Classes of "bases" of an ideal of polynomials

Disclaimer. This answer is far from complete. Some pointers are provided (see [1] and [2]) that could help go beyond the results presented below.

Rings are supposed to be commutative and unital. Given such a ring $R$, we denote by $R^{\times}$ its unit group. For $r, s \in R$, we denote by $(r:s)$ the ideal of $R$ consisting of the elements $\lambda$ such that $r$ divides $\lambda s$.

Claim. Let $R$ be a ring. Let $I$ be an ideal of $R$ be generated by two elements $a$ and $b$ which are not zero divisors. Let $$\operatorname{V}_2(I) = \{(r, s) \in R^2 \, \vert \, Rr + Rs = I\}$$ and let $J := \operatorname{Fitt}_1(I)$, that is, the second Fitting ideal of $I$. Then the following hold.

1. If $I$ is principal, then $\vert \operatorname{V}_2(I)/\operatorname{SL}_2(R) \vert = 1$.
2. The exterior product $\Lambda^2(I)$ is $R$-isomorphic to $R/J$ so that there is a unique $R$-alternating map $\det_{(a,b)}: I \times I \rightarrow R/J$ such that $\det_{(a, b)}(a, b) = 1 + J$.
3. The set $\operatorname{V}_2(I)/\operatorname{SL}_2(R)$ is equipotent with $\det_{(a, b)}(\operatorname{V}_2(I)) \subseteq (R/J)^{\times}$. The set $\det_{(a, b)}(\operatorname{V}_2(I))$ contains the image of $R^{\times}$ under the natural map $R \twoheadrightarrow R/J$.
4. If the natural map $(R/ (a:b))^{\times} \rightarrow (R/J)^{\times}$ is surjective, then $\det_{(a, b)}(\operatorname{V}_2(I)) = (R/J)^{\times}$.
5. If $R$ is a GCD domain, e.g. $R$ is unique factorization domain (UFD), then $J = \frac{1}{\gcd(a, b)} I$.

The above claim proved useful for rings of Krull dimension at most $1$ [1]. It has nonetheless some modest bearing on the case $R = k[x, y]$; see the propositions below.

Proof.

1. This follows from the straightforward and well-known identity $\vert \operatorname{V}_2(R)/\operatorname{SL}_2(R) \vert = 1$.
2. By Exercise 20.9.i of [3], the ideal $J$ is the annihilator of $\Lambda^2(I)$, which is clearly a cyclic $R$-module. Hence there is a unique $R$-isomorphism $\phi: \Lambda^2(I) \rightarrow R/J$ such that $\phi(a \wedge b) = 1 + J$. The map $\det_{(a, b)}$ defined by $\det_{(a, b)}(c, d) = \phi(c \wedge d)$ where $c, d \in I$ has the desired properties.
3. Apply [1, Theorem A].
4. Apply [1, Lemma 4.1].
5. Observe that $J = (a:b) + (b:a)$ and that $(a:b) = Ra$ if $\gcd(a, b) = 1$.

Proposition 1. Let $k$ be an algebraically closed field, let $R = k[x, y]$ and let $I$ be a maximal ideal of $R$. Then $\vert \operatorname{V}_2(I)/\operatorname{GL}_2(R) \vert = 1$.

Proof. Reason as in [1, Example 3.9] to infer that $\operatorname{V}_2(I)/\operatorname{SL}_2(R)$ identifies with $k^{\times}$.

Proposition 2. Let $k$ be a field and let $R = k[x, y]$. Let $f \in k[y]$ and let $I$ be the ideal of $R$ generated by $x$ and $f$. Then $\operatorname{V}_2(I)/\operatorname{SL}_2(R)$ is equipotent with $(R/I)^{\times} = (k[y]/(f))^{\times}$.

Proof. Use the assertions (4) and (5) of the above claim.

Example. Let $R = \mathbb{R}[x, y]$ and let $I$ be the ideal of $R$ generated by $x$ and $y^2 + 1$. By Proposition 2, the orbit set $\operatorname{V}_2(I)/\operatorname{SL}_2(R)$ is equipotent with the unit group of $R/I$. As $R/I \simeq \mathbb{R}[y]/(y^2 + 1) \simeq \mathbb{C}$, we infer that $\operatorname{V}_2(I)/\operatorname{SL}_2(R) \simeq \mathbb{C}^{\times}$ and it easily follows that $$\operatorname{V}_2(I)/\operatorname{GL}_2(R) \simeq \mathbb{C}^{\times} / \mathbb{R}^{\times}.$$

• [1] L. Guyot, "Equivalent generating pairs of an ideal of a commutative ring", 2020.
• [2] M. Murthy, "Generators of a general ideal", 2003.
• [3] D. Eisenbug, "Commutative Algebra", 1995.