Let $(R,\mathfrak m,k)$ be an Artinian Gorenstein local ring such that $$\mu(\mathfrak m)=2, \quad\mathfrak m^2\ne 0,\quad\text{and}\quad \mathfrak m^3=0.$$
Then, is it true that every non-maximal ideal of $R$ is principal?


*Thoughts*: We have $\mathfrak m^2 \subseteq (0:\_R \mathfrak m)$. Since $R$ is Artinian Gorenstein, so $\dim\_k (0:\_R \mathfrak m)=1$, hence $\dim\_k \mathfrak m^2=1$. Thus, $\mathfrak m^2, (0:\_R \mathfrak m)$ are principal ideals. For an arbitrary ideal $I\subsetneq \mathfrak m$, we have
$$\begin{array}{r@{}l}
\mu(I)
&=\dim\_k(I/\mathfrak mI)\\
&\le \dim\_k(\mathfrak m/\mathfrak m I)-1\\
&=\dim\_k(\mathfrak m/\mathfrak m^2)+\dim\_k(\mathfrak m^2/\mathfrak m I)-1\\
&\le \mu(\mathfrak m)+1-1\\
&=2,
\end{array}
$$
which is just one off from being principal.


[Also note that since $\mu(\mathfrak m) -\dim R=2$ and $R$ is Gorenstein, so a result of Serre implies $R$ is complete intersection]



Ideals in Artinian Gorenstein local ring $(R,\mathfrak m)$ with $\mu(\mathfrak m)=2, \mathfrak m^2\ne 0$ and $\mathfrak m^3=0$
We use the fact that in an Artinian Gorenstein ring, any ideal contains the socle. The assumption tells us that the socle of $A$ is $\mathfrak m^2$, which is principal.


Let $I\neq (0)$ be a non-maximal ideal. If $I=\mathfrak m^2$, we are done. Otherwise, $I$ strictly contains $\mathfrak m^2$. Thus $\mathfrak mI\neq 0$, but then $\mathfrak mI \supset\mathfrak m^2$. On the other hand as $I\subset \mathfrak m$, $\mathfrak mI = \mathfrak m^2$.


We have $I/\mathfrak mI = I/\mathfrak m^2$ is a non-zero proper subspace of $\mathfrak m/\mathfrak m^2$ which has $k$-dimension 2, so it has $k$-dimension $1$.


