A key doubt about the resultant of two polynomials and their common zero locus

Question

I am studying algebraic geometry following Harris.
To prove that the projection of a variety $X\subset\mathbb{P}^n$ on an hyperplane $H\cong\mathbb{P}^{n-1}$ is a projective variety, the notion of resultant of two polynomials is used there. In particular let us fix:

• $F,G\in K[x_0,…,x_n]\cong K[x_0,…,x_{n-1}][x_n]$
• $p=[0:0:…:1]\in\mathbb{P}^n$, $q\in H$, $H\cong\mathbb{P}^{n-1}=V(x_n)$

Now it is clear to me that the resultant $R(F,G)(q)=0$ iff $F[q](x_n), G[q](x_n)$, the polynomials in $x_n$ resulting from the evaluation of $x_1,…,x_{n-1}$ in the coordinates of $q$, have a common zero or the leading coefficient of $F$ or $G$ vanishes in $q$, as a polynomial of $K[x_0,…,x_{n-1}]$.

What is not clear is why this is the case if and only if $F,G$ have a common zero on the line $l=\bar{pq}$.

Thanks in advance for help.

P.s. these are the two pages where Harris succintly adresses the problem.

Answer 1

Let's write down the claims:

• A) The line $l=\overline{pq}$ meets $X$.

• B) Every pair of homogeneous $F,G\in I(X)$ has a common zero on $l$

• C) $Res_{x_n}(F,G)$ vanishes at $q$ for all homogeneous pairs $F,G\in I(X)$.

You state that you understand that C) is equivalent to an intermediate result but aren't sure about why C) should be equivalent to A). We'll go through the steps carefully.

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First, we'll establish that A) and B) are equivalent (you didn't mention an issue with this, but I want to make sure we cover it anyways, plus it's short).

A) clearly implies B): any $F,G\in I(X)$ will have a common zero at every point in $l\cap X$ which is assumed nonempty.

For the other direction of the equivalence, we prove the contrapositive: if $X\cap l=\emptyset$, then there exist homogeneous $F,G\in I(X)$ so that $F,G$ have no common zero on $l$. Assume $X\cap l=\emptyset$. Up to a change of coordinates, we may assume $l=V(x_2,\cdots,x_n)\subset \Bbb P^n$. Now on $U_0=D(x_0)$, we have $X_0:= X\cap U_0$ and $l_0:= l\cap U_0$ are affine varieties which do not meet. Thus $I_{U_0}(X_0)+I_{U_0}(l_0)=(1)$, so we can find elements $a\in I_{U_0}(X_0)$ and $b\in I_{U_0}(l_0)$ so that $a+b=1$. Then $a$ vanishes on $X_0$ but not $l_0$, and after homogenizing it to $\widetilde{a}$ and multiplying by some power of $x_0$, we get that $x_0^p\widetilde{a}$ is a homogeneous element of $I(X)$ which vanishes only at $[0:1:0:\cdots:0]\in l$. Repeating this construction on $U_1=D(x_1)$, we get a homogeneous element of $I(X)$ which vanishes only at $[1:0:\cdots:0]\in l$. These two elements are our $F,G$ which do not share a common zero on $l$, so B implies A by contrapositive.

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Before we start tackling the equivalence of B) and C), let's recall some facts about the resultant:

1) The resultant of two polynomials with coefficients from an integral domain is zero iff they have a common divisor of positive degree.

2) If $A,B$ are two polynomials in $R[x]$ and $\varphi: R\to S$ is a ring homomorphism which extends to a ring homomorphism $\varphi:R[x]\to S[x]$ in the natural way, then:

• $Res_x(\varphi(A),\varphi(B))=\varphi(Res_x(A,B))$ if $\deg_x A = \deg_x \varphi(A)$ and $\deg_x B = \deg_x \varphi(B)$

• $\varphi(Res_x(A,B))=0$ if $\deg_x A > \deg_x \varphi(A)$ and $\deg_x B > \deg_x \varphi(B)$

• $\varphi(Res_x(A,B))=\varphi(a)^{\deg B-\deg \varphi(B)}Res_x(\varphi(A),\varphi(B))$ if $\deg A =\deg \varphi(A)$ and $\deg B > \deg \varphi(B)$ where $a$ is the top coefficient of $A$.

• $\varphi(Res_x(A,B))=\pm \varphi(b)^{\deg A-\deg \varphi(A)}Res_x(\varphi(A),\varphi(B))$ if $\deg B =\deg \varphi(B)$ and $\deg A > \deg \varphi(A)$ where $b$ is the top coefficient of $B$.

Each of these parts of 2) can be proven by noticing that $\varphi$ commutes with $\det$ since it's a polynomial.

(See the Wikipedia page if you care about when the $\pm$ is a $+$ versus a $-$.)

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Now let's look at the equivalence of B) and C). We'll quantify it as follows: for any pair $F,G\in I(X)$, their having a common zero on $l$ is equivalent to $Res_{x_n}(F,G)(q)=0$.

Suppose either $F$ or $G$ satisfies the condition that it's leading coefficient as a polynomial in $x_n$ doesn't vanish upon evaluation at $q$ (aka restriction to $l$). We apply the fact 2) about the resultant with $\varphi$ being the evaluation at $q$ map: the first, third, or fourth part of this fact applies, and we have that $Res_{x_n}(F,G)(q)=0$ iff $Res_{x_n}(F[q],G[q])=0$. But $Res_{x_n}(F[q],G[q]) = 0$ iff $F[q]$ and $G[q]$ have a common factor of positive degree by fact 1) about resultants, and this common factor is exactly equivalent to a common zero on $l$, so we see that B) and C) are equivalent in this case.

In the case where $F$ and $G$ both have leading coefficients as polynomials in $x_n$ which vanish upon plugging in $q$, we show that conditions B) and C) are automatically true. As $p\Rightarrow q$ is equivalent to $\neg p \vee q$, this will show that B) and C) are equivalent in this case.

If $F,G$ both have leading coefficients as polynomials in $x_n$ which vanish upon plugging in $q$, we are in the situation of the second part of fact 2), so $Res_{x_n}(F,G)(q)=0$. Similarly, the vanishing of the leading coefficient implies that $F,G$ both have a zero at $p$ because $\deg_{x_n} F < \deg F$. (To prove this last bit, it may be instructive to note that up to a change of coordinates leaving $p$ fixed, we may take $q=[1:0:\ldots:0]$, so that $F[q],G[q]$ are either zero or divisible by $x_0$ and therefore must have a zero at $p$.)

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I have to admit that I personally got a little turned around a few times attempting to write the last part of this answer - the key thing to note is that there's a case where the equivalence of B) and C) is automatic because they're both just true from the assumptions in this special case. Hope this helps!