3
$\begingroup$

I have a question concerning the following local ring:

$$R=K[X_1,...X_n,...]/(X_1,X_2^2-X_1,...,X^2_{n+1}-X_n,...).$$

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!

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.