Removal of 5 Terms from a Degree 21 Polynomial

Let $p(x)$ be a polynomial over a (not necessarily closed) field $K$. Then we have

for some coefficients $a_{1},\ldots,a_{n}\in k$ and roots $x_{1},\ldots,x_{n}\in\overline{K}$. Fixing some algebraic extension $L$ of $K$, a Tschirnhaus transformation is a polynomial $T\in L[x]$,

which we apply to the roots of $p$ to obtain a new polynomial

Note that the coefficients $A_{0},\ldots,A_{n}$ are symmetric functions of the transformed roots $T(x_{1}),\ldots,T(x_{n})$, and hence are symmetric in $x_{1},\ldots,x_{n}$. The coefficients $a_{0},\ldots,a_{n}$ generate the algebra of symmetric polynomials in $x_{1},\ldots,x_{n}$, so we can determine the $A_{i}$ polynomially in terms of $a_{0},\ldots,a_{n}$ and $b_{0},\ldots,b_{n-1}$. In particular, this implies $q\in L[y]$. The precise constraints on the field $L$ from which the coefficients of $T$ may be taken vary from author to author but generally one considers towers of extensions of bounded degree. Tschirnhaus’s original goal was to find a formula in radicals for the roots of $p$. More precisely, taking $L$ to be the solvable closure of $K$, his strategy was to determine a Tschirnhaus transformation $T\in L[x]$ which transforms $p$ into the solvable form $q(y)=y^{n}+A_{n}$. If $y_{i}$ is a root of $q$, we can find the corresponding root(s) of $p$ (i.e., those $x_{j}$ such that $T(x_{j})=y_{i}$) as the roots of

If there are $d$ roots of $p$ which map to $y_{i}$, this will be a degree $d$ polynomial. In particular, if $q$ has distinct roots, then the roots of $p$ can be recovered rationally from the roots of $q$. Thus if $q$ is solvable and $T$ has coefficients in a solvable extension of $K$, then $p$ will in general be solvable. (And at worst recovering the roots of $p$ will require the solution of an equation of degree $n-1$.) This strategy fails because it is not generally possible to determine $T$ solvably such that all the intermediate terms $A_{1},A_{2},\ldots,A_{n-1}$ of $q$ vanish. On the other hand, there are weaker versions of Tschirnhaus’s problem which are tractable. More precisely, we can ask: what conditions on $n$ and $L$ are sufficient for there to exist $T$ in $L[x]$ transforming the degree $n$ polynomial $p\in K[x]$ to a polynomial $q\in L[y]$ with $A_{1}=\ldots=A_{k}=0$? That is, we consider the problem of setting some (rather than all) of the intermediate coefficients to zero (thus reducing the number of parameters of the family of polynomials), and we relax the requirement that $L$ necessarily be a solvable extension.

We now consider in more detail what is necessary to determine $T$ such that $A_{1}=\ldots=A_{k}=0$. As we remarked above, $A_{i}$ is a polynomial function of $a_{1},\ldots,a_{n}$ and $b_{0},\ldots,b_{n-1}$. Treating the $b_{i}$’s as unknowns to be determined, we further observe that $A_{i}$ is homogeneous of degree $i$ in the variables $b_{0},\ldots,b_{n-1}$. For example,

we can ensure $A_{1}=0$ as long as the coefficients $b_{0},\ldots,b_{n-1}$ of $T$ are chosen to satisfy a single homogeneous linear equation. (The coefficients of this equation are symmetric in the $x_{i}$, so again can be expressed in terms of the coefficients $a_{1},\ldots,a_{n}$ of $p$; it is not necessary to know the roots $x_{i}$.) More generally, $A_{i}$ vanishes if and only if the $i$th symmetric function of $T(x_{1}),\ldots,T(x_{n})$ vanishes, so to find a Tschirnhaus transformation $T$ such that $A_{i}=0$ in the transformed polynomial we need to solve a degree $i$ polynomial in $b_{0},\ldots,b_{n-1}$ whose coefficients are polynomials in $a_{1},\ldots,a_{n}$. Thus, to determine $T$ we must find a point in $\mathbb{P}^{n-1}_{\mathbf{b}}$ on the intersection of the hypersurfaces $V(A_{1}),V(A_{2}),\ldots,V(A_{k})$, which are of degree $1,2,\ldots,k$, respectively.

