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In Sally's Paper stretched artinian local ring is defined as :

Let $(R, \mathfrak{m})$ be an Artin local ring of length $\lambda.$ If $\nu$ is the embedding dimension of $R$, that is, $\nu$ is the number of elements in a minimal basis of $\mathfrak{m}$, then $\mathfrak{m}^{\lambda-\nu+1} = 0.$ We will say that $R$ is stretched if $\lambda - \nu$ is the least integer $i$ such that $\mathfrak{m}^{i+1} = 0.$ If $R$ is not a field, $R$ is stretched if and only if $\mathfrak{m}^2$ is principal ideal.

$\textbf{I am thinking how to construct some examples of stretched artinian local ring}.$

$\textbf{It will be better if could get those with $\lambda \neq \nu$ and $\mathfrak{m}^3 \neq 0$}.$

Any help is highly appreciated.

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2 Answers 2

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Some of the easy examples are $k[x,y]/(x^2,xy,y^n)$.

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  • $\begingroup$ Thanks a lot sir. $\endgroup$
    – SKS
    Commented Apr 30, 2023 at 16:58
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Try $k[x_0,x_1,\ldots,x_{2n}]/(R)$ where $R$ consists of the relations $$x_0^{\lambda-2n-1}=x_1x_2=x_3x_4=\cdots=x_{2n-1}x_{2n},$$ together with $x_ix_j=0$ for the remaining products and squares (except $x_0^2$ of course) - which imply that $x_0^{\lambda-2n}=0$. This has length $\lambda$, $\nu=2n+1$, and the $(\lambda-2n)$th power of $\mathfrak{m}$ is the first to be zero. This is even a Gorenstein ring. There are many obvious variations on this example.

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    $\begingroup$ @Benson Thanks a lot sir. It's really a nice class of example. Greatly thankful. $\endgroup$
    – SKS
    Commented Apr 30, 2023 at 16:57

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