I have a question concerning the following local ring:


Is every ideal of $R$ a sum of a nilpotent ideal and an idempotent ideal?

As for any ideal of the form $(\bar X_n)$ which is a nilpotent ideal the answer is trivially in affirmative. Thanks for any suggestion!


Let $J$ be the ideal $(X_1, X_2^2-X_1,...,X_{n+1}^2-X_n,...)$ of $K[X_1,X_2,...]$, and let $x_i=X_i+J$ for all $i$. Thus, for each $i<j$ there exists $n$ such that $x_i=x_j^n\quad (*)$.
It follows that each term $x_{i_1}^{n_1}x_{i_2}^{n_2}\cdots x_{i_r}^{n_r}$ can be written as $x_k^m$ for $i_1,i_2,...,i_r\leq k$. Hence, if $f\in R$ there exists $k$ such that $$f=k_1x_k^{n_1}+k_2x_k^{n_2}+\cdots+k_{r}x_k^{n_r}$$ for ${n_1}<{n_2}<\cdots<{n_r}$, and so $f=x_k^{n_1}(k_1+k_2x_k^{n_2-n_1}+k_3x_k^{n_3-n_1}+\cdots+k_{r}x_k^{n_r-n_1})$, where $k_1+k_2x_k^{n_2-n_1}+k_3x_k^{n_3-n_1}+\cdots+k_{r}x_k^{n_r-n_1}$ is a unit in $R$.
It follows if $I$ is an ideal of $R$ such that $f\in I$ then $x_k^{n_1}\in I$. Therefore, if $I$ is an ideal of $R$ then there exists a family $\{x_\alpha\}_{\alpha\in \Gamma}$ and $\{n_\alpha\}_{\alpha\in\Gamma}\subseteq\mathbf{N}$ such that $I=(x_\alpha^{n_\alpha})_{\alpha\in\Gamma}$. Now, if $\Gamma$ is a finite set then $I$ is nilpotent, and if $\Gamma$ is an infinite set then $I=(x_\beta^{n_\beta})+(x_\alpha^{n_\alpha})_{\alpha\in\Gamma-\{\beta\}}$, and since $\Gamma-\{\beta\}$ is infinite, $(x_\alpha^{n_\alpha})_{\alpha\in \Gamma-\{\beta\}}$ is idempotent ideal of $R$ by $(*)$ property. (I hope this helps you.)

  • $\begingroup$ @E.Rostami I really could not reach the "idempotent-ness" of $I$ when it is infinitely generated. Please help! $\endgroup$ – karparvar Apr 10 '16 at 11:09

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