Syzygies of the residue field over Golod rings

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

We may write ${\rm syz}_{1}^{R}(JR)=\{(\overline{a},\overline{b})\mid ax+by\in I\}$ where $\overline{\phantom{a}}$ denotes images in $R.$ 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 $J\to R.$ ∎

Write $R=S/I$ and $(-)^{*}=\operatorname{Hom}_{S}(-,R).$ We denote images in $R$ by ${}^{-}.$

We will show that the restriction map $\rho:J^{*}\to I^{*}$ is 0 if and only if $a\cdot{\rm Fitt}_{2}(I)\subset J.$

Let $h^{\prime}_{1},\ldots,h^{\prime}_{n}$ be generators of $I^{\prime}$ so that the elements $h_{i}=ah^{\prime}_{i}$ generate $I$. Extending the ground field if necessary, we may assume that $h_{i}^{\prime},h_{j}^{\prime}$ form a regular sequence for every $i\neq j$ and that $h_{i}^{\prime},a$ form a regular sequence for every $i.$ Let

be a matrix with entries in $S$ satisfying $\begin{pmatrix}h^{\prime}_{1}&\dots&h^{\prime}_{n}\end{pmatrix}=\begin{pmatrix}x&y\end{pmatrix}\cdot\varphi^{\prime}$, and let $\varphi=a\varphi^{\prime}$.

Since

where $\Delta_{i,j}$ is the determinant of the submatrix of $\varphi$ involving columns $i,j,$ we see that $\rho=0$ if and only if $I_{2}(\varphi)\subset I,$ and this is the case if and only if $aI_{2}(\varphi^{\prime})\subset I^{\prime}$, as claimed.

Let $\Delta_{i,j}^{\prime}$ be the minor of $\varphi^{\prime}$ involving columns $i,j.$ Since $\begin{pmatrix}h_{i}^{\prime}&ah_{j}^{\prime}\end{pmatrix}=\begin{pmatrix}x&y\end{pmatrix}\cdot\begin{pmatrix}f_{i}^{\prime}&af_{j}^{\prime}\\ g_{i}^{\prime}&ag_{j}^{\prime}\end{pmatrix}$ and $h_{i}^{\prime},ah_{j}^{\prime}$ form a regular sequence, a theorem of Gaeta (see for instance [2, Example 3.2(b)]) gives

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

By the same theorem of Gaeta, the link $(h_{i}^{\prime},ah_{j}^{\prime}):I^{\prime}$ of $I^{\prime}$ is generated by $h_{i}^{\prime}$ and $a$ times the $n-2$ minors of the presentation matrix of $I^{\prime}\cong I$ with rows $i$ and $j$ deleted. This proves that $a\cdot I_{2}(\varphi^{\prime})\subset I^{\prime}$ if and only if $a\cdot{\rm Fitt}_{2}(I)\subset J.$ ∎

We apply the previous results with $J=\mathfrak{n}.$ If $I$ is principal, we use Example 2.3 and section 2. If $I$ is not principal, we may write $I=aI^{\prime}$ with $I^{\prime}$ perfect of grade 2. We see from section 2 and Theorem 2.4 that the result holds unless both ($a$) and ${\rm Fitt}_{2}(I)$ are unit ideals. If $a$ is a unit, then $I=I^{\prime}$ has grade 2. If in addition ${\rm Fitt}_{2}(I)=S,$ then $I$ is a zero dimensional complete intersection. ∎

If $I$ has codimension 1 then $I$ has a proper common divisor, which we may take to be $x.$ Writing $I=x(J+(y))$, we see that either $I=(xy)$ or $I=(xy,x^{\alpha})$ for some $\alpha\geq 1$ 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 $S/(xy,\mathfrak{n}I)$ is the image of an element of the form $f=ux^{\alpha}+vy^{\beta},$ where each of $u$ and $v$ is either 0 or a unit of $S$ and $\alpha,\beta$ are non-negative. Note that if $u\neq 0$ then, modulo $xy,$ every $x^{\mu}$ with $\mu>\alpha$ is a multiple of $f$ and similarly for $v$ and $y.$

