In an algebra test the following problem was presented:

Any commutative local ring $(R,\mathfrak m)$ with $\mathfrak m$ principal so that $⋂_{i≥0}\mathfrak m^i =0$ is Noetherian and each nonzero ideal of $R$ is a power of $\mathfrak m$.

It could be proven (assuming $R$ to be Noetherian) that each nonzero ideal, being finitely generated, is a subset of a power of $\mathfrak m$. Could anybody be so kind as to give some suggestions. Greateful!


Let $I$ be an ideal of $R$, $I\ne 0,R$. Then $I\subseteq\mathfrak m$, but $I$ can't be contained in all the powers of $\mathfrak m$, otherwise $I=0$, contradiction. So there is a greatest $i\ge 1$ such that $I\subseteq \mathfrak m^i$, and $I\not\subseteq\mathfrak m^{i+1}$. Let's show that $I=\mathfrak m^i$. Since $I\not\subseteq\mathfrak m^{i+1}$ there is $a\in I\setminus\mathfrak m^{i+1}$. If $\mathfrak m=(\pi)$, then $a=\pi^ia'$. Since $a\notin\mathfrak m^{i+1}$ we get $a'\notin\mathfrak m$, so $a'$ is invertible. From $a=\pi^ia'$ we get now that $\pi^i\in I$ and therefore $I=\mathfrak m^i$.


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.