# Do relatively prime polynomials $f$ and $g$ in $k[x,y]$ generate an ideal of finite codimension?

Let $$k$$ be a field and $$f,g \in k[x,y]$$ be two non-constant polynomials in two variables. Is it true that the following conditions are equivalent:

1. $$f$$ and $$g$$ are relatively prime in $$k[x,y]$$, in the sense that they have no non-constant common divisors, so if we have the identity $$\alpha f + \beta g = 0$$ in $$k[x,y]$$ for some polynomials $$\alpha,\beta\in k[x,y]$$, then $$\alpha = -g \gamma$$ and $$\beta = f \gamma$$ for some $$\gamma \in k[x,y]$$.

2. The ideal $$(f,g)$$ generated by $$f,g$$ is of finite codimension in $$k[x,y]$$ over $$k$$, i.e. $$\dim_{k} k[x,y] / (f,g) < \infty$$.

The following simple arguments prove that 2) implies 1).

Suppose $$f = a q$$ and $$g = b q$$ for some $$a,b,q\in k[x,y]$$ and $$q$$ is non-constant, so 1) fail. Exchanging if necessary $$x$$ and $$y$$, one can assume that $$q(x,y) \not= x^k$$ for all $$k$$. Then neither of the following infinite and linearly independent over $$k$$ family of polynomials $$\{x^i\}_{i\geq0}$$ belongs to $$(q) \supset (f,g)$$. Whence $$\dim_{k} k[x,y] / (f,g) \geq \dim_{k} k[x,y] / (q) = \infty$$, and thus 2) fails as well.

Thus the question is whether 1) implies 2)?

Perhaps one should assume that $$k$$ has characteristic $$0$$.

• This is also equivalent to: "$V(f)\cap V(g)$ is finite", where $V(f)$ is the set of zeros of $f$ in $\bar{k}^2$.
– YCor
Jun 27, 2021 at 23:43
• If I remember correctly you can find a proof in Fulton, Algebraic Curves. Notice that k is algebraically closed wlog. Jun 28, 2021 at 0:54
• Condition (1) means that the irreducible components of $V(f)$ and $V(g)$ are distinct. Now the intersection of two distinct irreducible closed subsets of dimension $1$ of $A^2_k$ consists of a finite set of closed points. This implies (2). Jun 28, 2021 at 12:41
• the usual way to show this would be to compute the resultants en.m.wikipedia.org/wiki/Resultant Jun 28, 2021 at 14:51
• @MarkSapir Sure, but there is an easy reduction (linear algebra) to the case k algebraically closed, which is presumably what Martin meant. Jun 28, 2021 at 21:44

A standard undergraduate maths approach is via resultants.

I am not going to survey resultants here (but see below), I'll just say that an immediate consequence of 1) is that there exists a nonzero monic $$u:=Res_y(f,g)\in k[x]\cap (f,g)$$ and a nonzero monic $$v:=Res_x(f,g)\in k[y]\cap (f,g)$$. Therefore, any monomial $$x^s y^t$$ in $$h\in R:=k[x,y]/(f,g)$$ satisfies $$s<\deg u$$, $$t<\deg v$$, as you can reduce $$x$$-degree below $$\deg u$$ by replacing $$x^{\deg u}$$ with $$u(x)-x^{\deg u}$$, and analogously for $$y$$-degree (using $$v$$). Hence $$R$$ is finite-dimensional.

EDIT: Remarks, definitions and explanations.

Degree $$d$$ polynomials form $$d+1$$-dimensional vectorspace $$k_d[t]\subset k[t]$$ over $$k$$. The resultant $$Res_t(p,q)$$ of $$p\in k_d[t]$$, $$q\in k_e[t]$$ is the determinant $$\det M$$ of the linear map $$M:k_d[t]\times k_e[t]\to k_{d+e}[t]$$ defined by $$M(w,z)=wq+zp$$ (here $$w\in k_d[t]$$, $$z\in k_e[t]$$) - I am cheating here a bit with dimensions, as $$\dim (k_d[t]\times k_e[t])=\dim(k_{d+e}[t])+1$$, but we don't want this extra 1, so we assume w.l.o.g. that $$z$$ is monic, i.e. $$z(t)=t^e+z_{e-1}t^{e-1}+\dots +z_0$$. Assuming that $$p,q$$ have a common root $$t^*$$ in the algebraic closure of $$k$$, we see that $$M$$ cannot be 1-1 in this case, as $$M$$ would be divisible by $$t-t^*$$ for any $$w,z$$, i.e. $$\det M=Res_t(p,q)=0$$. Usually $$M$$ is written using monomial bases of $$k_d[t]$$ and $$k_e[t]$$, as Sylvester_matrix.

