Proof. Let M be an MCM module over R. We want to show that for any $f \in S - R$ and any $m \in M$ , there is a well-defined element $f \cdot m \in M$ . Let $W = R - \{0\}$ . Since M is MCM, it is torsion-free over R and embeds in $W^{-1}M$ . It suffices to show that $f \cdot (m/1) \in M$ . Since any element in the $S_2$ -ification of R is multiplied back into R by an ideal in R of height at least $2$ (see [Reference Hochster and HunekeHH, (2.3)]), the height of the ideal $R :_{R} f$ is at least two. Thus, there exist u and v in $R :_{R} f$ such that the sequence $u, v$ is a part of a system of parameters for R. Since M is MCM, the sequence $u, v$ is a regular sequence on M. Then $v \cdot ((uf) \cdot (m/1)) = u \cdot ((vf) \cdot (m/1)) \in vM$ implies that $(vf)\cdot (m/1) \in vM$ . Since M is torsion-free over R, we have $f\cdot (m/1) \in M.$

Proof. By Lemma 4.1, any MCM module M over R is MCM over S. But S is regular. Hence, $M \cong S^{\oplus h}$ by the Auslander–Buchsbaum formula. The second statement follows immediately because $S^{\oplus h}$ is an Ulrich module of R if and only if S is an Ulrich module of R.

Proof. First, notice that $k[\![\underline {u}]\!] \subset k[\![\underline {x}]\!]$ is a module-finite extension. Then R is (Noetherian) local and $R \subset k[\![\underline {x}]\!] $ is a module-finite extension.

Let $\mathfrak {m}_R$ be the maximal ideal of R. From the construction of R, it is clear that $\underline {u} = u_1, \ldots , u_d$ is a system for parameters for R and, in fact, a minimal reduction of $\mathfrak {m}_R$ because all the other adjoined elements are integral over $(\underline {u}) S$ in S and thus integral over $(\underline {u}) R$ in R. Then for all $1 \leq j \leq d$ , the element $x_j$ is multiplied into R by $v_j$ and $w_j$ which generate a height $2$ ideal in R. Thus $x_j$ is in the $S_2$ -ification of R for all $1\leq j \leq d.$ But this means that $S = k[\![\underline {x}]\!]$ is the $S_2$ -ification of $R.$

By Theorem 4.2, it suffices to show that S is not an Ulrich module of R. But $(\underline {u})S \neq \mathfrak {m}_RS$ because $f \notin (\underline {u})S.$ Thus, R has no Ulrich modules.

Counterexample 4.6. This is the first counterexample we found, which led to the general construction in Theorem 4.3. The local domain

$$\begin{align*}R = k[x^n, x^{n+1}, x^ny, y^{n}, y^{n+1}, xy^{n}, xy]_{\mathfrak{m}},\end{align*}$$

where $\mathfrak {m}$ is the maximal ideal $(x^n, x^{n+1}, x^ny, y^{n}, y^{n+1}, xy^{n}, xy)$ , and its completion

$$\begin{align*}\widehat{R} = k[\![x^n, x^{n+1}, x^ny, y^n, y^{n+1}, xy^n, xy]\!]\end{align*}$$

do not have Ulrich modules for $n\geq 2.$

We can show that R and $\widehat {R}$ do not have Ulrich modules directly. The argument is essentially the same for R and $\widehat {R}$ . We will work with R. The $S_2$ -ification of R is $S = k[x,y]_{(x,y)}$ , and the ideal $(xy, x^n-y^n)R$ is a minimal reduction for $\mathfrak {m} R$ . But

$$\begin{align*}(xy, x^n-y^n)S \neq (x^n, x^{n+1}, x^ny, y^n, y^{n+1}, xy^n, xy)S\end{align*}$$

as ideals in S.

We can recover this counterexample from the general construction in Theorem 4.3 by setting

$$ \begin{align*} \underline{x} &= x, y \\ \underline{u} &= xy, x^n - y^n\\ f &= x^n\\ v_j &= (xy)^n \hspace{1em} \text{and } \hspace{1em} w_j = x^n-y^n \hspace{2em} 1\leq j \leq 2\\ g_1 &= x^{n+1} \hspace{1em} \text{ and } \hspace{1em} g_2 = y^{n+1}. \end{align*} $$

Counterexample 4.8 (Localization).

