Syzygies of the residue field over Golod rings

The last statement follows at once from condition (2). ∎

Suppose that $J$ is generated by $x,y$ and consider the maps

The composition is 0 if and only if $ax+by\in I$, and since $J$ is the cokernel of $\begin{pmatrix}y\\ -x\end{pmatrix}$, this is the condition that $(b,-a)$ induces a homomorphism $J\to R$. ∎

For part $(1)$, we use section 2 with $(a,b)=(x,y)$, and the condition that $J^{*}\cong(JR)^{*}$ from Theorem 2.2.

For parts $(2)$ and $(3)$ we must show that the restriction map $J^{*}\to I^{*}$ is 0 whenever $I$ is principal, and, when $I$ is not principal, that it is 0 if and only if $\gcd(I)\cdot{\rm Fitt}_{2}(I)\subset J$.

Let $F$ be the minimal $S$-free resolution of $R=S/I$, let $K$ be the Koszul complex resolving $S/J$, and let $\gamma:F\to K$ be the comparison map lifting the identity $K_{0}=S=F_{0}$. The restriction map is the left vertical map in the diagram

where $\kappa$ is the $R$-dual of the Koszul differential, and $\epsilon$ is the composition of $\gamma_{1}\otimes R$ with an isomorphism $K_{1}\otimes R\to K_{1}^{*}$ induced by the isomorphism $\wedge^{2}(K_{1}\otimes R)\to R$.

Since all the rows of this diagram are exact, the restriction map $J^{*}\to I^{*}$ is zero if and only if the composition $\gamma_{1}^{*}\circ(\kappa,\epsilon)$ is 0. Now $\gamma_{1}^{*}\circ\kappa$ is the dual of the map $(F_{1}\to S)\otimes R=0$, so the restriction map is 0 if and only if $\gamma_{1}^{*}\circ\epsilon=0$.

One checks using the cofactor expansion of the determinant that the composition $\gamma_{1}^{*}\circ\epsilon:F_{1}\otimes R\to F_{1}^{*}$ corresponds to the composition

so that $\gamma_{1}^{*}\circ\epsilon$ is 0 if and only if

Condition (*) clearly holds if $I$ is principal. Thus we may assume that $I$ is not principal. Write $a=\gcd(I)$ and suppose that $I^{\prime}:=I\cdot a^{-1}$ is minimally generated by $g_{1}^{\prime},\dots,g_{n}^{\prime}$. We may harmlessly assume that $g_{i}^{\prime},g_{j}^{\prime}$ form a regular sequence for any $i\neq j$ and that $g_{i}^{\prime},a$ form a regular sequence for any $i.$ Let $F^{\prime}$ be the minimal $S$-free resolution of $S/I^{\prime}$ and let $\gamma^{\prime}:F^{\prime}\to K$ be the comparison map lifting the identity $K_{0}=S=F_{0}^{\prime}$, which exists because $I^{\prime}\subset J$. We may assume that the map $\gamma_{1}$ is multiplication by $a$ composed with $\gamma^{\prime}_{1}.$ Let $\Delta_{i,j}^{\prime}$ be the minor of $\gamma_{1}^{\prime}$ involving columns $i,j.$ Condition (*) means that

Since $g_{i}^{\prime},ag_{j}^{\prime}$ form a regular sequence contained in $J$, we have

and as $I^{\prime}$ is perfect of grade 2, it follows that $a\cdot\Delta_{i,j}^{\prime}\in I^{\prime}$ if and only if $(g_{i}^{\prime},ag_{j}^{\prime}):I^{\prime}\subset J$ by the symmetry of linkage.

