# Does ACC for principal ideal plus Krull dimension equal 0 imply DCC for principal ideals

Assume the ring is commutative and with 1.

We know that ACC + $$\dim(R)=0$$ imply DCC. However, if we only insist on the condition for principal ideals, can we conclude the same?

We know that having DCC on the principal ideals is the same as being a perfect ring but I'm not sure how to proceed from here.

(Note: ACC/DCC= ascending/descending chain condition.)

No. Consider $$K$$ a field and $$R=K[x_n:n\ge 0]/(x_n^2:n\ge 0)$$.

Clearly $$R$$ is local and its nilradical equals its unique maximal ideal, so it has Krull dimension zero. It has the properly descending sequence of ideals $$R\supset (x_0)\supset (x_0x_1)\supset\cdots$$.

But it has ACC on principal ideals.

Indeed, consider $$(a_k)_{k\ge 0}$$ such that $$a_{k+1}$$ divides $$a_k$$ for all $$k$$ (say $$a_k=a_{k+1}d_k$$); we have to show that $$a_k$$ divides $$a_{k+1}$$ for large enough $$k$$. This is clear if some $$a_k$$ is invertible, hence assume they're all in the maximal ideal. Also the case all $$a_k$$ zero being trivial, we can suppose (extracting if necessary) $$a_0\neq 0$$.

Let $$I$$ be the (nonempty finite) set of $$n$$ such that $$x_n$$ occurs in $$a_0$$. We consider the linear decomposition $$R=R_I\oplus J_I$$, where $$R_I$$ is the subring $$K[x_i:i\in I]$$ and $$J_I$$ is the ideal generated by the $$x_n$$ for $$n\notin I$$. Then the projection $$R\to R_I$$ (with respect to this decomposition) is a $$K$$-algebra homomorphism.

Write $$a_k=b_k+c_k$$ and $$d_k=e_k+f_k$$ in the above decomposition. Then $$b_k=b_{k+1}e_k$$ for all $$k$$. Since it belongs to the noetherian subring $$R_I$$, for $$k$$ large enough, $$b_k$$ is a nonzero scalar multiple of $$b_{k+1}$$. Up to rescaling, we suppose henceforth that all $$b_k$$ are equal for large enough $$k$$, say to $$b$$. Since $$b$$ divides $$b_0=a_0\neq 0$$, we have $$b\neq 0$$.

So $$be_k=b$$ for all large enough $$k$$, that is, $$b(e_k-1)=0$$. Since $$b\neq 0$$, this implies that $$e_k-1$$ is a zero divisor, and hence belongs to the maximal ideal, which implies that $$e_k$$ is invertible. So $$d_k$$ is invertible, and hence the sequence $$(a_k)$$ is stationary.

• Why is $b$ nonzero? – auniket Mar 29 at 11:12
• The question is then why $b_0 \neq 0$? E.g. why can't $a_0$ be a linear combination of some binomials? – auniket Mar 29 at 11:37
• @auniket you're right, I overly simplified. I've edited. Thanks! – YCor Mar 29 at 11:44