Consider the ring

$$ \begin{align*} T = k[s^{n+1}, sx^{n}, x^{n+1}, x^ny, sy^n, y^{n+1}, xy^n, s^{n-1}xy]_{\mathfrak{n}}, \end{align*} $$

where $\mathfrak {n} = (s^{n+1}, sx^{n}, x^{n+1}, x^ny, sy^n, y^{n+1}, xy^n, s^{n-1}xy)$ and $n\geq 2$ . Let

$$\begin{align*}\varphi: T \hookrightarrow k[s,x,y]_{(s,x,y)}\end{align*}$$

be the inclusion map and $\mathfrak {p} = \varphi ^{-1}((x,y))$ . Then the localization $T_{\mathfrak {p}}$ is the ring

$$ \begin{align*} k(s^{n+1})\Big[\Big(\frac{x}{s}\Big)^n, \Big(\frac{x}{s}\Big)^{n+1}, \Big(\frac{x}{s}\Big)^n \Big(\frac{y}{s}\Big), \Big(\frac{y}{s}\Big)^n, \Big(\frac{y}{s}\Big)^{n+1}, \Big(\frac{x}{s}\Big)\Big(\frac{y}{s}\Big)^n, \Big(\frac{x}{s}\Big)\Big(\frac{y}{s}\Big) \Big] \end{align*} $$

localized at the obvious maximal ideal. But $T_{\mathfrak {p}}$ has no Ulrich modules by Counterexample 4.6.

It remains to show that T has an Ulrich module. Let S be the localization of the $(n+1)$ th Veronese subring of $k[s, x, y]$ at the homogeneous maximal ideal. One can compute $e_{T}(T) = (n+1)^2 = e_{S}(S)$ . The rings T and S have the same fraction field and so $\text {rank}_T(S) = 1$ . Now S has an Ulrich module M by Proposition 3.6 in [Reference HanesHa04]. Then M is MCM over T and $\text {rank}_{T}(M) = \text {rank}_{S}(M)$ . Then

$$ \begin{align*} (n+1) \cdot \text{rank}_{T}(M) = e_{T}(M) \geq \nu_{T}(M) \geq \nu_{S}(M) = e_{S}(M) = (n+1) \cdot \text{rank}_{T}(M). \end{align*} $$

Thus, $e_{T}(M) = \nu _{T}(M)$ and M is Ulrich over $T.$

Proof. Since $\nu _R(N_n) \sim _{\mathcal {M}} \nu _R(M_n)$ , we have $\lim _{n \to \infty } \frac {\nu _R(M_n)}{\nu _R(N_n)} = 1.$ Then

$$ \begin{align*} 0 = \lim_{n \to \infty} \frac{a_n-b_n}{\nu_R(M_n)} = \left(\lim_{n \to \infty} \frac{a_n-b_n}{\nu_R(M_n)} \right)\left(\lim_{n \to \infty} \frac{\nu_R(M_n)}{\nu_R(N_n)}\right) &= \lim_{n \to \infty}\left(\frac{a_n-b_n}{\nu_R(M_n)} \cdot \frac{\nu_R(M_n)}{\nu_R(N_n)}\right) \\ &= \lim_{n \to \infty} \frac{a_n-b_n}{\nu_R(N_n)}.\\[-32pt] \end{align*} $$

Proof. Let be a weakly lim Cohen–Macaulay sequence over R. Let $I = (\underline {x})$ be a system of parameters of the maximal ideal $\mathfrak {m}$ . Recall that $\nu _R(M_n)$ denotes the minimal number of generators of $M_n$ . First, we check that

$$ \begin{align*} \nu_R(M_n) \sim_{\mathcal{M}} \nu_R(\overline{M}_n). \end{align*} $$