By a theorem of Gaeta, the link of $I^{\prime}$ with respect to $g_{i}^{\prime},ag_{j}^{\prime}$ is $a$ times the ideal generated by the $n-2$ minors of the presentation matrix of $I^{\prime}\cong I$ with rows $i$ and $j$ deleted. This proves the equivalence of condition (*) with the condition $a\cdot{\rm Fitt}_{2}(I)\subset J$. ∎

We apply Theorem 2.3 with $JR=\mathfrak{m}$. If $I$ is principal then the result follows from Theorem 2.3, parts (1) and (2). If $I$ is not principal then the result holds by parts (1) and (3) unless both $\gcd(I)$ and ${\rm Fitt}_{2}(I)$ are unit ideals. If $\gcd(I)=(1)$ then $I$ has codimension 2. If in addition ${\rm Fitt}_{2}(I)=(1)$ then $I$ is a 0-dimensional complete intersection. ∎

If $I$ has codimension 1 then we can remove a common divisor, which we may take to be $x$. In this case either $I=(xy)$ or $I=(xy,x^{a})$ for some $a\geq 1$ because $S/(y)$ is a discrete valuation ring with parameter $x$.

Any element of an artinian local ring can be written as a polynomial in the generators of the maximal ideal with unit coefficients. In particular, if $I$ has codimension 2, then any element of $S/(xy,\mathfrak{n}I)$ can be written in the form $f=ux^{a}+vy^{b}$ where each of $u$ and $v$ is either 0 or a unit of $S$. Note that if $u\neq 0$ then, modulo $xy$, every $x^{m}$ with $a<m$ is a multiple of $f$ and similarly for $y$.

If $I/(xy)$ is principal, then we may write $I=(xy,ux^{a}+vy^{b})$, and we are done. Otherwise, modulo $xy$, we may write two generators for $I$ as $\{ux^{a}+vy^{b},px^{c}+qy^{d}\}$ where $u$ and $q$ are units and $a$ and $d$ are minimal. Thus we may assume that $p=0$, and $v=0$, and thus $I:=(xy,x^{a},y^{d})$. ∎

We may assume $\mathfrak{m}=xR\oplus yR$. In case (1) according to Theorem 3.1 we can write $I=(x^{a},xy,y^{b})$ and $xR\cong R/(x^{a-1},y)$ and $yR\cong R/(x,y^{b-1})$. Thus ${\rm syz}^{R}(\mathfrak{m})=(x^{a-1},y)R\oplus(x,y^{b-1})R$. Both $x^{a-1},y^{b-1}$ are in the socle of $R$, and the claim follows.

In case (2) Theorem 3.1 shows that we may write $R=S/(xy,x^{a})$. In this case $xR\cong R/(y,x^{a-1})$, and $yR\cong R/xR$, so ${\rm syz}_{1}^{R}(\mathfrak{m})=(y,x^{a-1})R\oplus xR\cong yR\oplus k\oplus xR\cong\mathfrak{m}\oplus k$ since $x^{a-1}R$ is the socle of $R$.

Finally, if $R$ is Cohen-Macaulay of dimension 1, then $I=(xy)$ and all of the modules ${\rm syz}_{i}^{R}(k)$ for $i\geq 1$ are isomorphic. ∎

Now suppose that $R$ is 0-dimensional. Suppose that $I$ contains an element $xy+f$ with order $\operatorname{ord}f\geq 3$ such that $\mathfrak{n}=(x,y)$. If $\mathfrak{n}^{p}\subset I$ and $\operatorname{ord}f\geq p$ then $xy\in I$, which is impossible, since then $\mathfrak{m}$ is decomposable by Theorem 3.1. Otherwise, suppose there is an expression $xy+f\in I$ such that $\mathfrak{n}=(x,y)$ with order $\operatorname{ord}f$ maximal and $<p$.

We may write $f=xf_{1}+yf_{2}+g$ with $\min\{\operatorname{ord}f_{1},\operatorname{ord}f_{2}\}\geq\operatorname{ord}f-1$ and $\operatorname{ord}g>\operatorname{ord}f$. Thus $xy+f=(x+f_{2})(y+f_{1})+(g-f_{1}f_{2})$. Note that $\operatorname{ord}g-f_{1}f_{2}>\operatorname{ord}f$. We may replace $x,y$ by $x+f_{2},y+f_{1}$, thus increasing the order of $f$, a contradiction.

