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

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]

• Sorry for this naive question, but what is wrong with letting $R$ be $k[x,y]/(x,y)^3$? It is Artinian, local, Gorenstein, its maximal ideal $m = (x,y)$ satisfies $m^2\neq 0$ but $m^3 = 0$, and if by $\mu(m)$ you mean the $k$-vector space dimension of $m/m^2$, then this appears to be two, as you asked for. But $R$ has nonprincipal nonmaximal ideals, like $(x^2,y^2)$. (I apologize if this off-the-cuff comment reveals that I have misunderstood something very simple here.)
– user164898
Jul 4, 2021 at 2:43
• $k[x,y]/(x,y)^3$ is not Gorenstein. Jul 4, 2021 at 3:13
• @ZachTeitler Aha, of course--I had been thinking of $k[x,y]/(x^3,y^3)$ instead of $k[x,y]/(x,y)^3$ when considering whether the ring was Gorenstein. Thanks for pointing out what I had wrong.
– user164898
Jul 4, 2021 at 3:48

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 $$1$$.

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, $$\mu(I) \le \mu(\mathfrak m) -1.$$