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Jun 23, 2020 at 18:44 comment added M.G. @R.vanDobbendeBruyn: thanks for making it obvious what to look at! If you post it as an answer, I will accept it.
Jun 22, 2020 at 4:51 review Close votes
Jul 1, 2020 at 9:46
Jun 21, 2020 at 19:49 comment added R. van Dobben de Bruyn In fact in the case $\mathfrak m^2 = 0$ where $R$ contains $k = R/\mathfrak m$, we must have $$R = k \oplus M = \operatorname{Sym}^*(M)/\operatorname{Sym}^{\geq 2}(M)$$ for some $k$-module $M$. This is an example of the type you're looking for if and only if $M$ is not finitely generated.
Jun 21, 2020 at 19:43 comment added R. van Dobben de Bruyn Take $R = k[x_1,\ldots]$ with $\mathfrak m = (x_1,\ldots)$ and look at $R/\mathfrak m^2$. (However, this does not work for any maximal ideal that is not finitely generated, because it is possible that $\mathfrak m = \mathfrak m^2$, in which case $R/\mathfrak m^2 = R/\mathfrak m$ is a field. An example of such an $\mathfrak m$ is the maximal ideal in a valuation ring with divisible value group.)
Jun 21, 2020 at 19:27 history asked M.G. CC BY-SA 4.0