}A_{i,j}B_{i,j}$$ Then the $27$ lines of the cubic surface $S_1:\{F=0\}$ are the intersection of $S_1$ with $S_2:\{R-4HT=0\}$. I think if we choose $5$ lines that only intersect at $(0,0,0,0)$, then they're multiples of $M_i$.

Since the matrix $A$ have linear entries, $H$ is a quartic, $B$ have cubic entries, $R$ is a $27$ degree, $T$ is a cubic, then $S_2$ has degree $27$, but computation of $S_2$ implies that it's a degree $9$ surface. And calculating the resultant of $S_2$ and $F$ yields a $27$ degree

Question: If there's a quintic equation that contains information about $\beta_{ij}$ or $M_i$, or there exist a nice way to solve, what is it?

Providing a non-singular, non-symmetric cubic surface example with their corresponding decomposition relating to quintics at most will also be accepted.

Answer 1

Given the equation: $$x^9+c_1x^5+c_2x^4+\dots+c_6=0$$ and the Tschirnhaus transformation: $$x^8+ax^7+bx^6+cx^5+dx^4+ex^3+fx^2+\color{red}mx+n-y=0$$ Yields: $$y^9+C_1y^8+C_2y^7+C_3y^6+\dots+C_9=0$$ But because the original nonic is missing $3$ terms $(x^8,x^7,x^6)$, then $C_1,C_2,C_3$ have the form: $$C_1=\alpha_1$$ $$C_2=\alpha_2m+\alpha_3$$ $$C_3=\alpha_4m^2+\alpha_5m+\alpha_6$$ with the first three $C_k$ also missing their leading terms $(m,m^2,m^3)$ respectively, and the $\alpha_k$ are in the other unknowns. The objective is to make all six $\alpha_k$ vanish, and that frees up $m$ to solve for $C_4$.

For example, consider the reduced nonic equation: $$x^9-x^5-3x^4+2x^3+x^2+x-3=0$$ After the transformation, we get the following explicit expressions for the six $\alpha_k$: $$\alpha_1=7a+12b-15c-4d-9n+4$$ $$\alpha_2=4a+7b+12c-15d-4e-54$$ $$\alpha_3=-105a^2+19ab-85ac+49ad-25ae-54af-56an+163a+82b^2-103bc-73bd-54be+4bf-96bn-211b+100c^2+6cd+4ce+7cf+120cn-125c+10d^2+7de+12df+32dn+4d+6e^2-15ef+77e-2f^2-25f+36n^2-32n+194$$ $$\alpha_4=-54a+4b+7c+12d-15e-4f-25$$ $$\alpha_5=163a^2-421ab-106ac-81ad+126ae-50af-28an+382a+19b^2+79bc-54bd-98be-108bf-49bn-350b-103c^2+127cd-48ce+8cf-84cn+340c+6d^2+24de+14df+105dn+26d+7e^2+24ef+28en+24e-15f^2+154f+378n-284$$ $$\alpha_6=-462a^3-144a^2b+1031a^2c+251a^2d-243a^2e-6a^2f+735a^2n-1202a^2-890ab^2-922abc+537abd+66abe-350abf-133abn+2207ab+332ac^2-449acd+673ace+341acf+595acn+322ac-103ad^2-369ade+170adf-343adn-1572ad-23ae^2-65aef+175aen-977ae+63af^2+378afn-527af+196an^2-1141an-2058a+172b^3-374b^2c-256b^2d-513b^2e-211b^2f-574b^2n-1209b^2+417bc^2+409bcd+340bce-106bcf+721bcn+1325bc-39bd^2+170bde+168bdf+511bdn-432bd+92be^2-26bef+378ben+572be-49bf^2-28bfn+460bf+336bn^2+1477bn-435b-330c^3+151c^2d-125c^2e-85c^2f-700c^2n+667c^2+cd^2-81cde-54cdf-42cdn-152cd-13ce^2+175cef-28cen-395ce-24cf^2-49cfn+673cf-420cn^2+875cn-2008c-12d^3+49d^2e-73d^2f-70d^2n+332d^2-49de^2-48def-49den-10de+12df^2-84dfn-370df-112dn^2-28dn+875d-18e^3+4e^2f-42e^2n-235e^2+7ef^2+105efn-190ef-539en-376e+4f^3+14f^2n+12f^2+175fn-808f-84n^3+112n^2-1358n+1100$$ Let $\alpha_1=\alpha_2=\alpha_4=0$ and solve for $(e,f,n)$, $$e=(4a+7b+12c-15d-54)/4$$ $$f=(-276a-89b-152c+273d+710)/16$$ $$n=(7a+12b-15c-4d+4)/9$$ Substitute into $\alpha_3=\alpha_5=\alpha_6=0$, $$A=517648a^2+289608ab+724416ac-1206920ad-3077296a+4329b^2+203184bc-315618bd-761196b+283968c^2-950064cd-2381088c+741049d^2+3768268d+4814404$$ $$B=-952048a^2-478584ab-1480512ac+2333496ad+5954256a-58383b^2-416976bc+616334bd+1563188b-588224c^2+1835856cd+4686048c-1446207d^2-7338708d-9321756$$ $$C=-110646640a^3-941644584a^2b-899045712a^2c+1047198072a^2d+2373426528a^2-519339303ab^2-1857906072abc+2795671494abd+6823612620ab-1115458560ac^2+3222557208acd+7710590160ac-2162947359ad^2-10313261148ad-12157798476a-77234661b^3-487774305b^2c+745307352b^2d+1832449041b^2-869860944bc^2+2688176322bcd+6629799132bc-2055308913bd^2-10107354942bd-12394513008b-389765952c^3+1789379856c^2d+4372233120c^2-2674272033cd^2-13038155100cd-15830900100c+1280208574d^3+9351795597d^2+22642682196d+18165319028$$

