2. The relationship of an ideal to its Fitting ideals

In this section we compare the Fitting ideals of an ideal I with powers of I, the radical and the trace of I. As the first result we have

For $K_1$ we choose the basis $e_1,\ldots,e_m$ with $\partial_1(e_i)=f_i$ for $i=1,\ldots, m$ . For any subset $A\subseteq [m]$ , $A=\{a_1 \lt a_2 \lt \cdots \lt a_{\ell}\}$ we set $e_A=e_{a_1}\wedge \cdots \wedge e_{a_{\ell}}$ . Then the elements $e_A$ with $|A|={\ell}$ form a basis of $K_{\ell}$ .

Let $1 \leq j \lt m$ , and let $f=f_{i_1}f_{i_2}\cdots f_{i_{m-j}}$ with $1 \leq i_1 \leq i_2 \leq \cdots \leq i_{m-j} \lt m$ be a generator of $I^{m-j}$ . By induction on $m-j$ one shows that there exist integers $1\leq k_1\lt\cdots \lt k_{m-j}\leq m$ such that $k_\ell\neq i_\ell$ for $\ell=1,\dots, {m-j}$ . Let $z=z_1\wedge\cdots \wedge z_{m-j}$ , where $z_\ell=f_{i_\ell} e_{k_\ell}-f_{k_\ell }e_{i_\ell}$ is an element of $Z_1$ . Then $z\in Z_1^{m-j}$ , and the coefficient of $e_{k_1}\wedge\cdots\wedge e_{k_{m-j}}$ in z is equal to $\pm f$ . This proves that $I^{m-j}\subseteq I_{m-j}(A)=\mathrm{Fitt}_j(I)$ .

Now, suppose that $f_1,\ldots,f_m$ is a regular sequence. Then $Z_1$ is generated by the 1-boundaries $z_{i,j}=f_{i} e_j-f_{j }e_{i}$ with $1 \leq i \lt j \leq m$ . This implies that the coefficients of $Z_1^{m-j}$ belong to $I^{m-j}$ .

Finally, assume that R is local with maximal ideal ${\mathfrak m}$ and $I={\mathfrak m}$ . By the previous part ${\mathfrak m}^{m-j}\subseteq\mathrm{Fitt}_j({\mathfrak m})$ for all $1\le j\le m$ . On the other hand, any nonzero entry of a minimal presentation matrix A of ${\mathfrak m}$ belongs to ${\mathfrak m}$ . Hence $\mathrm{Fitt}_j({\mathfrak m})=I_{m-j}(A)\subseteq{\mathfrak m}^{m-j}$ for all $1\le j\le m$ , and this completes the proof.

(b) Let ${f}=f_1,\dots,f_m$ be a system of generators of I, and let as before $Z_1$ be the module of 1-cycles of the Koszul complex $K_{{\large\bf.}}({\mathbf f};R)$ . As observed in (a), the coefficients of the elements $z\in Z_1^{m-j}$ (with respect the canonical basis of $K_{m-j}$ ) generate $\mathrm{Fitt}_{j}(I)$ . Since we assume that $\mathrm{grade}\, I=j+1$ , it follows that $H_{m-j}({f};R)=0$ , see [ Reference Bruns and Herzog1 , theorem 1·6·17(b)]. Therefore, any $z\in Z_1^{m-j}$ is a boundary, since $Z_1^{m-j}\subseteq Z_{m-j}=B_{m-j}$ , and so all coefficients of z belong to I. This shows that $\mathrm{Fitt}_{j}(I)\subseteq I$ .

Corollary 2·3. Let R be a Noetherian ring, and $I\subset R$ be an ideal with $\mathrm{grade}\, I=j$ . Then

\begin{align*} {}\sqrt{\mathrm{Fitt}_i(I)}=\sqrt{I} \quad \text{for}\quad i=1,\ldots, j-1. {}\end{align*}

In theorem 5·1 it is shown that for the equality in corollary 2·3, $\mathrm{grade}\, I$ can be replaced by height $I$ .