Tschirnhaus, in his 1683 paper introducing these transformations, claimed to be able to solve any degree $n$ polynomial by removing all intermediate terms.[17] That is, he claimed one could find a transformation $T$ such that

and the transformed polynomial has the solvable form

The problem with this strategy is that finding the necessary $T$ requires the solution of a system of polynomial equations of degrees $1,2,\ldots,n-1$. In general this leads to an equation of degree $(n-1)!$, so for $n>3$ finding $T$ is a priori more complicated than finding the roots of the original degree $n$ polynomial. Tschirnhaus did show explicitly how to find $T$ in the $n=3$ case, and his proof readily generalizes to removing the first two intermediate terms from any degree $n$ polynomial.

Finding a Tschirnhaus transformation $T$ yielding $A_{1}=A_{2}=\ldots=A_{k}=0$ requires solving a system of $k$ equations in $n$ variables, of degree $1,2,3,\ldots,k$. In the worst case this leads to an equation of degree $k!$, but when the system is sufficiently undetermined (i.e., $n$ much larger than $k$), a solution can be found without solving any polynomial of degree greater than $k$. The first result of this kind was given in 1786 by Bring, who showed that when $n\geq 5$, we can find $T$ such that $A_{1}=A_{2}=A_{3}=0$ without solving any equation of degree higher than three.[4]

In 1834, Jerrard recovered independently Bring’s result that removal of three terms from a degree 5 polynomial was possible by means of a Tschirnhaus transformation.[10] In fact, Jerrard went on to claim that his methods could yield a reduction of a general quintic to a solvable form.[11] (Jerrard was aware of, but did not accept, Abel’s 1824 work on the insolvability of the quintic.)

Shortly thereafter Hamilton was commissioned by the British Association for the Advancement of Science to investigate the validity of Jerrard’s methods, issuing his report in 1837.[8] He showed that Jerrard’s reductions were in many cases “illusory”. Jerrard allowed Tschirnhaus transformations of degree potentially as high or higher than the original polynomial $p$, and Hamilton demonstrated that in the “illusory” cases the transformation $T$ found by Jerrard was a multiple of $p$. In this case the transformed polynomial is always $q(y)=y^{n}$ (the same as if the “transformation” $y=T(x)=0$ were permitted) for any degree $n$ polynomial $p$, so there is no hope of determining the roots of $p$ by studying $q$. On the other hand, Hamilton showed that Jerrard’s methods can be a made to work –and the “illusory” cases avoided – in higher degrees. More precisely, for any fixed $k$, one can find a (nonzero, degree $\leq n-1$) Tschirnhaus transformation $T$ yielding $A_{1}=A_{2}\ldots=A_{k}=0$ without solving any equation of degree higher than $k$, provided the degree $n$ of the original polynomial $p$ is large enough relative to $k$. In particular, for $k=4$, Hamilton showed $n\geq 11$ suffices, and for $k=5$, $n\geq 47$.

In 1886, Sylvester published a geometric explanation of the Hamilton/Jerrard method, sharpened some of the bounds (finding, in particular, $n\geq 10$ suffices for $k=4$, and $n\geq 44$ for $k=5$).[16] Sylvester claimed that these sharpened bounds are optimal if no “elevation of degree” is allowed – that is, if no equations of degree higher than $k$ are to be solved. This claim requires careful interpretation – Sylvester does not consider the possibility of higher degree polynomials arising but factoring into lower degree terms. This is known to happen for $k=3$: if one attempts to remove three terms from a degree $n$ polynomial, the system of equations of degree 1, 2, and 3 leads in general to an equation for degree 6. Sylvester’s method is able to avoid this elevation of degree only when $n\geq 5$. For $n=4$, Sylvester’s method is inapplicable, but Lagrange had shown (in 1771, more than 100 years prior) that the degree 6 equation that arises in this case always factors into a pair of cubics, so that the necessary Tschirnhaus transformation can in fact be determined without solving any equation of degree higher than 3.[12]

Further reductions have been achieved, but all loosen or ignore the “elevation of degree” restriction in one way or another. Viewing the degree 9 polynomial

as a multi-valued “algebraic function” sending $(a_{0},\ldots,a_{9})$ to the set of roots of $p$, Hilbert asked whether this could be written as a composition of algebraic functions of at most 4 variables.[9] As part of his investigation, Hilbert sketched a method for finding a Tschirnhaus transformation such that $A_{1}=A_{2}=A_{3}=A_{4}=0$ when $n\geq 9$. Hilbert’s method requires finding a line on a cubic surface. This in turn requires the solution of an equation of degree 27. However, by putting the equation of the cubic surface into a four-parameter canonical form (the “pentahedral form”, originally suggested by Sylvester), Hilbert was able to show that the coefficients of the necessary line were themselves algebraic functions of four variables, so this was sufficient for his purposes.

