In the middle of page 34 in Bruns and Herzog, Cohen-Macaulay Rings, the authors present the following situation: Let $k$ be a field and let $R=k[X,Y]$ be a graded ring with grading induced by $\operatorname{deg}X=0, \operatorname{deg} Y=1$. Then define the quotient ring $S=R/(XY)$, which is also graded. Then the authors claim that the ideal $(x,y)$, where i suppose lowercase $x,y$ mean the residue classes of $X,Y$ mod $(XY)$, has grade $1$, i.e. according to my understanding the maximal length of an $S$-regular sequence in $(x,y)$ is $1$. Indeed, i can see that $x+y \in (x,y)$ is $S$-regular and so the grade of $(x,y)$ on $S$ is at least $1$.

However by the definition of the grade of an ideal $\operatorname{grade}((x,y),S)= \inf \left\{i: \operatorname{Ext}^i_S(S/(x,y),S) \neq 0 \right\}$. But $S/(x,y) = (k[X,Y]/(XY)) / ((X,Y)/(XY)) \cong k[X,Y]/(X,Y) \cong k$ and certainly $\operatorname{Hom}(k,k[X,Y]/(XY)) \neq 0$, which shows that the grade of $(x,y)$ on $S$ is 0. What am i missing?

Remark: The "unusual" grading of $R$ seems to be irrelevant to my actual question. I am just giving it here for completeness of the context.

  • 2
    $\begingroup$ Are you sure that $\operatorname{Hom}_{\color{red} S}(k,k[X,Y]/(XY)) \neq 0$? $\endgroup$ – user119598 Jan 10 '14 at 22:42

The $Hom$ is $0$:
Let $g$∈ $Hom$ and be nonzero so $g(r+m)≠0$ for some $r∈R$. Let $u∈m:=(x,y)$ and $u$ be $S-regular$. We have
$u.g(r+m)≠0$ so $g(u.r+m)≠0$. contradicts $ur∈m$.


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