On the other hand, when $\mathrm{grade}\, I=j$ , we cannot expect that $\sqrt{\mathrm{Fitt}_j(I)}= \sqrt{I}$ . Indeed, consider the ideal $I=(x_1x_2,x_1x_3)\subset S=K[x_1,x_2,x_3]$ and K a field. Then $\mathrm{grade}\, I=1$ , $\sqrt{\mathrm{Fitt}_1(I)}=(x_2,x_3)$ and $\sqrt{I}=I$ . More examples follow later.

When I is the maximal ideal in a local ring, theorem 2·1 gives the answer to our general question of when $\mathrm{Fitt}_j(I)=I$ .

The next result describes another situation for which a Fitting ideal of an ideal I coincides with I.

Proof. From the inequalities

\begin{align*}\mathrm{grade}\, I\leq \mathrm{grade}\, IR_P\leq\ \text{height}\ IR_P\leq\mu(IR_P)=\mathrm{grade}\, I,\end{align*}

we obtain that $IR_P$ is a complete intersection for any $P\in V(I)$ . Hence, by theorem 2·1(a) , we conclude that $\mathrm{Fitt}_{j-1}(I)R_P=\mathrm{Fitt}_{j-1}(IR_P)=IR_P$ for any $P\in V(I)$ . This together with theorem 2·1(b) implies that $\mathrm{Fitt}_{j-1}(I)=I$ .

Using Proposition 2·5 and under the additional assumption that R is a regular ring, corollary 2·4 can be extended as follows.

The next result shows that if I is a 2-generated ideal in a domain, then $\mathrm{Fitt}_1(I)$ is the trace of I.

3. Fitting ideals of the canonical module

Let $(R,{\mathfrak m})$ be a Cohen-Macaulay local ring with canonical module $\omega_R$ . The Cohen-Macaulay type of a Cohen-Macaulay module M will be denoted by r(M). We have

In the next two results we assume that $(R,{\mathfrak m})$ is a local Cohen–Macaulay ring which is generically Gorenstein. We furthermore assume that R admits a canonical module $\omega_R$ . By [ Reference Bruns and Herzog1 , Proposition 3·3·18], $\omega_R$ can be identified with a Cohen-Macaulay ideal of grade 1. Assume that $\mathrm{Fitt}_j(\omega_R)\cong \omega_R$ for some j. Then there exists a non-zerodivisor $t\in R$ such that $\mathrm{Fitt}_j(\omega_R) =t\omega_R$ . Then $\mathrm{Fitt}_j(t\omega_R)=\mathrm{Fitt}_j(\omega_R)=t\omega_R$ . Since the ideal $t\omega_R$ is also a canonical ideal of R, the condition $\mathrm{Fitt}_j(\omega_R)\cong \omega_R$ can always be replaced by $\mathrm{Fitt}_j(\omega_R)= \omega_R$ for a suitable choice of $\omega_R$ .

If R is not Gorenstein, then it follows from our assumption that $\omega_R$ is a proper ideal of R. The ideal $\omega_R$ is of height one, since it is a Cohen-Macaulay module. This together with the first part of the proof completes the proof.

Assume that R is not Gorenstein, and let P be a minimal prime ideal of the canonical ideal $\omega_R$ . The proof of corollary 3·3 shows that if $\mathrm{Fitt}_1(\omega_R)=\omega_R$ , then $R_P$ is a one-dimensional Cohen-Macaulay ring but not Gorenstein. So far we could not find a one-dimensional non-Gorenstein ring with canonical ideal $\omega_R$ for which $\mathrm{Fitt}_1(\omega_R)=\omega_R$ . However, we have

Proof. Suppose that R is not Gorenstein. Then $\mu(\omega_R)=2$ and it follows from Proposition 2·7 and our hypothesis that $\omega_R=\omega_R^{-1}\omega_R$ . This implies that

\[ {}R=\omega_R:\omega_R=\omega_R:(\omega_R^{-1}\omega_R)=(\omega_R:\omega_R):\omega_R^{-1}=R:\omega_R^{-1}=(\omega_R^{-1})^{-1}. {}\]

Therefore, $R=((\omega_R^{-1})^{-1})^{-1}=\omega_R^{-1}$ .