In 1945, B. Segre strengthened Hilbert’s result, describing an algorithm for finding a Tschirnhaus transformation removing 4 terms from a degree $n\geq 9$ polynomial, without solving any equation of degree higher than 5.[13] The same result was later proven by Dixmier using different methods.[6] Also building on Hilbert’s work were Wiman, who sketched an alternative proof of the $n=9$ bound for $k=4$, and G. Chebotarev, who extended Wiman’s ideas to sketch a proof that $n=21$ for $k=5$.[3, 18, 14] In 1975, Brauer showed that, when $n>k!$, a degree $n$ polynomial can be written as a composition of algebraic functions of $(n-k-1)$ variables; in this framing of the problem it is permissible to simply solve the degree $k!$ to find a Tschirnhaus transformation removing $k$ terms, since this will be an algebraic function of fewer variables than the original degree $n$ polynomial.[1] For $k\geq 7,$ this $n>k!$ bound is lower than the corresponding bounds for removal of $k$ terms found by Sylvester, though it should again be emphasized that this ignores the “no elevation of degree“ constraint which is explicit in Sylvester, and implicit in pre-Sylvester work on the problem.

In 2020, Wolfson described a function $F(k)$ such that a degree $n$ polynomial can be put in an $(n-k-1)$-parameter form whenever $n\geq F(k)$, and which improved significantly on Brauer’s bounds.[19] Wolfson frames the problem in terms of the theory resolvent degree, which allows bounds on the number of necessary parameters for a family of algebraic functions to be computed based on the dimensions of certain spaces. For $k=5$, Wolfson shows $n\geq 41$, which also improves on Sylvester’s bound of $n\geq 44$. To find the Tschirnhaus transformation accomplishing the reduction in this case requires, among other things, finding a 2-plane inside of a cubic hypersurface in $\mathbb{P}^{6}$. Informally, since the dimension of the moduli space of cubic hypersurfaces in $\mathbb{P}^{6}$ is 35, finding a 2-plane inside such a hypersurface is at worst a 35-parameter problem, so is permissible when putting a degree 41 polynomial into 35-parameter form. This approach does not produce the degrees of the equations which must be solved to find the Tschirnhaus transformation (which may be higher than $n$, as in Hilbert’s proof that $n=9$ suffices for $k=4$, where it was necessary to solve a degree 27 equation to find a line on a cubic surface).

Most recently, Sutherland used a combination of ideas from the classical theory of polarity together with optimizations to the moduli-dimension-counting method to improve on the $F(k)$ bound for $k=5,\ldots,15$ and all $k\geq 18$.[15] In particular, for $k=5$, Sutherland shows that $n=21$ suffices, giving the first rigorous proof of the bound first claimed by Chebotarev.

In the current note, we show that the $n=21$ result can be recovered from Sylvester’s method if partial elevation of degree is allowed. More precisely, one can find a Tschirnhaus transformation removing 5 terms (i.e., such that the coefficients of the transformed polynomial satisfy $A_{1}=A_{2}=A_{3}=A_{4}=A_{5}=0$) from a degree $n$ polynomial whenever $n\geq 21$, by solving no equation of degree higher than 20. A key ingredient is that there exists an explicit algorithm to find a 2-plane on a cubic hypersurface in $\mathbb{P}^{9}$, without solving any equation of degree higher than 5.