If $I/(xy)$ is principal, then we may write $I=(xy,ux^{\alpha}+vy^{\beta}),$ and we are done. Otherwise, modulo $xy,$ we may write two of the generators of $I$ as $ux^{\alpha}+vy^{\beta},\ px^{\gamma}+qy^{\delta},$ where $u$ and $q$ are units and $\alpha$ and $\delta$ are minimal. Thus we may assume that $p=0,$ and $v=0,$ so $I=(xy,x^{\alpha},y^{\delta}).$ Notice that $\alpha,\beta$ have to be positive. ∎

We apply the analysis of Theorem 3.1. This shows that $I\subset\mathfrak{n}^{2}$ and we may assume $\mathfrak{m}=\overline{x}R\oplus\overline{y}R.$ Furthermore, if $\dim R=0$ then we can write $I=(x^{\alpha},xy,y^{\beta}),$ where $\alpha,\beta$ are $\geq 2$. In this case $\overline{x}R\cong R/(\overline{x}^{\alpha-1},\overline{y})$ and $\overline{y}R\cong R/(\overline{x},\overline{y}^{\beta-1}).$ Thus ${\rm syz}_{1}^{R}(\mathfrak{m})=(\overline{x}^{\alpha-1},\overline{y})\oplus(\overline{x},\overline{y}^{\beta-1})\cong k\oplus\overline{y}R\oplus\overline{x}R\oplus k\cong\mathfrak{m}\oplus k^{2}$ since $\overline{x}^{\alpha-1},\overline{y}^{\beta-1}$ are in the socle of $R.$

If $\operatorname{depth}R=0$ and $\dim R=1,$ then Theorem 3.1 shows that we may write $R=S/(xy,x^{\alpha})$ with $\alpha\geq 2.$ Now $\overline{x}R\cong R/(\overline{y},\overline{x}^{\alpha-1})$ and $\overline{y}R\cong R/\overline{x}R,$ so ${\rm syz}_{1}^{R}(\mathfrak{m})=(\overline{y},\overline{x}^{\alpha-1})\oplus\overline{x}R\cong\overline{y}R\oplus k\oplus\overline{x}R\cong\mathfrak{m}\oplus k$ because $\overline{x}^{\alpha-1}$ is in the socle of $R.$

Finally, if $R$ is Cohen-Macaulay of dimension 1, then $I=(xy)$ by Theorem 3.1 and all the modules ${\rm syz}_{i}^{R}(k)$ for $i\geq 1$ are isomorphic. ∎

Since $R$ is not a complete intersection, we must have $I\subset\mathfrak{n}^{2}$.

Now assume that $\dim R=0.$ Suppose that $I$ contains an element $xy+f$ with $\operatorname{ord}f\geq 3$ 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 by Theorem 3.1(2). 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.

We may assume that $I\not=0.$ We fix generators $x,y$ of $\mathfrak{n}$ and the corresponding embedding $Z:={\rm syz}_{1}^{R}(\mathfrak{m})\subset R^{2}\,,$ and we use the notation introduced before section 4.

Notice that

Thus it suffices to prove that

Thus it remains to prove

The right hand side is in the left hand side, because $\frac{Z}{RL_{0}}=H_{1}(x,y;R)$ and therefore $\mathfrak{m}Z\subset RL_{0}.$ As to the converse, the indecomposibility of $\mathfrak{m}$ implies that $I\not=\mathfrak{n}^{2},$ hence $\mathfrak{m}L_{0}\not=0.$ Therefore $RL_{0}\cap\operatorname{Soc}Z\subset\mathfrak{m}L_{0}\subset\mathfrak{m}Z.$

Since $Z=N\oplus N^{\prime}$ and $\mathfrak{m}N^{\prime}=0,$ we have $\mathfrak{m}N=\mathfrak{m}Z.$ As shown above $\mathfrak{m}Z\subset RL_{0}.$ But $\overline{L_{0}}$ is a minimal generator of $Z,$ hence $\mathfrak{m}Z\subset\mathfrak{m}L_{0},$ and therefore $\mathfrak{m}Z=\mathfrak{m}L_{0}.$ Finally, $RL_{0}\cong R/(0:\mathfrak{m}),$ so $\mathfrak{m}L_{0}\cong\mathfrak{m}/(0:\mathfrak{m}).$ Putting these facts together, we have

