Homogeneous Resultants and Projective Intersection Length

The map ${\mathrm{\Phi }}_{uF,vG}$ sends $(A,B)$ to $uAF+vBG$. In the ordered bases of [L1] its matrix is obtained from the matrix of ${\mathrm{\Phi }}_{F,G}$ by multiplying the $e$ columns of the $F$-block by $u$ and the $d$ columns of the $G$-block by $v$. Column multilinearity [L2] therefore gives $\det {\mathrm{\Phi }}_{uF,vG}=u^e{v}^d\det {\mathrm{\Phi }}_{F,G},$ which is the scaling formula.

Let $\phi \text{\hspace{0.17em}⁣}:R\to R^{\mathrm{\prime }}$ be a unital ring homomorphism to a commutative ring $R^{\mathrm{\prime }}$. By [L3] applied to the polynomial rings $R[X]$, $R[X,Y]$ and to $R^{\mathrm{\prime }}$, there is a unique ring homomorphism $\tilde{\phi }\text{\hspace{0.17em}⁣}:R[X,Y]\to R^{\mathrm{\prime }}[X,Y]$ extending $\phi$ with $X\mapsto X$ and $Y\mapsto Y$; it sends every basis monomial $X^{m-i}{Y}^i$ to the corresponding basis monomial and every entry of the matrix of ${\mathrm{\Phi }}_{F,G}$ to the corresponding entry of the matrix of ${\mathrm{\Phi }}_{\phi (F),\phi (G)}$. The Leibniz formula exhibits the determinant as a polynomial with integer coefficients in the matrix entries, so it commutes with $\tilde{\phi }$, and $\phi ({\mathrm{Res}}_{d,e}(F,G))={\mathrm{Res}}_{d,e}(\phi (F),\phi (G))$.

By [L4] every point of ${\mathbf{P}}^1(K)$ is a class $[a_0:a_1]$ with $(a_0,a_1)\mathrm{\ne }(0,0)$. If $a_1\mathrm{\ne }0$ then $[a_0:a_1]=[a:1]$ with $a=a_0\mathrm{/}{a}_1\in K$, and if $a_1=0$ then $a_0\mathrm{\ne }0$ and $[a_0:0]=[1:0]$. Thus every point is $[a:1]$ for some $a\in K$ or is $[1:0]$, and the two kinds are disjoint.

Fix $a\in K$. The point $[a:1]$ is a common zero of $F$ and $G$ exactly when $F(a,1)=G(a,1)=0$. Indeed a general representative of the class is $(\lambda a,\lambda )$ with $\lambda \in K^{\times }$, and by [L5] $F(\lambda a,\lambda )={\lambda }^{d}F(a,1),G(\lambda a,\lambda )={\lambda }^{e}G(a,1);$ as ${\lambda }^d,{\lambda }^e\mathrm{\ne }0$, both vanish exactly when $F(a,1)=G(a,1)=0$. By definition $F(a,1)=f(a)$ and $G(a,1)=g(a)$.

Evaluating the defining linear combinations at $(X,Y)=(1,0)$ shows that $F(1,0)$ is the coefficient of $X^d$ in $F$ (all remaining monomials have a factor $Y$) and $G(1,0)$ is the coefficient of $X^e$ in $G$. Hence the point $[1:0]$ is a common zero of $F$ and $G$ exactly when both of these coefficients vanish.

Steps 1.3, 1.4 and 1.5 describe every point of ${\mathbf{P}}^1(K)$ and decide when it is a common zero, so the common zeros are exactly the affine zeros $[a:1]$ with $f(a)=g(a)=0$ together with the possible point $[1:0]$ detected by the two vanishing top coefficients. Nothing in the argument replaces $d$ or $e$ by the actual degrees of $f$ or $g$: the zero forms have all coefficients zero, so they contribute the point $[1:0]$ as claimed, and a form whose top coefficient vanishes contributes no condition beyond the vanishing already recorded.

The inclusion $k\to K$ is a unital ring homomorphism, and an element of the field $k$ is zero exactly when its image in $K$ is zero, so ${\mathrm{Res}}_{d,e}(F,G)=0$ in $k$ if and only if its image in $K$ is zero; by the specialization clause of [L2] that image is ${\mathrm{Res}}_{d,e}(F_K,G_K)$, where $F_K,G_K\in K[X,Y]$ are the coefficientwise images, that is, the same forms viewed over $K$, homogeneous of the same nominated degrees. Evaluating $F$ and $G$ at a point of ${\mathbf{P}}^1(K)$ gives the same field elements as evaluating $F_K$ and $G_K$, so also the common-zero condition is unchanged. Hence it suffices to prove the equivalence for forms over the algebraically closed field $K$, and from here on we work over $K$.