Since $\mathrm{tr}_R(R/\omega_R, R)=0$ , the exact sequence $0\rightarrow \omega_R\rightarrow R\rightarrow R/\omega_R\rightarrow 0$ , induces the exact sequence

\[ {}0\rightarrow R\rightarrow \omega_R^{-1} \rightarrow \mathrm{Ext}^1_R(R/\omega_R, R)\rightarrow 0. {}\]

It follows that $\mathrm{Ext}^1_R(R/\omega_R, R)= 0$ . Since R is not Gorenstein, $R/\omega_R\neq 0$ , and since $\mathrm{grade}\ \omega_R =1$ , the theorem of Rees (cf. [ Reference Bruns and Herzog1 , theorem 1·2·5]) implies that $\mathrm{Ext}^1_R(R/\omega_R, R)\neq 0$ , a contradiction.

4. When is $\mathrm{Fitt}_j(I)=I$ ?

Let R be a Noetherian local ring or a finitely generated graded K-algebra. The Hilbert-Burch theorem implies that if $I\subset R$ is a perfect ideal of grade 2, then $\mathrm{Fitt}_1(I)=I$ . In this section we are interested in a converse of this theorem, and ask more generally that if $\mathrm{grade}\, I\geq j$ and $\mathrm{Fitt}_{j-1}(I)=I$ , then what can be said about I. The following theorem gives an answer for $j=2$ .

Proof. We show that the projective dimension of I is 1. Let

\begin{align*}\cdots\longrightarrow R^m\stackrel{\varphi_2}\longrightarrow R^n\stackrel{\varphi_1}\longrightarrow R\rightarrow R/I\longrightarrow 0\end{align*}

be a minimal free resolution of $R/I$ . Since $\mathrm{grade}\, I\geq 2$ , the first two steps of the dual of this resolution is a resolution of $\mathrm{Coker}(\varphi_2^*)$ . By [ Reference Buchsbaum and Eisenbud3 , theorem 3·1], there are maps $a_0, a_1$ and $a_2$ making the following diagrams commute:

For a map $\varphi$ of free modules, let $I_j(\varphi)$ be the ideal of j-minors of $\varphi$ . Since $\mathrm{grade}\, I\geq 2$ , the map $a_0$ must be a unit, so the ideal $I_1(a_1)=I_1(a_1^*)$ is equal to I. Since, by assumption, $I_{n-1}(\varphi_2)=I=I_1(a_1)$ , it follows that $a_2^*$ is surjective. Thus one of the usual basis elements of $\wedge^{n-1} R^m$ must map by $a_2^*$ to a unit, and it follows that if $\varphi'_2$ is the restriction to an appropriate summand of rank $n-1$ in $R^m$ , then $I_{n-1}(\varphi'_2)$ has grade 2. Thus by [ Reference Buchsbaum and Eisenbud2 , corollary 1],

\begin{align*}0\longrightarrow R^{n-1}\stackrel{\varphi'_2}\longrightarrow R^n\stackrel{\varphi_1}\longrightarrow R\rightarrow R/I\longrightarrow 0\end{align*}

is a resolution.

Now we consider radical ideals I of grade $\geq j\geq 2$ which satisfy the condition $\mathrm{Fitt}_{j-1}(I)=I$ . Recall that an ideal I is called unmixed, if all the associated prime ideals of I have the same height, equal to height $I$ .

Proof. Let $P\in \mathrm{Ass} I$ . Then $PR_P=IR_P$ , since I is reduced. Let $m=\mu(PR_P)$ . theorem 2·1 and our assumptions imply that

\[ {}PR_P=IR_P= \mathrm{Fitt}_{j-1}(I)R_P=\mathrm{Fitt}_{j-1}(IR_P)=\mathrm{Fitt}_{j-1}(PR_P)=P^{m-j+1}R_P. {}\]

It follows that $\mu(PR_P)=j$ . Krull’s generalized principal ideal theorem implies that height $P=\ \text{height} PR_P\leq j$ . On the other hand, $j\leq \mathrm{grade}\, I\leq\ \text{height}\ P$ . Therefore, $\mathrm{grade}\ I=j$ and height $P=j$ for all $P\in \mathrm{Ass} I$ . This shows that I is an unmixed ideal of grade j.