Now consider the $2\times 2$ matrix

If $\lambda_{1}\in\mathfrak{n}$, or $\dim_{k}(I+\mathfrak{n}^{3})/\mathfrak{n}^{3}=0$ then $L_{0}+L_{1}^{\prime}\equiv L_{0}$ modulo $\mathfrak{m}^{2}R^{2}$, and the claim follows. Thus we can assume that $\lambda_{1}$ is a unit and the determinant of $L^{\prime}$ is a minimal generator of $I$ not in $\mathfrak{n}^{3}$. This is also the determinant of the $2\times 2$ matrix $(L_{0}+L_{1}^{\prime}\ \ L_{1}^{\prime})$. If the entries of $L_{0}+L_{1}^{\prime}$ were linearly dependent, then the determinant would factor; and by part (2) it would have to be the perfect square $\ell^{2}=x_{0}^{2}$. But then, modulo $\mathfrak{m}^{2}$ the matrix $L^{\prime}$ must be

Thus the entries of $L_{0}+L_{1}^{\prime}$ are linearly independent as claimed. ∎

If $R$ were Gorenstein then syzygies of indecomposable maximal Cohen-Macaulay modules are indecomposable, so we may assume that $R$ is not Gorenstein.

We write $\mu(I)$ for the minimal number of generators of $I$ and similarly for $\mu(N)$.

We fix generators $x,y$ of $\mathfrak{n}$ and the corresponding embedding $Z:={\rm syz}_{1}^{R}(\mathfrak{m})\subset R^{2}\,.$ Notice that

Thus it suffices to prove that

To this end we define an $R$-linear map $\psi$ as the composition of the maps

Notice that $\psi((f,g))=(xF+yG)+\mathfrak{n}I$, where $F$, $G$ are preimages of $f,g$ in $S$. As $\operatorname{Soc}Z=\operatorname{Soc}R^{2}$, it follows that $\operatorname{Im}\psi=\frac{\mathfrak{n}(I:\mathfrak{n})}{\mathfrak{n}I}$. Clearly $\operatorname{Ker}\psi=RL_{0}\cap\operatorname{Soc}Z$.

Thus it remains to prove

The right hand side is in the left hand side because $\mathfrak{m}Z\subset RL_{0}$. As to the converse, the indecomposibility of $\mathfrak{m}$ implies that $I\not=\mathfrak{n}^{2}$, hence $\mathfrak{m}L_{0}\not=0$. Therefore $RL_{0}\cap\operatorname{Soc}Z\subset\mathfrak{m}L_{0}\subset\mathfrak{m}Z$.

Suppose first that $\dim R=1$. In this case we may assume that $I:\mathfrak{n}=(xy,ux^{\alpha})$. Assume that $\alpha\geq 3$. Set $I^{\prime}:=\mathfrak{n}\,(I:\mathfrak{n})$. We have $I^{\prime}\subset I\subset(xy,ux^{\alpha})$. By Theorem 3.1 the ideal $I$ contains no product of two elements that generate the maximal ideal of $S$ so no element of the form $xy+\lambda x^{\alpha}$ can be in $I$. Thus $I\subset\mathfrak{n}(xy)+(ux^{\alpha})=(x^{2}y,xy^{2},ux^{\alpha})=:I^{\prime\prime}$. If $u=0$ then $I=I^{\prime}$. If $u$ is a unit then $I^{\prime\prime}/I^{\prime}\cong k$. Because $x^{\alpha-1}\notin I:\mathfrak{n}$, the ideal $I$ cannot equal $I^{\prime\prime}$ , so in any case $I=I^{\prime}$. In this case $a=\dim_{k}(I^{\prime}/\mathfrak{n}I)$ is the minimal number of generators of $I$. Thus $N$ is cyclic, hence indecomposable.

Now, suppose that $\dim R=0$ or $I:\mathfrak{n}=(xy,x^{2})$. We first show that the module $N$ can be generated by 2 elements.

