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Let $B:=k[r,s]/(r^2,s^2,rs)$ be a polynomial ring , $k$ a field and $m={\ }_BB$. What is the radical of $m$?

Thanks for the help.

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Do we know sth else about $m$? For example, $m\subseteq B$ or $m\subseteq k[r,s]$? And, $\{1,r,s\}$ basis over $k$ or over $B$? – Berci Oct 10 '12 at 23:48
up vote 1 down vote accepted

Your subspaces are not closed under the action of $B$ (e.g. $r\cdot 1=r\notin <\{1,s\}>$). Each submodule of $B$ containing $1$ is already $B$ itself.

The only maximal submodule of $m$ is the two-dimensional subspace spanned by $r$ and $s$. Hence this is the radical.

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Since $r$ and $s$ are nilpotent, $r,s\in rad(B)$, and $(r,s)\subseteq rad(B)$.

Since $B/(r,s)\cong k$, we have that $(r,s)$ is maximal, an hence $rad(B)\subseteq (r,s)$.

Hence, $rad(B)=(r,s)$.

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