Removing four terms from the decic using a Tschirnhausen transformation?

Question

I. Transformation

In this 2021 paper, one can remove four terms from an equation of degree $n$ using a Tschirnhausen transformation of degree $n-1$ (and with radical coefficients), but only if $n\geq10$. (This high bound allows enough "free" parameters.) Specifically, in the paper, given the equation with $n\geq10$,

$$x^n+c_1x^{n-1}+c_2x^{n-2}+\dots c_n=0\tag1$$

and Tschirnhausen transformation,

$$y = a_0+a_1x+\dots+a_{n-1}x^{n-1}\tag2$$

yields,

$$y^n+C_1y^{n-1}+C_2y^{n-2}+\dots+C_n = 0\tag3$$

But with appropriate radicals $a_n$, then,

$$C_1 = C_2 = C_3 = C_4 = 0$$

Of course, to reverse the degree $n-1$ transformation for $n=10$ is no longer in radicals, unlike the Bring quintic.

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II. Example

Given a sample decic missing three terms,

$$x^{10} + x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1 = 0$$

and the deg-$9$ transformation,

$$x^9+ax^8+bx^7+cx^6+dx^5+ex^4+fx^3+gx^2+\color{red}mx+n-y=0$$

yields,

$$y^{10}+C_1y^9+C_2y^8+C_3y^7+\dots+C_{10} = 0$$

Since the original decic is missing three terms $(x^9, x^8, x^7)$, then,

\begin{align} C_1 &= \alpha_1\\ C_2 &= \alpha_2m+\alpha_3\\ C_3 &= \alpha_4m^2+\alpha_5m+\alpha_6 \end{align}

with the first three $C_n$ also missing their leading terms $(m, m^2, m^3)$, respectively, and the $a_k$ are in the other unknowns.

My theory is since one has 6 equations $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = \alpha_5 = \alpha_6 = 0$ but 8 unknowns $(a,b,c,d,e,f,g,n),$ then 2 unknowns may be used to reduce the quadratics $(\alpha_3, \alpha_5)$ to linear equations like $(\alpha_1, \alpha_2, \alpha_4)$. But this is just a theory, since my slow computer can't handle the deg-$9$ transformation. Once the system is solved, then $m$ becomes a "free" parameter to solve,

$$C_4 = \beta_1m^4+\beta_2m^3+\beta_3m^2+\beta_4m+\beta_5 = 0$$

analogous to the method used to derive the Bring-Jerrard quintic.

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III. Question

1. Anybody knows how to solve $C_1 = C_2 = C_3 = C_4 = 0$ in radicals? Or if you have the newer Mathematica, what are the explicit expressions for $(C_1, C_2, C_3)$ using the sample decic above? ($C_4$ is not needed.)

P.S. In Heberle's 2021 paper, "Removal of 5 Terms from a Degree 21 Polynomial", he takes the next step of five terms. There are other recent papers on Tschirnhausen transformations. With computer algebra systems like Mathematica or Maple, it makes things easier.

Answer 1

I. Coefficients $\pmb{C_k}$

Given the sample decic above, $$x^{10} + x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1 = 0$$ and the deg-$9$ Tschirnhausen transformation,

$$y=x^9+ax^8+bx^7+cx^6+dx^5+ex^4+fx^3+gx^2+mx+n$$

Using resultants to eliminate $x$ between the two yields a new decic,

$$y^{10}+C_1y^9+C_2y^8+C_3y^7+C_4y^6+\dots+C_{10} = 0$$

Collecting coefficients with respect to $\color{red}m$, the $C_k$ above have the form,

$$C_1=\alpha_1$$ $$C_2=\alpha_2m+\alpha_3$$ $$C_3=\alpha_4m^2+\alpha_5m+\alpha_6$$

The objective is to make all six $a_k = 0$, and that frees up $m$ to solve $C_4 = 0$. I obtained the following explicit expressions for the six $\alpha_k$,

