Homogeneous Resultants and Projective Intersection Length: Examples

The domain $k[X,Y{]}_0\oplus k[X,Y{]}_0$ has basis $(1,0),(0,1)$, and $\mathrm{\Phi }(1,0)=F=aX+bY$, $\mathrm{\Phi }(0,1)=G=cX+dY$ are the two columns of the matrix in the target basis $X,Y$; hence the Sylvester matrix is $\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)$ and its determinant is ${\mathrm{Res}}_{1,1}(F,G)=ad-bc$.

If $ad-bc=0$ and $(c,d)\mathrm{\ne }(0,0)$, then the nonzero vector $(d,-c)\in K^2$ satisfies $a\cdot d+b\cdot (-c)=ad-bc=0$ and $c\cdot d+d\cdot (-c)=0$, so $[d:-c]\in {\mathbf{P}}^1(K)$ is a common zero of $F$ and $G$. If $(c,d)=(0,0)$ then $G=0$ vanishes at every point. When also $(a,b)=(0,0)$, $F=0$ too and any point, such as $[1:0]$, is common; otherwise $(b,-a)\in K^2$ is nonzero and $F(b,-a)=ab-ba=0$, so $[b:-a]\in {\mathbf{P}}^1(K)$ is common.

Conversely, if $F$ and $G$ vanish together at a point $[x:y]$ of ${\mathbf{P}}^1(K)$, then $(x,y)\mathrm{\ne }(0,0)$ is a nonzero vector orthogonal to both coefficient vectors $(a,b)$ and $(c,d)$, so those two vectors are linearly dependent and $ad-bc=0$.

Steps 1.1-1.3 show that $ad-bc=0$ if and only if $F$ and $G$ have a common zero in ${\mathbf{P}}^1(K)$; this agrees with the general criterion [L2], and the zero-form cases are covered by step 1.2.

The Sylvester map has $\mathrm{\Phi }(X,0)=XY$, $\mathrm{\Phi }(Y,0)=Y^2$ and $\mathrm{\Phi }(0,1)=XY$, so the first and the third column of its matrix are equal; the matrix is therefore singular and ${\mathrm{Res}}_{1,2}(F,G)=0$.

One has $F(1,0)=0$ and $G(1,0)=0$, so $[1:0]\in {\mathbf{P}}^1(K)$ is a common zero of $F$ and $G$, in agreement with the vanishing of the resultant by [L2]; in the chart description of [L3] this is the extra point $[1:0]$, because the coefficient of $X^1$ in $F=Y$ is $0$ and the coefficient of $X^2$ in $G=XY$ is $0$.

The dehomogenisations are $F(T,1)=1$ and $G(T,1)=T$, and a common affine root would satisfy $T=0$ from $G(T,1)=0$ and $1=0$ from $F(T,1)=0$, which is impossible in $k$; so there is no affine common root even though the resultant vanishes.

Hence the vanishing of the resultant detects the common projective zero $[1:0]$ that naive affine dehomogenisation misses: the affinely dehomogenised forms $1$ and $T$ are coprime, and the nominated degrees $1$ and $2$ are still the degrees used in the matrix of step 1.1.

The two forms $x_0^2$ and $x_1^3$ are coprime in $k[x_0,x_1,x_2]$ because they involve distinct variables, and have degrees $2$ and $3$, so [L2] gives ${\mathrm{HS}}_S(t)=(1-t^2)(1-t^3)\mathrm{/}(1-t{)}^3=(1+t)(1+t+t^2)\mathrm{/}(1-t)$, using $(1-t^2)=(1-t)(1+t)$ and $(1-t^3)=(1-t)(1+t+t^2)$.

Equivalently, the monomials $x_0^a{x}_1^b{x}_2^c$ with $0\le a\le 1$, $0\le b\le 2$, $c\ge 0$ form a $k$-basis of $S$ by [L3], and their generating function by total degree is $(1+t)(1+t+t^2)(1+t+t^2+\cdots )=(1+t)(1+t+t^2)\mathrm{/}(1-t)$.

By step 1.2, ${\dim }_k{S}_n=\mathrm{\#}\{(a,b):0\le a\le 1,0\le b\le 2,a+b\le n\}$, which is $1,3,5,6,6,6,\dots$ for $n=0,1,2,3,4,5,\dots$; equivalently these are the partial sums of the coefficients $1,2,2,1$ of $(1+t)(1+t+t^2)$, in agreement with [L1].

Since $S_n\mathrm{\ne }0$ for every $n$ by step 2.1, the $k$-vector space $S$ is infinite-dimensional, so $S$ is not a finite-dimensional $k$-algebra; no Artinian claim is made.

If the Axiom of Choice is assumed, then as a consistency check [L4] gives ${\mathrm{len}}_k(\mathrm{Proj}S)=2\cdot 3=6$, which is exactly the eventual value of the Hilbert function computed in step 2.1. The Hilbert-series and Hilbert-function claims above hold over every field without this additional assumption.

In $S$ one has $x_2=0$ and hence $x_1^2=x_0{x}_2=0$, so $x_1,x_2$ are nilpotent in $S$ and the localisations $S_{x_1},S_{x_2}$ are the zero ring: the charts $D_{+}(x_1)$ and $D_{+}(x_2)$ are empty; moreover $S\cong k[x_0,x_1]\mathrm{/}(x_1^2)$ by $x_2\mapsto 0$, whose only homogeneous prime not containing $(x_0,x_1)$ is $(x_1)$, so $X$ has exactly one point, namely the point $P=[1:0:0]$ cut out by $(x_1,x_2)$.

In the chart $D_{+}(x_0)$ the dehomogenised equations are $v-u^2$ and $v$ with $u=x_1\mathrm{/}{x}_0$, $v=x_2\mathrm{/}{x}_0$, so $A_0\cong k[u,v]\mathrm{/}(v-u^2,v)\cong k[u]\mathrm{/}(u^2)$, and this is the chart through $P$ by step 1.1; this ring has the unique prime $(u)$ with $A_0\mathrm{/}(u)\cong k$, so ${\mathcal{O}}_{X,P}\cong A_0$ and $\kappa (P)=k$, that is $[\kappa (P):k]=1$.

In $A_0=k[u]\mathrm{/}(u^2)$ the chain $0⊊(u)⊊A_0$ is a composition series: $(u)=k\cdot u$ is a simple module (it is annihilated by $(u)$, so it is the simple $A_0$-module $k$) and $A_0\mathrm{/}(u)\cong k$ is simple, so by [L4] the length is ${\mathrm{\ell }}_{{\mathcal{O}}_{X,P}}({\mathcal{O}}_{X,P})={\mathrm{\ell }}_{A_0}(A_0)=2$.

Since $X$ has the single point $P$ with local length $2$ and residue degree $1$, [L3] gives ${\mathrm{len}}_k(X)=2\cdot 1=2$, which agrees with the Bezout value $\mathrm{deg}F\cdot \mathrm{deg}G=2\cdot 1=2$ for the coprime forms $F,G$.