For $m\ge 0$ let ${\delta }_m\text{\hspace{0.17em}⁣}:K[X,Y{]}_m\to K[T]$ be the substitution $H\mapsto H(T,1)$, a $K$-linear map by [L4]. It carries the ordered monomial basis $X^m,X^{m-1}Y,\dots ,Y^m$ to the ordered list $T^m,T^{m-1},\dots ,1$, and it is bijective onto the polynomials of degree $\le m$, with two-sided inverse $p(T)\mapsto Y^{m}p(X\mathrm{/}Y)$; it is also multiplicative in the sense ${\delta }_{m+n}(H{H}^{\mathrm{\prime }})={\delta }_m(H){\delta }_n(H^{\mathrm{\prime }})$. Writing $f:={\delta }_d(F)$ and $g:={\delta }_e(G)$, of degrees $\le d$ and $\le e$, the map $\mathrm{\Psi }\text{\hspace{0.17em}⁣}:(u,v)\mapsto uf+vg$ from $P_{\le e-1}\oplus P_{\le d-1}$ to $P_{\le d+e-1}$ (polynomials in $K[T]$ of the indicated degree bounds) has, in the bases transported by $\delta$, exactly the matrix $M$ of the resultant map $(A,B)\mapsto AF+BG$; hence ${\mathrm{Res}}_{d,e}(F,G)=\det M$ by [L1].

By [L6] the coordinates of $\mathrm{\Psi }(u,v)$ are $M$ times the coordinates of $(u,v)$; coordinates are unique, so $\mathrm{\Psi }$ is bijective if and only if the multiplication map $x\mapsto Mx$ on column vectors is bijective, which by [L6] holds if and only if $M$ is invertible. By [L5] and the fact that $K$ is a field, $M$ is invertible if and only if $\det M\mathrm{\ne }0$. Since domain and target of $\mathrm{\Psi }$ both have dimension $d+e$ by [L1], [L7] gives that $\mathrm{\Psi }$ is bijective if and only if $\mathrm{\Psi }$ is injective. Chaining these equivalences, ${\mathrm{Res}}_{d,e}(F,G)=0$ if and only if $\mathrm{\Psi }$ is not injective.

Suppose that $f$ and $g$ have a common root $\alpha \in K$; if $f=g=0$ every $\alpha$ qualifies, and then $\mathrm{\Psi }$ is the zero map on a nonzero space by [L1], so it is not injective. Otherwise, by [L8] there are $f_1,g_1\in K[T]$ with $f=(T-\alpha )f_1$ and $g=(T-\alpha )g_1$, where $f_1=0$ if $f=0$ and $g_1=0$ if $g=0$; by [L9] and $\mathrm{deg}f\le d$, $\mathrm{deg}g\le e$ we get $\mathrm{deg}{f}_1\le d-1$ and $\mathrm{deg}{g}_1\le e-1$. Then $(g_1,-f_1)$ lies in $P_{\le e-1}\oplus P_{\le d-1}$ and is nonzero, because if $f\mathrm{\ne }0$ then $f_1\mathrm{\ne }0$ by [L9], and if $f=0$ then $g\mathrm{\ne }0$ and $g_1\mathrm{\ne }0$ by [L9]; and $\mathrm{\Psi }(g_1,-f_1)=g_{1}f-f_{1}g=g_1(T-\alpha )f_1-f_1(T-\alpha )g_1=0$. So $\mathrm{\Psi }$ is not injective.