We also have seen that $\mu(PR_P)=\ \text{height}\ PR_P=\dim R_P$ for all $P\in \mathrm{Ass} I$ . It follows that $R_P$ is regular for all $P\in \mathrm{Ass} I$ .

In the following theorem we answer the main question of this section for squarefree monomial ideals. Before stating this result we need to recall some concepts and notation.

For a finite simple graph G, we denote the vertex set and the edge set of G by V(G) and E(G), respectively. The edge ideal of G is the squarefree monomial ideal in the polynomial ring $S=K[x_i:\ i\in V(G)]$ generated by all monomials $x_ix_j$ such that $\{i,j\}\in E(G)$ . The complementary graph of G is the graph $G^c$ on vertex set V(G) whose edges are the non-edges of G. A cycle C in G is called an induced cycle if no non-adjacent vertices of C form an edge in G. The graph G is called chordal if it has no induced cycles of length bigger than 3.

For a monomial ideal I, the relation matrix of I arising from the Taylor relations is quite useful in the study of the Fitting ideals of I. Let $\{u_1,\ldots,u_m\}$ be the minimal set of monomial generators of I. Recall that the Taylor relations of I are of the form $(u_j/\gcd(u_i,u_j))e_i-(u_i/\gcd(u_i,u_j))e_j$ for any distinct indices i and j. So any nonzero entry of the relation matrix is of the form $u_j/\gcd(u_i,u_j)$ for some i and j.

(i) $\implies$ (iii). It follows from Proposition 4·2 that I is an unmixed ideal of grade j. First assume that $j=2$ . Then I is unmixed of height 2. It is enough to show that $S/I$ is a Cohen-Macaulay ring. Let G be the graph on [n] such that $\{i,j\}\in E(G)$ if and only if $(x_i,x_j)$ is a minimal prime ideal of I. Then $I=\bigcap_{\{i,j\}\in E(G)} (x_i,x_j)$ . Since $I^{\vee}=I(G)$ , by Eagon-Reiner [ Reference Herzog and Hibi8 , theorem 8·1·9], $S/I$ is a Cohen-Macaulay ring if and only if I(G) has a linear resolution. Thus by [ Reference Fröberg6 ], (see, also, [ Reference Herzog and Hibi8 , theorem 9·2·3]), we need to show that $G^c$ is a chordal graph. By contradiction assume that $G^c$ has an induced cycle $C_k$ with $k\ge 4$ . Then $C_k^c$ is an induced subgraph of G. Consider the prime ideal $P=(x_i:\ i\in V(C_k))$ in S and set $J=\bigcap_{\{i,j\}\in E(C_k^c)} (x_i,x_j)$ . Then $\mathrm{Fitt}_1(JS_P)=\mathrm{Fitt}_1(IS_P)=IS_P=JS_P$ . The minimal monomial generators of J are of the form $x_A$ , where A is a minimal vertex cover of $C_k^c$ . One can see that each minimal vertex cover A of $C_k^c$ is of the form $A=V(C_k)\setminus e$ , where $e\in E(C_k)$ . Therefore $JS_P$ is generated in degree $k-2$ and $\mu(JS_P)=k$ . On the other hand, $\mathrm{Fitt}_1(JS_P)$ is the ideal which is generated by $(k-1)$ -minors of a relation matrix of $JS_P$ . So each monomial generator of $\mathrm{Fitt}_1(JS_P)$ is of degree at least $k-1$ . This contradicts to the equality $\mathrm{Fitt}_1(JS_P)=JS_P$ . Hence $G^c$ is chordal, as claimed.

