If $F,G \in k[X,Y]$ have no common factors, then $V(F,G)$ is finite.

Question

Let $k$ be a field and suppose $F,G$ do not have common factors in $k[X,Y]$.

Then $V(F,G) = \{P \in k^2\mid F(P) = 0 = G(P)\}$ is finite.

The proof my book provides does the following:

Because $F,G$ have no common factors in $k[X,Y] = k[X][Y]$, they don't have common factors in $k(X)[Y]$ (here $k(X)$ is the fraction field of $k[X]$). Because $k(X)[Y]$ is a PID, we have

$$(F,G) = 1$$ in $k(X)[Y]$

Questions:

(1) I can see that if $F,G$ have no non-constant common factors in $k[X,Y] = k[X][Y]$, they don't have common non-constant factors in $k(X)[Y]$. Shouldn't this be included in the proof?

(2) Why does it follow that $(F,G) = 1?$ I know that a gcd always exists in a PID.

Do you mean non-constant in $k[X,Y]$ or in $k[X][Y]$? conventionally a constant in the former is some $c \in k$ but in the latter is any polynomial $f \in k[X]$. Saying there are no common factors in $k[X,Y]$ implies there are no nonconstant common factors in $k[X][Y]$ — Badam Baplan, Oct 31 '18 at 11:59
No common factors in $k[X,Y]$. But if there is a common constant factor in $k(X)[Y]$ that can violate that the gcd is 1 right? — , Oct 31 '18 at 12:01
Two related posts: Irreducible polynomials and affine variety and Bezout in $\mathbb C [x,y]$ — Viktor Vaughn, Oct 31 '18 at 14:05

score 1 · Accepted Answer · answered Oct 31 '18 at 12:28

A constant in $k(X)[Y]$ is an element of $k(X)$ and hence a unit. Units are excluded from the definition of factor because they divide everything. Greatest Common Divisors and Least Common Multiples are only meaningfully defined up to unit multiples. With this in mind, knowing that $F,G$ have no common factors in $k(X)[Y]$, you know that their $GCD$ is $1$. In a PID, the GCD of two elements generates their ideal.

score 0 · Answer 2 · answered Oct 31 '18 at 11:56

0

The first claim (1) is Gauss' lemma. Second claim follows from the fact $(F,G)$ is an ideal, and it must be principal. It's generator is the gcd of $F$ and $G$, which is in this case $1$ because we assumed $F$ and $G$ have no common factors in $\mathbb k[X,Y]$ and by Gauss' lemma this transfers to $\mathbb k(X)[Y]$.

answered Oct 31 '18 at 11:56

Pedro

122,002

But doesn't Gauss lemma say anything about non-constant common factors (the non-constant part is left out here). – Oct 31 '18 at 11:57
@Math_QED I am not sure I understand your question. – Pedro Oct 31 '18 at 12:26
Perhaps the first section of these notes helps. – Pedro Oct 31 '18 at 12:29

If $F,G \in k[X,Y]$ have no common factors, then $V(F,G)$ is finite.

2 Answers2