Let $(R,m)$ be a Cohen-Macaulay local ring, I and J are ideals of height $r.$ Then we say $I$ is directly linked to $J$, i.e. $I \sim J$ if there exists an ideal K generated by a regular sequence $x_1,\ldots,x_r$ such that $K\subset I\cap J,$ $I=K:J$ and $J=K:I.$

We say $I$ is linked to $J$ if there exist ideals $I_1,\ldots,I_s$ of height $r$ and generated by regular sequences such that $I\sim I_1, I_1\sim I_2,\ldots, I_s\sim J.$

Question: Is the operation "$I$ is linked to $J$" an equivalence relation? Particularly I do not understand how it is reflexive?

enter link description here [Linkage and the Koszul Homology of Ideals - C. Huneke]

• A functorial definition of linkage for stable modules over Noetherian rings (i.e. in the module category modulo projectives) is presented in the following paper: [arxiv.org/pdf/1505.01112v3.pdf] ... a module $M$ over a Noetherian ring is linked if there is an isomorphism between $M$ and the 'syzygy transpose syzygy transpose' of $M$, for lack of a better name. Not sure if that's too high-flown to be of any help – Samantha Y Sep 15 '16 at 16:07
• The relation is clearly symmetric, so if $I\sim J$, then $J\sim I$ and thus putting them together, one gets $I\sim I$. – Mohan Sep 15 '16 at 16:24
• @Mohan Your comment presupposes that a given $I$ is directly linked to at east one $J$. – Andreas Blass Sep 15 '16 at 18:13
• @AndreasBlass Yes, I am supposing $R/I$ is Cohen-Macaulay, which is where linkage really plays a role. If this is the case, then such an $I$ is always directly linked to a $J$ with $R/J$ Cohen-Macaulay. – Mohan Sep 15 '16 at 18:38
• @Mohan It will be extremely helpful for me if you explain your last comment that $R/I$ is Cohen-Macaulay. – Cusp Sep 16 '16 at 3:01

If $I$ is an ideal with $R/I$ CM, pick a regular sequence $x_1,\ldots, x_r\in I$, where $r$ is the height of $I$. If we call $K$ the ideal generated by the $x_i$s, we can define $J=(K:I)$. One checks that $I=K:J$ and then we are done once we also check that $R/J$ is also CM.