Obstructions to polynomial and rational quadratic compression

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Status and relevance

These are proved limitations of a proposed transfer of the quartic quaternion argument. They do not prove that any additional integers are missing from $S_r$.

The quartic proof maps certain four-variable quartics to positive quaternary quadratic forms. It is tempting to compress a higher-dimensional diagonal form similarly, perhaps using rational functions rather than polynomials. The proposition below excludes this for every $r\ge5$.

Rational compression into at most four squares with fixed coefficients

Proposition. Let $r\ge5$ and $d_i\in\mathbb Q_{>0}$. Set

$$F(X_1,\ldots,X_r)=\sum_{i=1}^r d_iX_i^r.$$

There is no rational-function identity

$$F=Q(G_1,\ldots,G_m) \tag{1}$$

with $m\le4$, a fixed positive-definite rational quadratic form $Q$, and $G_i\in\mathbb Q(X_1,\ldots,X_r)$. There is no restriction on denominators.

Proof. If necessary pad $Q$ to dimension four by positive square terms with zero inputs. Its determinant is positive. Over the real number field $L=\mathbb Q(\sqrt{\det Q})$, its determinant squareclass is trivial, so $Q$ is similar to a quaternion norm form. For completeness, after diagonalization and division by the first coefficient, write $Q\sim\langle1,u,v,w\rangle$. The condition $uvw\in L^{\times2}$ makes $w$ equivalent to $uv$ modulo squares, giving $\langle1,u,v,uv\rangle$, the norm of $(-u,-v)$. Positivity at a real embedding implies that this quaternion algebra $A$ is division.

The projective hypersurface

$$Z=\{F=0\}\subset\mathbb P_L^{r-1}$$

is smooth and geometrically integral, with dimension $r-2\ge3$. Indeed, its gradient vanishes only at the zero vector, and smooth positive-degree hypersurfaces of this dimension are geometrically integral. Standard Brauer-group results give an isomorphism $\operatorname{Br}(L)\to\operatorname{Br}(Z)$ and an injection $\operatorname{Br}(Z)\to\operatorname{Br}(L(Z))$. The exact inputs are Colliot-Thélène--Skorobogatov, The Brauer--Grothendieck group, Corollary 4.4.5 and Theorem 3.5.4, printed pages 117 and 88; their hypotheses and full passages were checked.

Consequently $A$ remains division over $L(Z)$. It remains division after adjoining one independent transcendental variable: an isotropic rational-function vector for its norm could be specialized at a field element avoiding its finitely many poles and nonzero-coordinate zeros. Thus $Q$ is anisotropic over the function field $L(Z)(t)$ of the affine cone $F=0$.

Suppose (1) held. The irreducible polynomial $F$ defines a discrete valuation $v_F$ on $L(X_1,\ldots,X_r)$, with residue field $L(Z)(t)$. Put

$$e=\min_{G_i\ne0}v_F(G_i),\qquad H_i=F^{-e}G_i.$$

All $H_i$ are integral at this valuation, and at least one has nonzero reduction. Homogeneity gives

$$Q(H_1,\ldots,H_m)=F^{1-2e}.$$

The left side is integral, so $1-2e\ge0$. This integer is odd and hence positive. Reducing the identity modulo $F$ supplies a nonzero isotropic vector of $Q$ over $L(Z)(t)$, a contradiction. $\square$

For odd $r$ the proposition also follows directly from signs over the reals. Its additional content concerns even $r\ge6$.

A stronger dimension obstruction for $r\equiv2\pmod4$

If $r=2k$ with $k$ odd, the same $F$ cannot be represented rationally by any positive-definite rational quadratic form of dimension $m<r$.

Let $Y_i=X_i^k$ and $\varphi=\langle d_1,\ldots,d_r\rangle$. The extension

$$\mathbb Q(X_1,\ldots,X_r)/\mathbb Q(Y_1,\ldots,Y_r)$$

has odd degree $k^r$. A putative identity makes $Q\perp\langle-\varphi(\mathbf Y)\rangle$ isotropic over the larger field. Springer's theorem descends isotropy to the smaller field. Since $Q$ stays anisotropic over a real rational-function field, the last coordinate of the descended isotropic vector is nonzero. Thus $Q$ represents $\varphi(\mathbf Y)$ over $\mathbb Q(\mathbf Y)$. The generic representation theorem then makes $\varphi$ a subform of $Q$, contradicting $m<r$.

The checked inputs are Elman--Karpenko--Merkurjev, The Algebraic and Geometric Theory of Quadratic Forms, Corollary 18.5 and Theorem 17.12, printed pages 77 and 73.

Two elementary polynomial limitations

More generally, a smooth homogeneous form in $r$ variables cannot be written as $H(G_1,\ldots,G_m)$ with $m<r$, all $G_i$ homogeneous of positive degree, and $H$ homogeneous of degree at least two. Over an algebraic closure the $G_i$ have a common nonzero zero, by the projective dimension theorem. At this point $H(G)=0$ and the chain rule makes every derivative vanish, contradicting smoothness.

Also, a nondegenerate diagonal degree-$r$ form in $r\ge3$ variables is absolutely irreducible, whereas the norm form of a degree-$r$ field or étale algebra splits into linear factors over the algebraic closure. Hence an invertible linear change cannot identify these forms. This rules out a linear norm-form shortcut, not all uses of norm arithmetic.