We see that it readily generalises to rings, e.g. if we replace $$k$$ with $$k[x]$$ then $$\det M\in k[x]$$, and we can think of $$Res_y(f,g)\in k[y]$$, vanishing at common roots of $$f$$ and $$g$$ in the algebraic closure of $$k$$.

• It looks like you've typed $(x,y)$ for $(f,g)$ twice? Jun 28, 2021 at 22:14
• I did, thanks for catching this. Fixed. Jun 28, 2021 at 22:37
• The answer is nice but one can add two short phrases: that $u=Resultant(f,g,y)$ and $v=Resultant(f,g,x)$. The fact that $u,v\ne 0$ is immediate from 1) and the fact that $u.v\in (f,g)$ follows from a definition of resultant. I am not sure about undergraduate math, though. Jun 29, 2021 at 1:46
• I might be spoilt by Oxford; here is what they teach in 3rd year undergraduate: courses.maths.ox.ac.uk/node/48825 Jun 29, 2021 at 9:00
• For resultants, you can't assume that the polynomials in question are monic when the coefficients are only assumed to be elements of a ring. And it does not help to pass to the algebraic closures of the fraction fields $k(x)$ and $k(y)$. Instead, one needs to exploit the fact that $k[x,y]$ is a UFD, i.e. the first point in the question, to see that the resultant map $(u,v)\mapsto uf+vg$ in question is injective.
– Z. M
Jun 29, 2021 at 15:32

There's perhaps a more elementary answer than this, but here's what I have so far. Define $$A := k[x,y]$$ and $$I := (f,g)$$. I'll be using the following fact:

Fact: $$A$$ is a Cohen-Macaulay UFD.

To prove that $$\dim_k A/I < \infty$$, it suffices to show that the Krull dimension of $$A/I$$ (denoted simply by $$\dim A/I$$) is zero. For then $$A/I$$ is an Artinian $$k$$-algebra of finite type, a fortiori finite-dimensional over $$k$$.

Now recall in a Cohen-Macaulay ring $$A$$ that we have the inequality $$\dim A \geq \dim A/I + \operatorname{ht}(I)$$, where $$\operatorname{ht}(I)$$ is the height of $$I$$, defined as $$\operatorname{ht}(I) = \min \{\operatorname{ht}(\mathfrak{p}) | I \subseteq \mathfrak{p} \text{ is prime} \}.$$ Therefore, it is enough to show that there is no prime ideal $$\mathfrak{p}$$ containing $$I$$ of height $$0$$ or $$1$$.

Suppose $$I$$ is contained in some $$\mathfrak{p}$$ of height zero, i.e. a minimal prime. Since $$A$$ is a domain this means that $$\mathfrak{p} = 0$$, i.e. that $$(f,g) = 0$$ which is ridiculous. If $$I \subset \mathfrak{q}$$ where $$\operatorname{ht}(\mathfrak{q}) = 1$$, then since $$A$$ is a UFD, we must have $$\mathfrak{q} = (h)$$ for some polynomial $$h$$. Then we have $$(f,g) \subset (h)$$, contradicting the fact that $$f,g$$ are relatively prime.

Edit: As Bogdan Zavyalov pointed out to me, it's not necessarily true for an ideal $$I$$ in a Cohen-Macaulay ring $$A$$ that $$\operatorname{ht}(I) + \dim A/I = \dim A$$. Consider $$A = \mathbf{Z}_p[X]$$ and $$I = (1 - pX)$$. By the Krull Hauptidealsatz, $$\operatorname{ht}(I) = 1$$. On the other hand, $$A/I = \mathbf{Q}_p$$ which is zero-dimensional. Hence $$\operatorname{ht}(I) + \dim A/I = 1$$ which is strictly less than $$2$$ (the Krull dimension of $$A$$).

• Just out of curiosity, what does "f.t." stand for? Is it finite type? Jun 28, 2021 at 13:43
• @Malkoun Yes that's right. Jun 28, 2021 at 14:42
• Is it true that in a 2-dim UFD, any coprime pair $(f,g)$ generates an ideal of finite colength? or is CM necessary?
– YCor
Jul 22, 2021 at 6:36