Suppose that $f$ and $g$ have no common root in $K$, that they are not both zero, and that $\mathrm{deg}f=d$ or $\mathrm{deg}g=e$. By [L8] take $A,B$ with $Af+Bg=\gcd (f,g)$; the gcd is monic and divides both $f$ and $g$. If the gcd were not $1$, it would be nonconstant, hence by [L3] would have a root $\alpha \in K$, and by [L8] applied to the divisibility that $\alpha$ would be a common root of $f$ and $g$; so the gcd is $1$ and $Af+Bg=1$. Now let $\mathrm{\Psi }(u,v)=uf+vg=0$. Multiplying $1=Af+Bg$ by $v$ gives $v=A(vf)+B(vg)=A(vf)-B(uf)=f(Av-Bu)$, so $f$ divides $v$; since $\mathrm{deg}v\le d-1$ and, when $\mathrm{deg}f=d$, a nonzero multiple $v=fw$ would have $\mathrm{deg}v=d+\mathrm{deg}w\ge d$ by [L9], we get $v=0$ in that case. Symmetrically, multiplying $1=Af+Bg$ by $u$ gives $u=g(Bu-Av)$, so $g$ divides $u$; since $\mathrm{deg}u\le e-1$ and, when $\mathrm{deg}g=e$, a nonzero multiple $u=gw$ would have $\mathrm{deg}u=e+\mathrm{deg}w\ge e$ by [L9], we get $u=0$ in that case. At least one of the two cases holds. If both cases hold, the two conclusions just displayed give $u=v=0$. If only $\mathrm{deg}f=d$ holds, then $v=0$ and therefore $\mathrm{\Psi }(u,0)=uf=0$, which forces $u=0$ because $K[T]$ is a domain by [L9] and $f\mathrm{\ne }0$; if only $\mathrm{deg}g=e$ holds, the same argument with the roles of $f$ and $g$ interchanged gives $u=0$, and then $vg=0$ forces $v=0$ because $g\mathrm{\ne }0$. Hence $(u,v)=(0,0)$ and $\mathrm{\Psi }$ is injective.

Suppose the $X^d$-coefficient of $F$ and the $X^e$-coefficient of $G$ both vanish. In the homogeneous expansion only the pure terms $X^d$ and $X^e$ avoid a factor $Y$, so $Y$ divides $F$ and $Y$ divides $G$; write $F=Y{F}^{\mathrm{\prime }}$ and $G=Y{G}^{\mathrm{\prime }}$ with $F^{\mathrm{\prime }},G^{\mathrm{\prime }}$ homogeneous of degrees $d-1$ and $e-1$ (the zero polynomials allowed). Then $f=F^{\mathrm{\prime }}(T,1)$ and $g=G^{\mathrm{\prime }}(T,1)$, so $\mathrm{\Psi }(G^{\mathrm{\prime }}(T,1),-F^{\mathrm{\prime }}(T,1))=G^{\mathrm{\prime }}(T,1)F^{\mathrm{\prime }}(T,1)-F^{\mathrm{\prime }}(T,1)G^{\mathrm{\prime }}(T,1)=0$: if $(F,G)\mathrm{\ne }(0,0)$ this displays a nonzero kernel vector because $F^{\mathrm{\prime }}$ or $G^{\mathrm{\prime }}$ is nonzero, and if $F=G=0$ then $\mathrm{\Psi }$ is the zero map on a nonzero space by [L1]. Either way $\mathrm{\Psi }$ is not injective, so ${\mathrm{Res}}_{d,e}(F,G)=0$ by 1.3; and conversely $[1:0]$ is a common zero of $F$ and $G$ in ${\mathbf{P}}^1(K)$ by the chart clause of [L2], because the two top coefficients vanish. Thus in this case the resultant vanishes and a common projective zero exists.

Suppose now that the $X^d$-coefficient of $F$ or the $X^e$-coefficient of $G$ is nonzero, so that $\mathrm{deg}f=d$ or $\mathrm{deg}g=e$ and $f,g$ are not both zero. If $f$ and $g$ have a common root in $K$, then ${\mathrm{Res}}_{d,e}(F,G)=0$ by 1.4 and 1.3; if they do not, then $\mathrm{\Psi }$ is injective by 1.5 and ${\mathrm{Res}}_{d,e}(F,G)\mathrm{\ne }0$ by 1.3. Hence ${\mathrm{Res}}_{d,e}(F,G)=0$ if and only if $f,g$ have a common root in $K$, and by the chart clause of [L2] the common roots of $f$ and $g$ are exactly the affine common zeros $[a:1]$, the point $[1:0]$ being excluded because it is a common zero only when both top coefficients vanish, which is not the case here. So ${\mathrm{Res}}_{d,e}(F,G)=0$ if and only if $F$ and $G$ vanish together at a point of ${\mathbf{P}}^1(K)$.

Every pair $(F,G)$ satisfies the hypothesis of 2.2 or the hypothesis of 2.1, and in both cases ${\mathrm{Res}}_{d,e}(F,G)=0$ is equivalent to the existence of a common zero in ${\mathbf{P}}^1(K)$, which proves the equivalence over the algebraically closed field $K$; undoing the coefficient extension of 1.1 gives the equivalence over the original field $k$. No $k$-rational point was produced anywhere: the points obtained are points of ${\mathbf{P}}^1(K)$ over the supplied algebraic closure.

For $r=0$ the ring is the field $k$ by [L6]. Every nonzero element of a field is a unit by [L7], so there are no nonzero nonunits at all; the existence and uniqueness clauses of [L1] hold vacuously, and by [L2] there is no irreducible element, so the statement about irreducibles is vacuous too. This is the base case of the induction on $r$.