Let $j\gt 2$ . Let $\mathcal{G}(I)=\{u_1,\ldots,u_m\}$ be the set of minimal monomial generators of I. Without loss of generality we may assume that $\bigcup_{i=1}^m \mathrm{supp}(u_i)=[n]$ . Let $k_I$ be the number of minimal prime ideals of I. By induction on $k_I$ we show that if $\mathrm{Fitt}_{j-1}(I)=I$ and $\mathrm{grade}\, I=j$ , then $u_1,\ldots,u_m$ form a regular sequence. If $k_I=1$ , then I is a prime ideal generated by variables and there is nothing to prove. So we may assume that $k_I\ge 2$ . We claim that $m=j$ . By contradiction assume that $m\gt j$ . Any monomial generator of $\mathrm{Fitt}_{j-1}(I)$ has degree at least $m-j+1\ge 2$ . So $\deg(u_i)\geq 2$ for all i. For each $1\le i\le n$ , let $J_i$ be the monomial localisation of I at $x_i$ , that is the monomial ideal obtained from I by substituting the variable $x_i$ by 1. Then $J_i$ is a monomial ideal of grade j with $k_{J_i}\lt k_I$ and $\mathrm{Fitt}_{j-1}(J_i)=J_i$ . So by induction hypothesis we may assume that $J_i$ is generated by a regular sequence of monomials of length j. This implies that for distinct p and q if $\gcd(u_p,u_q)\neq 1$ , then it has degree at most one. Hence any nonzero entry of a relation matrix A of I has degree at least $d-1$ , where $d=\min\{\deg(u_p):\ 1\le p\le m\}$ . Therefore any $(m-j+1)$ -minor of A has degree at least $(m-j+1)(d-1)$ . Since $\mathrm{Fitt}_{j-1}(I)=I$ , we obtain $(m-j+1)(d-1)\le d$ . Since $d\ge 2$ , we get $m-j+1=2$ and $d=2$ . So $m=j+1$ . Without loss of generality let $u_1=x_1x_2$ be a monomial of degree 2 in I. Since $x_1x_2\in \mathrm{Fitt}_{j-1}(I)$ , there exists $u_2,u_3\in \mathcal{G}(I)$ such that $x_1|u_2$ and $x_2|u_3$ . The equality $m=j+1$ together with height $I=j$ implies that any variable $x_i$ appears in at most two of the $u_i$ ’s. Note that $m\ge 4$ . Since $m=j+1$ and $(u_1,u_2,u_3)\subset (x_1,x_2)$ we conclude that $u_4,\ldots,u_m$ form a regular sequence. Therefore, from $u_4\in \mathrm{Fitt}_{j-1}(I)$ , we have $u_4=(u_2/x_1)(u_3/x_2)$ . Since $\gcd(u_2,u_4)$ has degree at most one, we conclude that $u_2=x_1x_r$ for some r. Similarly, $u_3=x_2x_s$ for some s. Hence $u_4=x_rx_s$ . If $m\gt 4$ , then $u_5=x_rx_s$ as well, which is not possible. Hence $m=4$ and we may write $I=(x_1x_2,x_2x_3,x_1x_4,x_3x_4)$ . Hence height $I=2$ , a contradiction.

(iii) $\implies$ (i) follows from the Hilbert-Burch theorem together with theorem 2·1(a).

For an ideal I in a one-dimensional local ring, it may very well happen that $\mathrm{Fitt}_1(I)=I$ , as we will see below.

An ideal I in a ring R is called a trace ideal, if $I=\mathrm{tr}(M)$ for some R-module M. By [ Reference Lindo10 , Proposition 2·8], I is a trace ideal if and only if $I=\mathrm{tr}(I)$ .

Even when the multiplicity of a one-dimensional local domain R is not 2, there may exist a proper ideal $I\subset R$ with $\mathrm{Fitt}_1(I)=I$ , as the above example shows. We expect that for any one-dimensional local Cohen-Macaulay ring there always exists an ideal I with $\mathrm{Fitt}_1(I)=I$ .