Let $k$ be an algebraically closed field and let $X\to Y$ be a generically finite dominant map of $k$-varieties. Following Buhler and Reichstein [2], we define the essential dimension of this map, $\mathrm{ed}(X\to Y)$, to be the minimal $d$ such that there exists a pullback square

with $\dim Z\leq d$. The idea of resolvent degree is to extend this notion to allow towers of pullbacks; essentially, a map is of resolvent degree at most $d$ if it can be written as a composition of maps each of essential dimension at most $d$. More precisely, following Farb and Wolfson [7], the resolvent degree, $\mathrm{RD}(X\to Y)$, is defined to be the minimal $d$ such that there exists a tower

where $U$ is an open subset of $Y$, $E_{r}\to U$ factors through a dominant map $E_{r}\to X$, and $\mathrm{ed}(E_{i}\to E_{i-1})\leq d$ for each $i$. It follows from these definitions that

It is convenient, in proving results on resolvent degree, to construct towers of maps of bounded resolvent degree (rather than essential dimension). Lemma 2.7 in Farb and Wolfson [7] shows that this is sufficient.

The connection to polynomials and their roots is as follows. Let $\mathcal{P}_{n}$ denote the space of monic degree $n$ polynomials with coefficients in $k$ and let

the space of monic polynomials together with a choice of root. We then define

where the map is “forgetting the root”. $\mathrm{RD}(n)$ captures the complexity of the root cover in the sense that if there is a formula in terms of algebraic functions of at most $d$ variables for finding the roots of a degree $n$ polynomial in terms of its coefficients, then $\mathrm{RD}(n)\leq d$.

It is necessary to extend the definition of resolvent degree to dominant maps of $k$-varieties which are not necessarily generically finite. Although we are principally interested in $\mathrm{RD}(n)=\widetilde{\mathcal{P}_{n}}\to\mathcal{P}_{n}$, which is generically finite, a number of maps which arise naturally in the study of Tschirnhaus transformations are not. For example, a key step in Bring’s proof that a general quintic can be reduced to a one-parameter form involves finding a line on a quadric surface in $\mathbb{P}^{3}$. Thus we would like to study

where $\mathcal{H}_{2,3}$ is the parameter space of quadric surfaces in $\mathbb{P}^{3}$ and $\mathcal{H}_{2,3}^{1}$ is the space of such quadrics together with a choice of line on the surface, and the map is “forgetting the line”. Since any quadric surface contains an infinite family of lines, this map is not generically finite. We would like to extend the notion of resolvent degree so that $\mathrm{RD}(\mathcal{H}_{2,3}^{1}\to\mathcal{H}_{2,3})$ captures the complexity of finding lines on quadric surfaces in the same way that $\mathrm{RD}(\widetilde{\mathcal{P}_{n}}\to\mathcal{P}_{n})$ captures the complexity of finding roots.

This is done in [19] in the following way: given a dominant map $\pi:X\to Y$, define the resolvent degree $\mathrm{RD}(X\to Y)$ to be the minimum $d$ for which there exists a dense collection of subvarieties $\{U_{\alpha}\subset X\}$ with

a generically finite dominant map and $\mathrm{RD}(U_{\alpha}\to Y)\leq d$ for all $\alpha.$ (Such a subvariety $U_{\alpha}$ is called a rational multi-section for $\pi$.)

Requiring the collection of multisections to be dense in $X$ is necessary to ensure that resolvent degree behaves well with respect to composition of dominant maps: given $X\to Y$ and $Y\to Z$ dominant, for the composition $X\to Z$ we have $\mathrm{RD}(X\to Z)\leq\max\{\mathrm{RD}(X\to Y),\mathrm{RD}(Y\to Z)\}$ by [19, Lemma 4.10]. Having multi-sections $V\to Y$ for $X\to Y$ and $U\to Z$ for $Y\to Z$ is not generally enough to product a multisection for $X\to Z$ (the range of $V\to Y$ could entirely miss $U$). On the other hand, if $V\to Y$ is surjective, then we do obtain a multi-section of $X\to Z$.

Returning to the example of lines on quadrics, an algorithm for determining a line (or finite set of lines) on each quadric surface in terms of algebraic functions of degree $d$ gives a multi-section $U\to\mathcal{H}_{2,3}$ with $\mathrm{RD}(U\to\mathcal{H}_{2,3})\leq\mathrm{RD}(d)$, but this is somewhat less than what is required to show that $\mathrm{RD}(\mathcal{H}_{2,3}^{1}\to\mathcal{H}_{2,3})\leq\mathrm{RD}(d)$. On the other hand, as we shall see in the next section, determining any one line is sufficient to yield a Tschirnhaus transformation reducing the quintic to a one-parameter form, so that $\mathrm{RD}(5)=1$.