---

Currently we have $4$ parameters, to extend that into $7$ we do the following substitution: $$ a=a\\ b=b_1+b_2\\ c=c_1+c_2\\ d=d_1+d_2$$ to get $1+6=7$. Now split $A=B=0$ into the following parts: $$A=A_{20}+A_{11}+A_{02}+A_{10}+A_{01}+A_{00}\\ B=B_{20}+B_{11}+B_{02}+B_{10}+B_{01}+B_{00}$$ where $A_{ij},B_{ij}$ has $(i,j)$ being the corresponding degree of $(b_1,c_1,d_1),(a,b_2,c_2,d_2)$.

First pair: $$A_{20}=4329b_1^2+203184b_1c_1-315618b_1d_1+283968c_1^2-950064c_1d_1+741049d_1^2=0$$ $$B_{20}=-58383b_1^2-416976b_1c_1+616334b_1d_1-588224c_1^2+1835856c_1d_1-1446207d_1^2=0$$ yields the solution $\color{blue}{(c_1,d_1)}$: $$c_1\approx(-0.2942035624831439143 - 1.0155324937347195244 i) b_1\\ d_1\approx(0.1586424132629645759 - 0.5849006781087639042 i) b_1$$ with "$b_1$" to be determined later.

Second pair: $$M_1=A_{11}+A_{10}=289608ab_1+724416ac_1-1206920ad_1+8658b_1b_2+203184b_1c_2+203184b_2c_1-315618b_1d_2-315618b_2d_1-761196b_1+567936c_1c_2-950064c_1d_2-950064c_2d_1-2381088c_1+1482098d_1d_2+3768268d_1=0$$ $$N_1=B_{11}+B_{10}=-478584ab_1-1480512ac_1+2333496ad_1-116766b_1b_2-416976b_1c_2-416976b_2c_1+616334b_1d_2+616334b_2d_1+1563188b_1-1176448c_1c_2+1835856c_1d_2+1835856c_2d_1+4686048c_1-2892414d_1d_2-7338708d_1=0$$

Since this is only linear, it is easy to solve $M_1=N_1=0$ for $\color{blue}{(c_2,d_2)}$. What remains to find of the 7 parameters are $(a,b_1,b_2)$.

Third pair:

$$M_2=A_{02}+A_{01}+A_{00}=517648a^2+289608ab_2+724416ac_2-1206920ad_2-3077296a+4329b_2^2+203184b_2c_2-315618b_2d_2-761196b_2+283968c_2^2-950064c_2d_2-2381088c_2+741049d_2^2+3768268d_2+4814404=0$$ $$N_2=B_{02}+B_{01}+B_{00}=-952048a^2-478584ab_2-1480512ac_2+2333496ad_2+5954256a-58383b_2^2-416976b_2c_2+616334b_2d_2+1563188b_2-588224c_2^2+1835856c_2d_2+4686048c_2-1446207d_2^2-7338708d_2-9321756=0$$

We try to find $\color{blue}{(b_1,b_2)}$. Unfortunately, when substituting the expressions $(c_2,d_2)$ then the values $(c_1,d_1)$ which contain $b_1$ into $M_2=N_2=0$, the variables $(b_1,b_2)$ vanish leaving two equations in only one variable "$a$",

$$M_2 = \gamma_1 a^2+ \gamma_2a+\gamma_3=0$$ $$N_2 = \gamma_4 a^2+ \gamma_5a+\gamma_6=0$$

where the $\gamma_k$ are numerical constants. This is a problem since this generally does not lead to a common root $a$.

Conclusion: In the nonic transformation, it ended with 2 quadratics in 2 variables which is solvable. In this octic version, it ended in 2 quadratics in just 1 variable (with no common root), so the octic transformation is illusory and can't remove four terms $C_1=C_2=C_3=C_4=0$ at once.