$$\alpha_1=4a+14b+6c+5d+4e-10n$$ $$\alpha_2=4b+14c+6d+5e+4f-5$$ $$\alpha_3=34a^2+61ab-18ac-19ad-22ae-33af-5ag-36an+102a+77b^2+45bc+32bd+23be-5bf-126bn+52b-c^2-3cd+19ce+4cg-54cn+5c+10d^2+20de+4df+14dg-45dn-42d+10e^2+14ef+6eg-36en-39e+3f^2+5fg-38f+2g^2-33g+45n^2+93$$ $$\alpha_4=-5a+4c+14d+6e+5f+4g-33$$ $$\alpha_5=102a^2+120ab+66ac-60ad-58ae-60af-66ag+352a+61b^2+136bc+26bd+10be-10bf-10bg-32bn+134b+45c^2+30cd+20ce+14cf-112cn+74c-3d^2+39de+20df+8dg-48dn-15d+20e^2+24ef+28eg-40en-104e+14f^2+12fg-32fn-78f+5g^2-76g+40n+190$$ $$\alpha_6=-130a^3-10a^2b-195a^2c+49a^2d+123a^2e+58a^2f+166a^2g-272a^2n-487a^2+147ab^2-308abc-115abd-40abe-110abf+134abg-488abn+168ab-117ac^2-76acd+52ace+6acf+90acg+144acn-380ac+31ad^2-62ade-45adf+41adg+152adn-120ad-60ae^2-66aef-80aeg+176aen+322ae-30af^2-58afg+264afn+126af-30ag^2+40agn+248ag+144an^2-816an-291a+175b^3+55b^2c+43b^2d+32b^2e+32b^2f+52b^2g-616b^2n+222b^2-58bc^2-93bcd+116bce+74bcf+66bcg-360bcn-80bc+52bd^2+67bde+41bdf+112bdg-256bdn-258bd+13be^2+92bef+6beg-184ben-60be+3bf^2-6bfg+40bfn-152bf-5bg^2-90bg+504bn^2-416bn+407b-16c^3-101c^2d+18c^2e+5c^2f-18c^2g+8c^2n+65c^2-53cd^2-27cde-60cdf+26cdg+24cdn+51cd-2ce^2+6cef-24ceg-152cen+166ce-20cf^2-36cfg+144cf+7cg^2-32cgn+6cg+216cn^2-40cn-176c+10d^3-2d^2e-19d^2f+32d^2g-80d^2n-20d^2+11de^2+10def+20deg-160den+9de-18df^2+15dfg-32dfn+14df+10dg^2-112dgn-61dg+180dn^2+336dn-41d+6e^3+23e^2f+19e^2g-80e^2n-54e^2+7ef^2+20efg-112efn-48ef+12eg^2-48egn-122eg+144en^2+312en+141e+4f^2g-24f^2n+5f^2+14fg^2-40fgn-84fg+304fn+35f+2g^3-16g^2n-39g^2+264gn+139g-120n^3-744n-18$$

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II. Equations

Let $\alpha_1=\alpha_2=\alpha_4=0$ and solve for $(f,g,n)$,

$$f=(5 - 4 b - 14 c - 6 d - 5 e)/4$$ $$g=(107 + 20 a + 20 b + 54 c - 26 d + e)/16$$ $$n=(4 a + 14 b + 6 c + 5 d + 4 e)/10$$

Substitute into $\alpha_3=\alpha_5=\alpha_6=0$,

$$A=15152a^2+23904ab+37776ac+13200ad+2904ae+17480a-4048b^2-5744bc-6640bd-17896be+27400b-2068c^2+6040cd-16764ce+13740c-3700d^2-7900de+33420d-6413e^2+9570e-25125=0,$$ $$B=6992a^2+21920ab+10992ac+26736ad+7656ae-40200a+16720b^2+43632bc+16880bd+12584be+6328b+21172c^2+16808cd+21596ce-4876c+7156d^2+1980de-492d+3253e^2+17166e-17907=0,$$ $$C=2504128a^3+10113984a^2b+4899296a^2c+12687200a^2d+3815184a^2e-18183120a^2+11793984ab^2+24113472abc+21200960abd+8719968abe-12778720ab+13099984ac^2+16734240acd+13819952ace-9046640ac+13752400ad^2+4777200ade-26629680ad+1585844ae^2+6063800ae+10060500a+3108928b^3+11201376b^2c+8948960b^2d+1896784b^2e-1469200b^2+15713904bc^2+16419680bcd+6089232bce+9416240bc+10844400bd^2+2858960bde-18766480bd-915556be^2+5860520be+18776700b+8220392c^3+11387160c^2d+6543524c^2e+9772460c^2+9307000cd^2+3419240cde-10792520cd-1787954ce^2+15593620ce+17469950c+3152200d^3+47300d^2e-5395700d^2-649050de^2+3888420de+1648150d-643437e^3+1551195e^2+5968225e+3104625=0.$$

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III. Radicals

To solve this in radicals using similar techniques in this paper by Hamilton, we 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 $2$ tuples that we'll determine later. The first tuple can be used to solve: $$A_{20}=B_{20}=0$$ which requires a quartic at most and leaves out a free parameter, thus the tuple contain $3$ parameters. Then the second tuple can be used to solve: $$A_{11}+A_{10} = B_{11}+B_{10}=0$$ $$A_{02}+A_{01}+A_{00} = B_{02}+B_{01}+B_{00}=0$$

which also requires a quartic at most and leaves out no free parameter, thus the tuple contain $4$ parameters. Then the remaining parameter can be used to solve $C=0$, a cubic. In total, we need $7$ parameters, however, we only have $5$, so the following substitution: $$d=d_1+d_2\\ e=e_1+e_2$$ will suffice and the $2$ tuples are $(a,d_1,e_1),(b,c,d_2,e_2)$ which is chosen as to not have $(d_1,d_2)$ or $(e_1,e_2)$ being in the same tuple.