As an illustrative example, we now give a detailed proof that (as observed in [7]) Bring’s work on Tschirnhaus transformations of quintic polynomials implies $\mathrm{RD}(5)=1$. (Recall that by definition, $\mathrm{RD}(5)=\mathrm{RD}(\widetilde{\mathcal{P}_{5}}\to\mathcal{P}_{5})$, where $\mathcal{P}_{5}$ is the parameter space of monic degree 5 polynomials and $\widetilde{\mathcal{P}_{5}}$ is the space of such polynomials together with a choice of root.)

Recall that given a polynomial $p(x)=x^{5}+a_{1}x^{4}+a_{2}x^{3}+a_{3}x^{2}+a_{4}x+a_{5}$ in $\mathcal{P}_{5}$ and a Tschirnhaus transformation of the form $T(x)=b_{0}x^{4}+b_{1}x^{3}+b_{2}x^{2}+b_{3}x+b_{4}$, we can set $y=T(x)$ and eliminate $x$ to obtain a polynomial

where the coefficient $A_{i}$ is a degree $i$ homogeneous polynomial in $b_{0},b_{1},b_{2},b_{3},b_{4}$, whose coefficients are integral functions of $a_{0},a_{1},a_{2},a_{3},a_{4},a_{5}$.

Our strategy will be to construct a tower of maps

such that the resolvent degree of each map is 1 and there is a dominant rational map $X_{5}\to\widetilde{\mathcal{P}_{5}}$. Informally, in $X_{1}$ there will be associated to each quintic polynomial $p$ the equation of a hyperplane determining the set of Tschirnhaus transformations such that $A_{1}=0$. In $X_{2}$, we will further have the information of a quadric $Q$ contained in this hyperplane, corresponding to $A_{1}=A_{2}=0$, and the equation of a line $l$ contained in $Q$. In $X_{3}$ we will further have a choice of Tschirnhaus transformation $T$ which lies on $l$ while also satisfying $A_{3}=0$. Using this information we can construct a map from $X_{3}$ to the one-dimensional space of quintic polynomials of the form $z^{5}+Az+1$, and we will take $X_{4}$ to be the pullback of the root cover of this space, so that $X_{4}$ associates to $p$ a root of the polynomial obtained by applying $T$ to $p$. Finally, in constructing $X_{5}$, we recover the roots of $p$ itself from the roots of the transformed polynomial. Each step of this process is of bounded resolvent degree.

More precisely: viewing the $b_{i}$ as homogeneous coordinates on $\mathbb{P}^{4}$, there is associated to each polynomial $p\in\mathcal{P}_{5}$ a hyperplane $H\subset\mathbb{P}^{4}_{\mathbf{b}}$ consisting of all Tschirnhaus transformations that send $p$ to a polynomial $q(y)$ with $A_{1}=0$. Let $X_{1}$ denote the space of ordered pairs $(p,H)$ with $p$ and $H$ as just described. Then the map $X_{1}\to\mathcal{P}_{5}$ is birational, hence has resolvent degree 1.

Next, we consider the set of all Tschirnhaus transformations such that $A_{1}=0$ and $A_{2}=0$ in the transformed polynomial. The latter is a degree 2 homogeneous polynomial in $b_{0},b_{1},b_{2},b_{3},b_{4}$, so to each polynomial $p$ there is associated a quadric surface $Q\subset H\cong\mathbb{P}^{3}$ whose points are Tschirnhaus transformations satisfying both conditions. This defines a map

where $\mathcal{H}_{2,3}$ is the parameter space of quadric surfaces in $\mathbb{P}^{3}$. Letting $\mathcal{H}_{2,3}^{1}$ denote the space of such surfaces together with a choice of line, we consider the map

An algorithm exists for finding a line on a quadric surface that requires solving only a single quadratic equation, so there is a rational multi-section $U\to\mathcal{H}_{2,3}$ with

(In fact there is a dense collection of such multi-sections, so $\mathrm{RD}(\mathcal{H}_{2,3}^{1}\to\mathcal{H}_{2,3})=1$, but this is stronger than we require.)

