If $R$ is a commutative ring, $\alpha: R \to R$ an automorphism of $R$, and $M$ a maximal ideal of $R$, then $\alpha(M)$ is also a maximal ideal of $R$ with the same quotient field. So the group of automorphisms of $R$ acts on the set of maximal ideals of $R$.

My question is: when is this action transitive? (I'm asking because I've proved something about certain non-commutative ring extensions of a commutative base ring if we assume that the base ring has this property).

Some examples:

If $R = k[x_1, ..., x_n]$ where $k$ is an algebraically closed field, then this action is transitive - the automorphism $x_i \mapsto x_i - \lambda_i$ sends $(x_1, ..., x_n)$ to $(x_1 - \lambda_1, ..., x_n - \lambda_n)$.

Similarly, if $R = k[x_1^{\pm 1}, ..., x_n^{\pm 1}]$, $k$ still an algebraically closed field, then the automorphism $x_i \mapsto \lambda_i^{-1}x_i$ (all $\lambda_i$ nonzero) sends $(x_1 - 1, ..., x_n - 1)$ to $(x_1 - \lambda_1, ..., x_n - \lambda_n)$, and so again the action is transitive.

On the other side, any automorphism of $k[x, y]/(xy)$ must fix the maximal ideal $(x, y)$, so it's not always the case that the action is transitive. (And obviously, if $R$ is a nontrivial algebra over a non algebraically closed base field then the action definitely won't be transitive).

So can we classify such rings in any way? Do we even have a name for such rings? Or anything in between, really. I'm only really interested in the case where $R$ is finitely generated (as an algebra) over some algebraically closed field $k$, if that makes things easier.

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    $\begingroup$ It's easy to see that such a ring $R$ must satisfy that for any two maximal ideals $M_1,M_2$ of $R$, $R/M_1\cong R/M_2$. This constraint implies that $R$ must be a $\mathbb F_p$ algebra or $\mathbb Q$ algebra or $\mathbb Z_{(p)}$ algebra such that $(p)$ is contained in the Jacobson radical (otherwise we can always find two maximal ideals of $R$ of which the quotient rings are of different characteristics). $\endgroup$ – Censi LI May 6 '15 at 21:35

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