Let $R$ be any UFD and $p\in R$ irreducible. Then $p$ is prime: if $p\mid ab$ with $a,b\mathrm{\ne }0$, factor $a={\prod }_i{a}_i$, $b={\prod }_j{b}_j$ and write $ab=pc$ with $c\mathrm{\ne }0$ factored as ${\prod }_k{c}_k$, as [L1] allows whenever the element in question is a nonzero nonunit, the unit cases being immediate from $p\mid ab$; comparing the two products of irreducibles for $ab$ and $pc$ by the uniqueness clause of [L1] shows that $p$ is associate to one of the $a_i$ or one of the $b_j$, hence divides $a$ or $b$.

Now let $R$ be a UFD, $K=\mathrm{Frac}(R)$ by [L4], and $0\mathrm{\ne }f\in R[x]$. For each associate class $C$ of irreducibles in $R$, let $v_C(a)$ be the exponent of that class in the factorization of a nonzero $a\in R$; it is well-defined by the uniqueness clause of [L1]. Let $\mathcal{C}(f)$ be the finite set of such classes appearing in the factorizations of the nonzero coefficients of $f$, and for each $C\in \mathcal{C}(f)$ choose one representative $p_C$. Set $e_C=\min \{v_C(a):a\mathrm{\ne }0\text{ is a coefficient of }f\}$ and $c(f)={\prod }_{C\in \mathcal{C}(f)}{p}_C^{e_C}$. This finite product is a common divisor of the coefficients, every common divisor divides it up to associates, and $f^{\ast }:=f\mathrm{/}c(f)$ is primitive in the sense of [L3]; thus $f=c(f)f^{\ast }$. For $\lambda \in K^{\times }$, define $v_C(\lambda )=v_C(a)-v_C(b)$ when $\lambda =a\mathrm{/}b$ with nonzero $a,b\in R$; this is well-defined, and $v_C(\lambda )=0$ for every associate class exactly when $\lambda$ is a unit of $R$. Note also that an irreducible $f$ of positive degree is primitive: a nonunit constant $r\in R$ dividing all coefficients would write $f=r\cdot (f\mathrm{/}r)$ with both factors nonunits, since $f\mathrm{/}r$ has positive degree by [L7]. The content $c(f)$ factors into irreducibles of $R$ by [L1], and each such constant is irreducible in $R[x]$, because a factorization of a constant in $R[x]$ has both factors constant by [L7] and so is a factorization in $R$. For the primitive part, [L5] gives a factorization $f^{\ast }=u{g}_1\cdots g_n$ in $K[x]$, with $u\in K^{\times }$ and each $g_j$ irreducible in $K[x]$; if $n=0$, then $f^{\ast }$ is a nonzero constant in $R$ and, being primitive, is a unit of $R$. For each $j$ choose $a_j\in R\setminus \{0\}$ with $a_j{g}_j\in R[x]$ and write $a_j{g}_j=d_j{h}_j$, where $d_j\in R$ is a content and $h_j\in R[x]$ is primitive; then $g_j=(d_j\mathrm{/}{a}_j)h_j$, so $h_j$ is a nonzero scalar multiple of $g_j$ in $K[x]$, hence irreducible in $K[x]$ and in $R[x]$ by [L3]. By [L3] the product $h_1\cdots h_n$ is primitive, and $f^{\ast }=\lambda h_1\cdots h_n$ with $\lambda =u{\prod }_j(d_j\mathrm{/}{a}_j)\in K^{\times }$. Since both $f^{\ast }$ and $h_1\cdots h_n$ are primitive, comparison of the least exponents in each associate class gives $v_C(\lambda )=0$ for every $C$, so $\lambda$ is a unit of $R$. Thus $f$ is a product of irreducibles of $R[x]$.

Assume now that $r\ge 1$ and that $k[x_1,\dots ,x_{r-1}]$ is a UFD in which every irreducible element is prime; this is the induction hypothesis.

For uniqueness, let $p_1\cdots p_m=q_1\cdots q_n$ be two factorizations of the same element $f\mathrm{\ne }0$ of $R[x]$ into irreducibles. Multiply the positive-degree factors of each side together, using the primitivity noted in 1.3; the result is primitive by [L3] on each side, so comparing contents as in 1.3 shows that the two sides' constant factors are associates and that the products of positive-degree factors are associates of one another. Each positive-degree factor is irreducible in $K[x]$ by [L3], and $K[x]$ is a UFD by [L5], so the two products of positive-degree factors agree up to order, associates in $K[x]$, and a unit scalar; that scalar is a unit of $R$ by the exponent comparison of 1.3, so they agree up to order and associates in $R[x]$. The constant factors are products of irreducibles of $R$ and agree up to order and associates by the uniqueness clause of [L1]. Hence $R[x]$ satisfies the uniqueness clause of [L1], and with 1.3 it is a UFD.