The first pair:

$$A_{20}=15152a^2+13200ad_1+2904ae_1-3700d_1^2-7900d_1e_1-6413e_1^2=0$$ $$B_{20}=6992a^2+26736ad_1+7656ae_1+7156d_1^2+1980d_1e_1+3253e_1^2=0$$

We use Mathematica's Resultant function to eliminate $e_1$ between $(A_{20}, B_{20})$ to get,

$$3181286163456 a^4 + 34419039798528 a^3 d_1 + 41730391294544 a^2 d_1^2 + 13753474610232 a d_1^3 + 1744056320441 d_1^4$$

and eliminate $d_1$ between $(A_{20}, B_{20})$,

$$41460804249344 a^4 - 16237920052224 a^3 e_1 - 24426455644896 a^2 e_1^2 + 12822251336256 a e_1^3 - 1744056320441 e_1^4=0$$

then solve for $\color{blue}{(d_1,e_1)}$. Since each quartic has four roots, the correct pairing should be chosen such that $A_{20} = B_{20} = 0$. One pair is,

$$d_1\approx-1.09630780731440469241846556\,a \\ e_1\approx1.376776690176728225648890218\,a $$

with "$a$" to be determined later.

The second pair:

$$M_1 = A_{11}+A_{10}=23904ab+37776ac+13200ad_2+2904ae_2+17480a-6640bd_1-17896be_1+6040cd_1-16764ce_1-7400d_1d_2-7900d_1e_2-7900d_2e_1+33420d_1-12826e_1e_2+9570e_1=0$$

$$N_1 = B_{11}+B_{10}=21920ab+10992ac+26736ad_2+7656ae_2-40200a+16880bd_1+12584be_1+16808cd_1+21596ce_1+14312d_1d_2+1980d_1e_2+1980d_2e_1-492d_1+6506e_1e_2+17166e_1=0$$

We then solve for $\color{blue}{(d_2,e_2)}$ in $M_1 = N_1 = 0$ which is easily done since they are just linear.

The third pair:

$$M_2=A_{02}+A_{01}+A_{00}=-4048b^2-5744bc-6640bd_2-17896be_2+27400b-2068c^2+6040cd_2-16764ce_2+13740c-3700d_2^2-7900d_2e_2+33420d_2-6413e_2^2+9570e_2-25125=0$$

$$N_2=B_{02}+B_{01}+B_{00}=16720b^2+43632bc+16880bd_2+12584be_2+6328b+21172c^2+16808cd_2+21596ce_2-4876c+7156d_2^2+1980d_2e_2-492d_2+3253e_2^2+17166e_2-17907=0$$

We first substitute our expressions for $\color{blue}{(d_2,e_2)}$ into $M_2 = N_2 = 0$. They remain quadratic in $(b,c)$. Then eliminate $c$ using resultants to get a quartic in $b$, then eliminate $b$ to get a quartic in $c$. Substituting $\color{blue}{(d_1,e_1)}$ into the two quartics, we can solve for $(b,c)$. As before, choose the correct pairing of roots so that $M_2 = N_2 = 0$. One pair is,

$$b\approx1.546090930591060861991324\\ c\approx-0.26002315195280406578959$$

which also yields,

$$d_2\approx-0.391286738563760310216691\\ e_2\approx-0.335941054168852721652646$$

where the variable "$a$" disappears from both pairs. Recall that,

$$d_1\approx-1.09630780731440469241846556\, a \\ e_1\approx1.376776690176728225648890218\, a$$

and we now have all unknowns except $a$. Substitute all values into $C=0$ (which is just a cubic in "$a$") defined in the Equations Section and any of its three roots can set $A=B=C=0$, one of which is,

$$a\approx-1.195803802373015534143974$$

Since the last variable $m$ is used to solve,

$$C_4 = m^4 + 3.07869451461021m^3 - 7.394031433568m^2 - 205.449166413545m - 181.437427587543 = 0$$ $$\implies m\approx 5.65540144869$$

then we eliminated four terms $C_1 = C_2 = C_3 = C_4 = 0$ of the decic using a nonic Tschirnhausen transformation with coefficients in the radicals and the reduced decic being,

$$y^{10} + 13728.2943025125y^5 + 434442.997136881y^4 + 4108739.12946482y^3 + 19681367.7315521y^2 + 49372512.1956258y + 57283977.8917708=0$$

also with coefficients in the radicals (approximate values above).