If $I:\mathfrak{n}=(xy,x^{2})$, as in the proof above $I^{\prime}=x(x,y)^{2}\subset I\subset(x^{2}y,xy^{2},x^{2})=x(y^{2},x)$. Thus either $I=I^{\prime}$ and as above $N$ is cyclic or $I=x(y^{2},x)$ in which case $a=1$ hence $\mu(N)=2$.

If $\alpha=\beta=2$ then $\mathfrak{n}^{3}\subset I\subset\mathfrak{n}^{2}$. By section 4, either $I=\mathfrak{n}^{3}$ hence $a=\mu(I)$, thus $N$ is cyclic, or $I=\mathfrak{n}^{3}+(q)$, where $q\in\mathfrak{n}^{2}\setminus\mathfrak{n}^{3}$. In the latter case $a=\mu(I)-1$, so that $\mu(N)=2$.

Finally, without loss of generality, we can assume that $\alpha\geq 3$. In this case following the same argument as in $\dim R=1$, we can show that $(x^{2}y,xy^{2},x^{\alpha+1},y^{\beta+1})\subset I\subsetneq(x^{2}y,xy^{2},x^{\alpha},y^{\beta})$. Thus $I=(x^{2}y,xy^{2},x^{\alpha+1},y^{\beta+1},ux^{\alpha}+vy^{\beta})$ where each of $u,v$ is a unit or 0, $\alpha\geq 3$, and $\beta\geq 2$. Thus $a\geq\mu(I)-1$, so $\mu(N)\leq 2$.

Since $\mathfrak{m}$ is indecomposable, $I\not=\mathfrak{n}^{2}$ and thus $0:\mathfrak{m}\subsetneq\mathfrak{m}$. Every minimal set of generators of ${\rm syz}_{1}^{R}(\mathfrak{m})$ contains an element of the form $L_{0}+\sum_{i>0}\lambda_{i}L_{i}$, whose annihilator is exactly $0:\mathfrak{m}$ by section 4(3). This generator cannot be among the minimal generators of $k^{a}$, so every minimal set of generators of $N$ contains such an element.

If $N=A\oplus B$ with $A,B\not=0$ then $A$ and $B$ must be cyclic. We may assume that $A$ is minimally generated by an element of the form $L_{0}+\sum_{i>0}\lambda_{i}L_{i}$ and thus $A\cong R/(0:\mathfrak{m})$. In particular $\mathfrak{m}A\oplus\mathfrak{m}B=\mathfrak{m}N\cong\mathfrak{m}A$. This implies that that $\mathfrak{m}B=0$ for instance because the number of generators of $\mathfrak{m}B$ is 0. Since $N$ does not have $k$ as a direct summand, $B=0$, and we are done. ∎

The formula for the differential $d$ in the minimal $R$-free resolution of $k$, in terms of the Massey operations $\mu$ [1, Theorem 5.2.2] has the form:

The only term that does not preserve the last tensor factor, $v_{i_{p}}$, is the one that maps to the Koszul complex alone. Thus the truncation of the complex beyond the Koszul complex splits as a direct sum of complexes, and the summands are the resolutions of

where $e$ is the embedding dimension. ∎

From Theorem 2.1 we have the first assertion. All indecomposable summands will be those appearing in $k,\mathfrak{m},{\rm syz}_{1}^{R}(m)$, so the last assertion follows from section 4 and Theorem 4.1. ∎

Part (1) follows from Theorem 3.1. For part (2), notice that if $\mathfrak{m}$ is decomposable then ${\rm syz}^{R}_{1}(\mathfrak{m})$ is decomposable. Now the assertion follows from section 4 and Theorem 4.1.

For part (3), we may assume that $R$ is Gorenstein: if ${\rm syz}_{2}^{R}(\mathfrak{m})$ is indecomposable then this is true by 1.2, and for the opposite implication, it is part of the assumptions. Since $\mathfrak{m}$ is a maximal Cohen-Macaulay $R$-module and $R$ is Gorenstein, $\mathfrak{m}$ is indecomposable if and only if one or all of its syzygies are indecomposable. Part (3) now follows from part (1). ∎