Since $N$ does not have $k$ as a direct summand, the indecomposability of $N$ follows from the indecomposability of $\mathfrak{m}N\cong\mathfrak{m}/(0:\mathfrak{m}),$ so it suffices to treat the cases where the maximal ideal $\mathfrak{m}/(0:\mathfrak{m})$ of $S/(I:\mathfrak{n})$ is decomposable. By Theorem 3.1(3) this is the case if and only if for a suitable choice of $x$ and $y$ one has $I:\mathfrak{n}=(xy,ux^{\alpha},vy^{\beta}),$ where each of $u,v$ is a unit or 0 and both of $\alpha,\beta$ are $\geq 2.$

We first show that in this case the module $N$ can be generated by 2 elements. Set $I^{\prime}=\mathfrak{n}\,(I:\mathfrak{n}).$ It suffices to prove that $\dim_{k}(I/I^{\prime})\leq 1$ because part (a) gives $\mu(N)=1+\mu(I)-\dim_{k}(I^{\prime}/\mathfrak{n}I)=1+\dim_{k}(I/I^{\prime}).$

Suppose first that $\dim R=1.$ In this case we may assume that $I:\mathfrak{n}=(xy,ux^{\alpha}),$ hence $I\subset(xy,ux^{\alpha}).$ By Theorem 3.1(2) the ideal $I$ contains no product of two elements that generate the maximal ideal of $S,$ so no element of the form $xy+\lambda x^{\alpha}$ with $\lambda\in S$ can be in $I.$ Thus $I\subset\mathfrak{n}(xy)+(ux^{\alpha})=(x^{2}y,xy^{2},ux^{\alpha})=:I^{\prime\prime}.$ As $I^{\prime}=(x^{2}y,xy^{2},ux^{\alpha+1}),$ it follows that $\dim_{k}(I^{\prime\prime}/I^{\prime})\leq 1.$ Now the containments $I^{\prime}\subset I\subset I^{\prime\prime}$ show that $\dim_{k}(I/I^{\prime})\leq 1.$

Now assume that $\dim R=0,$ thus $I:\mathfrak{n}=(xy,x^{\alpha},y^{\beta}),$ where $\alpha,\beta$ are $\geq 2.$ If $\alpha=\beta=2$ then $I:\mathfrak{n}=\mathfrak{n}^{2}$ and hence $I^{\prime}=\mathfrak{n}^{3}.$ Therefore $\dim_{k}(I/I^{\prime})\leq 1$ by section 4(b). Finally, without loss of generality, we can assume that $\alpha\geq 3.$ We have $I\subset I:\mathfrak{n}=(xy,x^{\alpha},y^{\beta}).$ Since $\alpha\geq 3$, section 4(a) shows that $I$ cannot contain an element of the form $xy+\lambda x^{\alpha}+\mu y^{\beta}.$ Thus $I\subset\mathfrak{n}(xy)+(x^{\alpha},y^{\beta})=(x^{2}y,xy^{2},x^{\alpha},y^{\beta})=:I^{\prime\prime}.$ As $x^{\alpha-1}\notin I:\mathfrak{n}$ but $x^{\alpha-1}\in I^{\prime\prime}:\mathfrak{n}$ because $\alpha\geq 3,$ the ideal $I$ cannot be equal to $I^{\prime\prime}.$ On the other hand $I^{\prime}=(x^{2}y,xy^{2},x^{\alpha+1},y^{\beta+1})$ and so $\dim_{k}(I^{\prime\prime}/I^{\prime})\leq 2.$ Since $I^{\prime}\subset I\subsetneq I^{\prime\prime},$ we see that, again, $\dim_{k}(I/I^{\prime})\leq 1.$ This concludes the proof of the inequality $\mu(N)\leq 2$ and the present choice of the elements $x,$ $y.$