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 $0$.

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

2 Answers 2


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 $k$.

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    $\begingroup$ It looks like you've typed $(x,y)$ for $(f,g)$ twice? $\endgroup$ Jun 28, 2021 at 22:14
  • $\begingroup$ I did, thanks for catching this. Fixed. $\endgroup$ Jun 28, 2021 at 22:37
  • $\begingroup$ 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. $\endgroup$
    – markvs
    Jun 29, 2021 at 1:46
  • $\begingroup$ I might be spoilt by Oxford; here is what they teach in 3rd year undergraduate: courses.maths.ox.ac.uk/node/48825 $\endgroup$ Jun 29, 2021 at 9:00
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    $\begingroup$ 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. $\endgroup$
    – 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 $A$).

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

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