5. On the radical of Fitting ideals

In this section we compare the ideal I with the radical of $\mathrm{Fitt}_j(I)$ .

Theorem 5·1 Let R be a Noetherian ring and $I\subset R$ be an ideal of height $I=h$ . Then for any $1\le j\le h-1$ we have

\begin{align*} {}\sqrt{\mathrm{Fitt}_j(I)}=\sqrt{I}. {}\end{align*}

Corollary 5·2. Let R be a Noetherian ring and let $I\subset R$ be an ideal of height $I=h$ . Then for any $1\le j\le h-1$ we have

\begin{align*} {}\sqrt{\mathrm{Fitt}_j(I)}=\sqrt{\mathrm{Fitt}_j(\sqrt{I})}. {}\end{align*}

Next, we determine radical of Fitting ideals of I, when I is the edge ideal of a graph. Let G be a finite simple graph with the vertex set $V(G)=[n]$ . The open neighbourhood of $i\in V(G)$ is the set $N(i)=\{j:\{i,j\}\in E(G)\}$ . Whereas, the closed neighbourhood of i is the set $N[i]=N(i)\cup\{i\}$ . For a subset A of V(G), we denote by $G\setminus A$ , the subgraph of G with the vertices of A and their incident edges deleted. An edge e of G is called a neighbour of i if $i\notin e$ and i is adjacent to an endpoint of e. The edge neighbourhood of i is the set of all edges in G which are neighbours of i and is denoted by E(i). Let $F=\{i_1,\ldots,i_k\}\subseteq V(G)$ . We call a subset $A=\{e_1,\ldots,e_t\}$ of E(G) an admissible cover of F of size t if:

1. (a) $A=\bigsqcup_{s=1}^k A_s$ , where $\emptyset\neq A_s\subseteq E(i_s)$ for all s.

2. (b) $\{i_s\}\cup e\neq \{i_t\}\cup e'$ for any distinct integers $s,t\in [k]$ , $e\in A_s$ and $e'\in A_t$ .

The minimum cardinality of a vertex cover of G is denoted by $c_G$ .

Theorem 5·3. Let G be a graph with m edges. Then

\begin{align*} {}\sqrt{\mathrm{Fitt}_j(I(G))}=\begin{cases} {} {}I(G)&\text{if}\ j \lt c_G,\\ {} {}I(G)+J&\text{if}\ j\ge c_G, {}\end{cases} {}\end{align*}

where J is the squarefree monomial ideal minimally generated by those monomials $x_{i_1}\cdots x_{i_k}$ for which $F=\{i_1,\ldots,i_k\}$ is an independent set of G, F has an admissible cover of size $m-j$ and no proper subset of F has such an admissible cover.

Note that any Fitting ideal of a monomial ideal is a monomial ideal. To see this, let $u_1,\ldots,u_m$ be the minimal set of monomial generators of I and A be a relation matrix of I whose rows correspond to $u_i$ ’s and whose columns correspond to generating relations of I, say $g_1,\ldots,g_s$ . Each $g_i$ can be chosen to be homogeneous with respect to the multigrading on the polynomial ring. Then any nonzero term in the $(m-j)$ -minor of A corresponding to the rows $r_1,\ldots,r_{m-j}$ and the columns $c_1,\ldots,c_{m-j}$ has multidegree equal to $\sum_{i=1}^{m-j}\mathrm{multideg}(g_{c_i})-\sum_{i=1}^{m-j}\mathrm{multideg}(u_{r_i})$ , where $\mathrm{multideg}(g)$ denotes the multidegree of a multigraded element g.

