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In the text below, when I mention curve defined by a polynomial $p \in k[x,y,z]$ I mean its set of zeros in $\Bbb P^2,$ i.e.,

$$ \text{ curve defined by $p$} = Z(p) = \{P \in \Bbb P^2 \, \colon \, p(P) = 0\}.$$

I wish to prove the following claim:

Claim. Given two coprime, nonzero and homogeneous polynomials $f,g \in k[x,y,z],$ the intersection of the curves defined by $f$ and $g$ is finite, i.e., $Z(f) \cap Z(g)$ is finite.

My attempt. We know we can partition the points of $\Bbb P^2$ into three sets of coordinates given by $(1:0:0), (a:b:1)$ and $(a:1:0).$ It is of trivial observation that $(1:0:0)$ is only one point and thus it doesn't bring problems of finitness. Furthermore, points with coordinates $(a:b:1)$ can be related to points of $\Bbb A^2,$ in which we know the intersection in cause is finite (there are several questions about this in MSE). With this in mind, we only worry about the points of coordinates $(a:1:0).$

In this case, we are looking for points such that $f(a,1,0) = g(a,1,0) = 0.$ So, we study the polynomials $f(x,1,0)$ and $g(x,1,0).$ Ideally, if one of this polynomials is nonzero (when treated as a function of $x$) we get the result wanted:

Suppose that (at least) one of the polynomials $f(x,1,0)$ or $g(x,1,0)$ is nonzero. We have a polynomial of one variable that's nonzero over a field, which has obviously only a finite number of roots. In particular, this means that there are only finitely many points of coordinates $(a:1:0)$ that vanish simultaneously on $f$ and $g$, proving what's wanted.

So, all we need to do is to show that indeed one of the polynomials $f(x,1,0)$ or $g(x,1,0)$ is nonzero. I am not sure how to reach this result: I think coprimality will be key here, since perhaps we can somehow relate this polynomials with having common factor of $z$ ?

This is my idea, but I don't know how to develop it formally.

Thanks for any help in advance.

xyz
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  • Hint: since $f(x,y,z), g(x,y,z)$ are homogeneous, $f(x,y,0)$ and $g(x,y,0)$ are also homogeneous... – Mor A. Jun 25 '23 at 19:38
  • @MorA. Thanks for your hint; I will try to use in my thinking: My goal is to show that (at least) one of the polynomials $f(x,1,0)$ or $g(x,1,0)$ is nonzero. For this, assume that they are both constantly zero. Then, since $f(x,y,0)$ and $g(x,y,0)$ are homogeneous, we have that $$ f(xy,y,0) = y^{\partial f} f(x,1,0) = 0, $$ and we can do the same for $g.$ I don't see how this helps thought... (here $\partial f$ represents the degree of $f$). – xyz Jun 25 '23 at 19:49
  • What do you know about the factors of homogeneous polynomials? What does $f(x,1,0)=0$ for all $x\in k$ tell you about $f(x,1,0)$ as a polynomial in $k[x]$? What does that tell you about $f(x,y,0)$ as a polynomial in $(k[x])[y]$? (note that we're assuming $k$ is infinite otherwise the result trivially holds) – Mor A. Jun 25 '23 at 20:02
  • @MorA. I know that factors of homogeneous polynomials are also homogeneous. I don't know how to deduce that $f(x,1,0) = 0 \implies f(x,y,0) = 0$ but I've tried several examples that seem to suggest this. Can you give me some more hints on how to prove this formally? – xyz Jun 25 '23 at 21:38
  • I feel like there could be an “elegant” solution, but not sure :) – julio_es_sui_glace Jun 25 '23 at 23:44
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    Does this answer your question? Intersection of two curves Or either of these? Irreducible polynomials and affine variety, Bezout in $\mathbb C [x,y]$ – Viktor Vaughn Jun 26 '23 at 01:29

1 Answers1

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Here's the hints I gave in the comments fleshed out:

We assume that $k$ is an infinite field.

First, let $f(x,y,z)$ be a homogeneous polynomial, therefore $f(x,y,0)$ is also a homogeneous polynomial.

Now suppose that $f(x,1,0)=0$ for all $x\in k$, since $k$ is infinite this implies that $f(x,1,0)=0$ as a polynomial in $k[x]$.

Now consider $f(x,y,0)$ as a polynomial in $(k[x])[y]$, then by the factor theorem (which holds for polynomials over any commutative unital ring), we get that $y-1$ divides $f(x,y,0)$.

Since $f(x,y,0)$ is a homogeneous polynomial, if it is nonzero then its factors are also homogeneous. But we've shown that $y-1$ divides it, and it isn't homogeneous, so $f(x,y,0)$ must be the zero polynomial, so again by the factor theorem we get that $z$ divides $f(x,y,z)$.

Now to conclude the proof of the statement in the question, suppose $f(x,y,z),g(x,y,z)$ are coprime, homogeneous polynomials, then since $z$ isn't a common factor of $f(x,y,z)$ and $g(x,y,z)$, we must have that at least one of the polynomials $f(x,y,0), g(x,y,0)$ is nonzero, and since the polynomials are homogeneous at least one of the polynomials $f(x,1,0), g(x,1,0)$ is nonzero by what we've shown above.

From here you can complete the proof as you outlined in the question.

Finally, note that the nonzero assumption in the claim can be dropped, since if one of the polynomials is zero then by coprimality the other polynomial must be invertible, and so has no roots.

Mor A.
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