Polarized split · K3 remainder — certificate

For the fixed natural numbers $n=5$, $d=6$, and $q=1$, work in the polynomial ring $R=\mathbb C[X_0,X_1,X_2,Y_0,Y_1,Y_2]$, with the six indeterminates corresponding, in that order, to the indices $0,1,2,3,4,5$. Set $A=X_0X_1X_2Y_0Y_1Y_2$ and set

The theorem asserts that there exist polynomials $D,Q\in R$ and six polynomials $N_0,\ldots,N_5\in R$ such that all of the following conditions hold simultaneously: $F\ne0$; every monomial of $F$ with nonzero coefficient has total degree $6$; every monomial of $A$ with nonzero coefficient has total degree $6$; $D\ne0$; writing $r=\deg D$ for the maximum of the sums of the exponents of monomials of $D$ with nonzero coefficient, every such monomial of $D$ has total degree $r$; for every $i\in\{0,1,2,3,4,5\}$ and every exponent tuple $(m_0,\ldots,m_5)\in\mathbb N^6$, if the coefficient of $X_0^{m_0}X_1^{m_1}X_2^{m_2}Y_0^{m_3}Y_1^{m_4}Y_2^{m_5}$ in $N_i$ is nonzero, then $m_0+\cdots+m_5=r+1$; every polynomial in $R$ that divides both $F$ and $D$ is a unit of $R$, that is, a nonzero constant; and the following is an equality of polynomials in $R$, where $(Z_0,\ldots,Z_5)=(X_0,X_1,X_2,Y_0,Y_1,Y_2)$ and $\partial_i$ denotes formal partial differentiation with respect to $Z_i$:

There are no hypotheses or free parameters in the assertion. The existential choices need not be unique. There is no upper bound on the nonnegative integer $r$; a nonzero constant denominator, with $r=0$, is permitted. Each $N_i$ may be zero, in which case its coefficient condition is vacuous. The polynomial $Q$ has no separately imposed degree or homogeneity condition and may be zero; the right-hand side is not required to vanish. The homogeneity degrees in these conditions arise from integer arithmetic: $6(1+1)-(5+1)=6$ for $A$ and $r+6\cdot1-5=r+1$ for each $N_i$.