First we show that for any minimal generator $x_{i_1}\cdots x_{i_k}\in \sqrt{\mathrm{Fitt}_j(I)}\setminus I$ of $\sqrt{\mathrm{Fitt}_j(I)}$ , $F=\{i_1,\ldots,i_k\}$ is an independent set of G and F has an admissible cover of size $m-j$ . The fact that F is an independent set of G follows from the assumption that $x_{i_1}\cdots x_{i_k}\notin I$ . Moreover, since $x_{i_1}\cdots x_{i_k}$ is a minimal generator of $\sqrt{\mathrm{Fitt}_j(I)}$ , we have $x_{i_1}^{r_1}\cdots x_{i_k}^{r_k}\in \mathrm{Fitt}_j(I)$ for some positive integers $r_1,\ldots,r_k$ with $r_1+\cdots+r_k\geq m-j$ . Considering the relation matrix B of I arising from the Taylor relations, one can see that the nonzero entries of B are either variables or monomials $x_ix_j$ for some edges $\{i,j\}$ of G. If $r_1+\cdots+r_k\gt m-j$ , then this means that the minor $x_{i_1}^{r_1}\cdots x_{i_k}^{r_k}$ is the product of $m-j$ entries of B such that at least one of these entries is of the form $x_ix_j$ with $\{i,j\}\in E(G)$ . So $x_ix_j$ divides $x_{i_1}\cdots x_{i_k}$ , contradicting the assumption that $x_{i_1}\cdots x_{i_k}\notin I$ . Hence $r_1+\cdots+r_k=m-j$ and all the entries of B whose product form the minor $x_{i_1}^{r_1}\cdots x_{i_k}^{r_k}$ are variables. Up to relabelling, there exists a subset $A=\{e_1,\ldots, e_{m-j}\}$ of E(G) with $A=\bigsqcup_{s=1}^k A_s$ such that for any $1\le s\le k$ , $|A_s|=r_s$ and for any $e_{\ell}\in A_s$ , $x_{i_s}f_{e_{\ell}}=x_pf_{e'}$ for some $p\in V(G)$ with $p\neq i_s$ and some $e'\in E(G)$ . Hence if $e_{\ell}=\{p,q\}$ , then $\{i_s,q\}=e'\in E(G)$ . Therefore $e_{\ell}\in E(i_s)$ and hence $A_s\subseteq E(i_s)$ .

To show property (b) of A to be an admissible cover of F, by contradiction assume that there exist distinct integers $s,t\in [k]$ , $e\in A_s$ and $e'\in A_t$ such that $\{i_s\}\cup e=\{i_t\}\cup e'$ . Note that $\{i_s\}\cup e=\{p\}\cup e''$ for some $p\in [n]$ and $e''\in A$ and $\{i_t\}\cup e'=\{q\}\cup e_0$ for some $q\in [n]$ and $e_0\in A$ . Thus $e''=e_0$ and $p=q$ . This means that $e''=\{i_s,i_t\}$ and hence $x_{i_1}\cdots x_{i_k}\in I$ , which is a contradiction. This shows that F has the admissible cover A of size $m-t$ .

Next we show that if $F=\{i_1,\ldots,i_k\}$ is an independent set of G and $A=\bigsqcup_{s=1}^kA_s=\{e_1,\ldots, e_{m-j}\}$ is an admissible cover of F, then $x_{i_1}\cdots x_{i_k}\in \sqrt{\mathrm{Fitt}_j(I)}\setminus I$ . To prove this, first by relabelling the $e_i$ ’s we may assume that if $s\lt t$ , then for any $e_i\in A_s$ and $e_{\ell}\in A_t$ one has $i\lt\ell$ . For any $e_t\in A_s$ , there exists a relation $x_{i_s}f_{e_t}=x_pf_{e_r}$ for some $p\neq i_s$ and some $e_r\in E(G)$ . Let $B_t$ be the $m\times 1$ -matrix corresponding to this relation, i.e., whose tth entry is $x_{i_s}$ , its r-th entry is $-x_p$ and its other entries are zero. Let B be the $m\times (m-j)$ -matrix whose i-th column is $B_i$ , and let B’ be the $(m-j)$ -minor of B obtained by deleting the last j rows of B. Then, we see that B’ is the determinant of the following matrix:

