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Let $A$ be a commutative ring, $B$ a commutative ring that is also an $A$-algebra of finite presentation. Let $f_1$ and $f_2$ be two elements of $B$ such that $(f_1) + (f_2) = B$ and as $A$-modules, $$B=A \oplus f_1 B \quad ; \quad B=A \oplus f_2 B.$$

In which cases do you know that there exists an $A$-automorphism of rings $\psi : B \to B$ such that $\psi(f_1) = f_2$ and $\psi(f_2) = f_1$ ?

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Please clarify the categories with which you work. Does $B = A \oplus f_1 B$ refer to the underlying $A$-modules? Is $\phi$ an automorphism of the $A$-algebra $B$? – Martin Brandenburg Feb 13 '13 at 20:05
Yes. And I would like to know if such a $\psi$ exists. – Damien L Feb 13 '13 at 20:14
Please edit the question to include all relevant information. – Mariano Suárez-Alvarez Feb 13 '13 at 20:48
Are $A,B$ commutative? What about other finiteness assumptions, for example is $A$ noetherian? Are $f_1,f_2$ regular? – Martin Brandenburg Feb 13 '13 at 21:55
$\rm A$ and $\rm B$ commutative yes but not noetherian. If you can say something in the case where $f_1$ and $f_2$ are regular, it is nice. – Damien L Feb 13 '13 at 22:04

A special case of this question with an algebro-geometric flavor is the following: Let $k$ be a field and let $X$ be an affine algebraic scheme over $k$. Does $\mathrm{Aut}(X)$ act transitively on the $k$-rational points of $X$ which correspond to principal maximal ideals? But this is far from being true since automorphisms preserve the local rings, which can be regular or not. A simple counterexample is given by $X = \mathrm{Spec}(k) \sqcup \mathrm{Spec}(k[\varepsilon]/\varepsilon^2)$. In purely algebraic terms and with the notation of the question, this means $A=k$, $B=k \times k[\varepsilon]/\epsilon^2$, $f_1=(0,1)$, $f_2 = (1,\varepsilon)$.

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