We define $X_{2}$ via the pullback square

so that $\mathrm{RD}(X_{2}\to X_{1})\leq\mathrm{RD}(U\to\mathcal{H}_{2,3})=1$. Points of $X_{2}$ are ordered tuples $(p,H,Q,l)$, where $p$ is a quintic polynomial, $H$ is the space of all Tschirnhaus transformations that send $p$ to a polynomial with $A_{1}=0$, $Q\subset H$ is the quadric surface whose points are Tschirnhaus transformations such that $A_{1}=0$ and $A_{2}=0$, and $l$ is a line contained in $Q$.

Finally, to find a Tschirnhaus transformation $T$ satisfying $A_{1}=0$, $A_{2}=0$, and $A_{3}=0$ requires intersecting the cubic hypersurface defined by $A_{3}=0$ with the line $l$. To find the three points of intersection requires only the solution of a cubic equation. Defining $X_{3}$ to be the space of tuples $(p,H,Q,l,T)$, with $p,H,Q,l$ as above and $T$ a Tschirnhaus transformation yielding $A_{1}=A_{2}=A_{3}=0$, the projection map

has resolvent degree $RD(3)=1$.

Now let $\mathcal{P}_{5}^{\prime}$ denote the space of quintic polynomials of the form $z^{5}+Az+1$. Given a point $(p,H,Q,l,T)$ in $X_{3}$, we can apply the transformation $T$ to $p$ to obtain a polynomial

Applying a change of variables $y=\sqrt[5]{A_{5}}z$ and dividing by $A_{5}$ yields

so this procedure defines a map $X_{3}\to\mathcal{P}_{5}^{\prime}$. Now, the root cover

has resolvent degree

so defining $X_{4}$ via the pullback square

we have $\mathrm{RD}(X_{4}\to X_{3})=1$. Points of $X_{4}$ are of the form $(p,H,Q,l,T,\lambda)$, where $(p,H,Q,l,T)\in X_{3}$ and $\lambda$ is a root of the transformed polynomial $z^{5}+\frac{A_{4}\sqrt[5]{A_{5}}}{A_{5}}z+1$. Let $X_{5}$ be the space whose points are tuples $(p,H,Q,l,T,\mu)$, where $\mu$ is a root of $p$. There is a map

defined by

On the other hand, given a root $\lambda$ of the transformed polynomial, we can find the corresponding roots of $p$ by solving the (at worst degree 4) polynomial equation

so $\mathrm{RD}(X_{5}\to X_{4})=\mathrm{RD}(4)=1$.

In all, we have constructed a tower of maps

each of which has resolvent degree 1. Since the root cover $\widetilde{\mathcal{P}_{5}}\to\mathcal{P}_{5}$ factors through the projection $X_{5}\to\widetilde{\mathcal{P}_{5}}$, $(p,H,Q,l,T,\mu)\mapsto(p,\mu)$, we have

Let $\mathcal{M}_{3,N}$ be the moduli space of smooth cubic hypersurfaces in $\mathbb{P}^{N}$ and let $\mathcal{M}_{3,N}^{r}$ be the moduli space of pairs $(C,P)$ where $C$ is a smooth cubic hypersurface and $P$ is an $r$-plane contained in $C$. We can then consider the “resolvent degree of finding an $r$-plane”, i.e.,

where the map forgets the choice of plane. For example, for the problem of finding a line on a cubic surface, we have

so that finding a line on a cubic surface in $\mathbb{P}^{3}$ requires the solution of algebraic equations of no more than 4 variables. Farb and Wolfson [7], using an argument due to Klein, show this bound can be improved to

In higher dimensions it is easier to find a line. In particular, by Sylvester’s obliteration method, as discussed in section 3, finding a line on a cubic surface in $\mathbb{P}^{5}$ requires the solution of no equation of degree higher than 3, and so is a resolvent degree 1 problem.

Sylvester’s method also extends to finding higher-dimensional linear subspaces of hypersurfaces, when the dimension of the ambient space is large enough. For example, for cubic hypersurfaces in $\mathbb{P}^{9}$, finding a 2-plane can also be done solvably.

The ambient dimension can be reduced slightly if some elevation of degree is allowed. From Segre [1945, pg 295, Sec 12], it is possible to determine a line on the intersection of two given quadrics by solving equations of degree no higher than 5, provided the ambient dimension is at least 4. Thus to find a linear solution of a system with 2 equations of degree 2, and 3 equations of degree 1, an ambient dimension of 7 is sufficient. Comparing to the last paragraph in the proof above we have