Every irreducible element of $R[x]$ is prime. Let $p\in R[x]$ be irreducible. If $\mathrm{deg}p=0$ then $p\in R$ is irreducible in $R$, hence prime in $R$ by 1.2; if $p\mid ab$ in $R[x]$, then the induced map $R[x]\to (R\mathrm{/}(p))[x]$ of [L7] kills $ab$, and $(R\mathrm{/}(p))[x]$ is a domain by [L7] because $R\mathrm{/}(p)$ is a domain for the prime element $p$; so all coefficients of $a$ or all coefficients of $b$ lie in $(p)$, that is, $p\mid a$ or $p\mid b$. If $\mathrm{deg}p\ge 1$, then $p$ is primitive by 1.3, hence irreducible in $K[x]$ by [L3], hence prime in the UFD $K[x]$ by 1.2; if $p\mid ab$ in $R[x]$, then $p\mid ab$ in $K[x]$, so after possibly swapping $a,b$ we have $a=pq$ for some $q\in K[x]$. Write $q=\lambda q^{\ast }$ with $\lambda \in K^{\times }$ and $q^{\ast }\in R[x]$ primitive, by the content construction of 1.3 applied to a polynomial clearing the denominators of $q$; then $a=\lambda (p{q}^{\ast })$, and the product $p{q}^{\ast }$ is primitive by [L3]. Comparing contents in the equality $a=\lambda (p{q}^{\ast })$ shows that $\lambda$ is associate to the content $c(a)\in R$ of step 1.3: clearing the denominators of $\lambda$ by some $u\mathrm{\ne }0$ gives $ua=(\lambda u)(p{q}^{\ast })$, whose left side has content $uc(a)$ and whose right side has content $\lambda u$ up to units because $p{q}^{\ast }$ is primitive, the exponents $v_p$ of step 1.3 being additive in a constant factor. Hence $\lambda \in R$ and $q=\lambda q^{\ast }\in R[x]$, so $a=pq$ with $q\in R[x]$ and $p\mid a$ in $R[x]$.

In any UFD $A$, every prime ideal of height one is generated by an irreducible element. By [L8], $\mathrm{ht}(\mathfrak{p})=1=\dim A_{\mathfrak{p}}$ means that inside $\mathfrak{p}$ there is a strict chain of primes of length one and none of length two; in particular $\mathfrak{p}$ contains a nonzero element, so choose $0\mathrm{\ne }f\in \mathfrak{p}$. Factoring $f$ into irreducibles and using that $\mathfrak{p}$ is prime, some irreducible factor $p$ lies in $\mathfrak{p}$ by [L1] and [L2]; then $p$ is prime by 1.2, so $(p)$ is a nonzero prime ideal contained in $\mathfrak{p}$. Were $(p)⊊\mathfrak{p}$, the strict chain $0⊊(p)⊊\mathfrak{p}$ of primes of $A$ would, by the inclusion-preserving bijection of [L8], give a chain of length two in $\mathrm{Spec}{A}_{\mathfrak{p}}$, contradicting $\dim A_{\mathfrak{p}}=1$. Hence $\mathfrak{p}=(p)$.

By [L6] the ring $k[x_1,\dots ,x_r]$ is the polynomial ring $A[x_r]$ over $A=k[x_1,\dots ,x_{r-1}]$, which is a UFD by step 1.4. Steps 1.3 and 2.1 therefore make $k[x_1,\dots ,x_r]$ a UFD, and step 2.2 shows that every irreducible element of it is prime, using the primitivity results quoted in those steps. So the UFD clause and the irreducible-is-prime clause hold for this $r\ge 1$ whenever they hold for $r-1$.

The base case 1.1 and the induction step 3.1 prove the UFD clause and the irreducible-is-prime clause for every $r\ge 0$. Finally, if $\mathfrak{p}\subseteq k[x_1,\dots ,x_r]$ is a prime ideal of height one, then $k[x_1,\dots ,x_r]$ is a UFD by 3.1, so 2.3 exhibits an irreducible element generating $\mathfrak{p}$; for $r=0$ the case is vacuous, a field having no nonzero prime ideal. This proves all three clauses of the statement.