Ideals in Artinian Gorenstein local ring $(R,\mathfrak m)$ with $\mu(\mathfrak m)=2, \mathfrak m^2\ne 0$ and $\mathfrak m^3=0$

Question

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]

Answer 1

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 $1$.

Answer 2

The answer by @Hailong Dao can be slightly generalized to easily show the following:

Let $(R,\mathfrak m,k) $ be an Artinian local Gorenstein ring with $\mathfrak m^3=0 $ and $\mathfrak m^2\ne 0.$

Then, it holds that $\mu(I) \le \max \{1, \mu(\mathfrak m)-1\}$ for every ideal $I\ne \mathfrak m.$

Proof: Enough to prove the claim for every non-zero ideal properly contained in $\mathfrak m$.
We will use that in an Artinian Gorenstein local ring, every non-zero ideal contains the socle $(0:\mathfrak m)$. Also, in our case, the non-zero ideal $\mathfrak m^2$ is inside $(0:\mathfrak m)$. Thus $\mathfrak m^2=(0:\mathfrak m)$, and this is a $1$-dimensional $k$-vector space. Now let $I $ be a non-zero ideal strictly contained in $\mathfrak m$. We know $I$ contains the socle $\mathfrak m^2.$ If $ I=\mathfrak m^2, $ then $\mu(I)=\mu(\mathfrak m^2)=\dim_k (\mathfrak m^2)=1 $ and we are done. Otherwise $I$ strictly contains $\mathfrak m^2$. In this case, $\mathfrak mI$ is non-zero (as otherwise, $\mathfrak mI=0$ implies $I $ is inside $(0:\mathfrak m)=\mathfrak m^2 $). So $\mathfrak mI $ contains the socle $\mathfrak m^2$. But also trivially, $\mathfrak mI $ is inside $\mathfrak m^2. $ Thus $\mathfrak mI=\mathfrak m^2. $ Thus $ I/\mathfrak mI=I/\mathfrak m^2 $ is a proper $k$-vector subspace of $\mathfrak m/\mathfrak m^2$. So, $\mu(I) \le \mu(\mathfrak m) -1.$