If $\det(B')\neq 0$ , then it is equal to $x_{i_1}^{r_1}\cdots x_{i_k}^{r_k}$ , where $r_i=|A_i|$ for all i. For any nonzero entry $x_t$ of B with $t\notin F$ replacing $x_t$ by zero in B’ does not change $\det(B')$ . Indeed, if $\det(B')\neq 0$ , since $x_t$ does not divide $\det(B')$ this is the case, and if $\det(B')=0$ this is obviously true, as well. Hence without loss of generality we may assume that each column of B’ has at most two nonzero entries which belong to $\{x_{i_1},\ldots,x_{i_k}\}$ . We claim that the columns of B’ are linearly independent. If this is not the case, then one can find two columns $B'_i$ and $B'_j$ of B’ whose r-th rows are equal up to sign to $x_{i_{\ell}}$ for some r. Then from the shape of B’ one can see that at least one of $B'_i$ and $B'_j$ has another nonzero entry, say $x_{i_t}$ . This means that we have the relation $x_{i_t}f_{e_d}=x_{i_{\ell}}f_{e_s}$ for some $\ell,s\in [m-j]$ . This contradicts to property (b) of being an admissible cover. Hence, the columns of B’ are linearly independent and so $\det(B')\ne0$ is the required monomial in $\mathrm{Fitt}_j(I)$ . Therefore $x_{i_1}\cdots x_{i_k}\in\sqrt{\mathrm{Fitt}_j(I)}\setminus I$ . Now, in addition assume that no proper subset of F has an admissible cover of size $m-j$ . Then $x_{i_1}\cdots x_{i_k}$ is a minimal generator of $\sqrt{\mathrm{Fitt}_j(I)}$ . Otherwise, there exists $F'\subsetneq F$ such that $\prod_{t\in F'} x_t$ is a minimal generator of $\sqrt{\mathrm{Fitt}_j(I)}$ . Then it follows from the first part of the proof that F’ has an admissible cover of size $m-j$ , a contradiction.

We complete the proof by showing that if $x_{i_1}\cdots x_{i_k}\in \sqrt{\mathrm{Fitt}_j(I)}\setminus I$ is a minimal generator of $\sqrt{\mathrm{Fitt}_j(I)}$ and $F=\{i_1,\ldots,i_k\}$ , then no proper subset of F has an admissible cover of size $m-j$ . On the contrary suppose that $F'\subsetneq F$ is a set which has an admissible cover of size $m-j$ . Then F’ is an independent set of G as well. As was shown above these imply that $\prod_{t\in F'} x_t\in \sqrt{\mathrm{Fitt}_j(I)}\setminus I$ , which contradicts to the minimality of $x_{i_1}\cdots x_{i_k}$ .

As a corollary, we characterise when the radical of Fitting ideal of an edge ideal I(G) is the graded maximal ideal, in terms of the combinatorics of G.

Let d be a positive integer. A graph G is called d-regular if $|N(i)|=d$ for all $i\in V(G)$ . For instance, a complete graph on n vertices is $(n-1)$ -regular, whereas any cycle is 2-regular.

Remark 5·5. The second statement of corollary 5·4 can be rephrased as follows, as well. We have $\sqrt{\mathrm{Fitt}_j(I)}=(x_1,\dots,x_n)$ if and only if

\begin{align*} {} {}\max\{|N(i)|+|E(G_i)|:\ i\in V(G)\}\le j \lt m, {} {}\end{align*}

where $G_i=G\setminus N[i]$ . In particular, when G is a d-regular graph, then $\sqrt{\mathrm{Fitt}_j(I)}=(x_1,\dots,x_n)$ if and only if $j\ge d+|E(G_i)|$ for each i.

Example 5·6. Let $G=K_n$ be the complete graph on [n] and $I=I(G)$ . Then,

\begin{align*}\sqrt{\mathrm{Fitt}_j(I)}=\begin{cases}I&\text{if}\ j \le n-1,\\(x_1,\dots,x_n)&\text{if}\ n-1\le j\lt\Big({\begin{array}{c} n \qquad\\[-5pt] 2 \end{array}\Big)},\\ S& \text{otherwise}.\end{cases}\end{align*}