# For a local complete intersection ring $(R,\mathfrak m)$ with $\mathfrak m^3=0\ne \mathfrak m^2$, $\mathfrak m$ can be generated by two elements

Let $$(R,\mathfrak m,k)$$ be a local complete intersection ring with $$\mathfrak m^3=0\ne \mathfrak m^2$$. As $$0\ne \mathfrak m^2 \subseteq \text{soc}(R)$$ and $$R$$ is Gorenstein, so we get $$\mathfrak m^2 =\text{soc}(R)\cong k$$.

Now according to the beginning of Section 3 in page 293 of https://doi.org/10.1016/0022-4049(85)90016-7, it holds that $$\dim_k (\mathfrak m/\mathfrak m^2)\le 2$$. However, no proof or reference to a proof is given. I have seen the same statement appear in other places as well, but again without any proof. Could someone please outline a proof or provide a reference to a proof?

• The associated graded ring is also a complete intersection. Its Hilbert series is the product from $i=1,\dots,n$ of $(1-t^{d_i})/(1-t)$, where the $d_i$ are the degrees of the defining equations. In particular, the socle is in degree $\sum_i (d_i-1)$. So, if the socle is in degree $2$, then the Hilbert series is either $(1-t^3)/(1-t)$ or $(1+t)^2$. Nov 26, 2022 at 13:39
• @JasonStarr: thank you for your comment, but why is the associated graded ring also a complete intersection? I thought that local complete intersection rings whose associated graded rings are also Complete Intersections, are called "strict complete intersections" . Definitely hypersurfaces are strict complete intersections, but this can fail in codimension $2$ as mentioned in page 2 (after Theorem 1) of arxiv.org/abs/2104.10140. Are you saying that Artinian local complete intersection rings with $\mathfrak m^3=0$ are also strict complete intersections? Nov 28, 2022 at 8:49
• Maybe I misunderstood your notation. Is $k$ a coefficient field for $R$? Nov 28, 2022 at 11:51
• @JasonStarr: No no, $k$ is just the residue field of the local ring $R$ ..... but let me also ask: If $R$ is a local complete intersection with a coefficient field, then the associated graded ring is also complete Intersection? Nov 28, 2022 at 12:02

By hypothesis, $$R$$ is a quotient $$A/I$$ of a complete local ring $$(A,\mathfrak{n},k)$$ that is regular, where $$I$$ is generated by an $$A$$-regular sequence in $$\mathfrak{n}^2$$. In particular, the induced $$k$$-linear map from $$\mathfrak{n}/\mathfrak{n}^2$$ to $$\mathfrak{m}/\mathfrak{m}^2$$ is an isomorphism. Let $$(x_1,\dots,x_d)$$ be an ordered $$d$$-tuple of elements of $$\mathfrak{n}$$ that map to a $$k$$-basis for $$\mathfrak{n}/\mathfrak{n}^2$$.
Thus, the ideal $$I$$ is generated by a regular sequence of length $$d$$ inside $$\mathfrak{n}^2$$. Now consider the $$d(d+1)/2$$ $$k$$-linearly independent elements $$x_ix_j\in \mathfrak{n}^2/\mathfrak{n}^3$$. The $$R$$-module $$\mathfrak{m}^2/\mathfrak{m}^3$$ equals the quotient of the $$A$$-module $$\mathfrak{n}^2/\mathfrak{n}^3$$ by $$I/(I\cap \mathfrak{n}^3)$$. Also $$I/(I\cap \mathfrak{n}^3)$$ is a quotient of $$I/I\cdot \mathfrak{n}$$, which is a $$k$$-vector space of dimension $$d$$. Thus, the $$k$$-vector space dimension of $$\mathfrak{m}^2/\mathfrak{m}^3$$ is at least $$d(d+1)/2 - d = d(d-1)/2$$.
Thus, if the socle of $$A$$ is $$\mathfrak{m}^2$$, so that $$\mathfrak{m}^2/\mathfrak{m}^3$$ has $$k$$-vector space dimension equal to $$1$$, then either $$d=1$$ and $$I\subset \mathfrak{n}^3$$ or $$d=2$$ and $$I\cap \mathfrak{n}^3$$ equals $$I\cdot \mathfrak{n}$$, i.e., the $$2$$ minimal generators of $$I$$ are elements of $$\mathfrak{n}^2$$ whose images in $$\mathfrak{n}^2/\mathfrak{n}^3$$ are $$k$$-linearly independent.