Consider the short exact sequence

$$ \begin{align*} 0 \to C_n \to M_n \to \overline{M}_n \to 0. \end{align*} $$

We know that $\nu _R(\overline {M}_n) \leq \nu _R(M_n) \leq \nu _R(\overline {M}_n) + \nu _R(C_n).$ So it suffices to show that

$$\begin{align*}\nu_R(C_n) \sim_{\mathcal{M}} 0.\end{align*}$$

From the short exact sequence, we get the long exact sequence of Koszul homology

$$ \begin{align*} 0 &\to H_2(\underline{x};C_n) \to H_2(\underline{x};M_n) \to H_2(\underline{x}; \overline{M}_n)\\ &\to H_1(\underline{x};C_n) \to H_1(\underline{x};M_n) \to H_1(\underline{x}; \overline{M}_n) \to H_0(\underline{x};C_n) \to H_0(\underline{x};M_n) \to H_0(\underline{x}; \overline{M}_n) \to 0. \end{align*} $$

Now $\overline {M}_n$ has no finite length torsion submodules, so $H_2(\underline {x};\overline {M}_n) = 0$ . We observe the following:

1. (a) $H_2(\underline {x};C_n) \cong H_2(\underline {x};M_n)$

2. (b) $h_1(\underline {x}; C_n) \leq h_1(\underline {x}; M_n)$

3. (c) $\chi _1(\underline {x};M) \geq 0$ for any finitely generated R-module M [Reference SerreS, Corollary on pg. 90]

4. (d) $\chi (\underline {x}; C_n) = e(\underline {x};C_n) = 0$ , where $e(\underline {x};C_n)$ is the multiplicity of $C_n$ with respect to $(\underline {x})$ , and the first equality is by [Reference SerreS, Theorem 1 on pg. 57]

5. (e) $0 \leq h_0(\underline {x}; M_n) - h_0(\underline {x}; \overline {M}_n) \leq h_0(\underline {x}; C_n)$

From (a), (b) and (c), it follows that

$$ \begin{align*} 0 \leq \chi_1(\underline{x}; C_n) \leq \chi_1(\underline{x}; M_n), \end{align*} $$

and because $\mathcal {M}$ is weakly lim Cohen–Macaulay, we have

(1) $$ \begin{align} \chi_1(\underline{x};C_n) \sim_{\mathcal{M}} 0. \end{align} $$

But $\chi (\underline {x};C_n) = 0$ and so, $\chi _1(\underline {x};C_n) = h_0(\underline {x};C_n) = \ell (C_n/(\underline {x})C_n).$ Then the inequality

$$ \begin{align*} \nu_R(C_n) = \ell(C_n/ \mathfrak{m} C_n) \leq \ell(C_n/(\underline{x})C_n) \end{align*} $$

yields

$$ \begin{align*} \nu_R(C_n) \sim_{\mathcal{M}} 0. \end{align*} $$

Next, we show that $\{\overline {M}_n\}$ is a lim Cohen-Macaulay sequence over R. We already know that $h_2(\underline {x}; \overline {M}_n) = 0.$ It remains to show

$$ \begin{align*} \lim_{n \to \infty} \frac{ h_1(\underline{x};\overline{M}_n)}{\nu_R(\overline{M}_n)} = 0. \end{align*} $$

By Lemma 5.2, it is enough to show that

$$ \begin{align*} \lim_{n \to \infty} \frac{ h_1(\underline{x};\overline{M}_n)}{\nu_R(M_n)} = 0 \end{align*} $$

because $\nu _R(\overline {M}_n) \sim _{\mathcal {M}} \nu _R(M_n)$ . Take the alternating sum of the lengths of the Koszul homology in the exact sequence

$$ \begin{align*} 0 \to H_1(\underline{x};C_n) \to H_1(\underline{x};M_n) \to H_1(\underline{x}; \overline{M}_n) \to H_0(\underline{x};C_n) \to H_0(\underline{x};M_n) \to H_0(\underline{x}; \overline{M}_n) \to 0. \end{align*} $$

This is the sum

$$ \begin{align*} h_1(\underline{x}; C_n) - h_1(\underline{x}; M_n) + h_1(\underline{x}; \overline{M}_n) - h_0(\underline{x}; C_n) + h_0(\underline{x}; M_n) - h_0(\underline{x};\overline{M}_n) = 0. \end{align*} $$

Then

$$ \begin{align*} h_1(\underline{x}; \overline{M}_n)& = -h_1(\underline{x}; C_n) + h_0(\underline{x}; C_n) + h_1(\underline{x}; M_n) - h_0(\underline{x}; M_n) + h_0(\underline{x};\overline{M}_n)\\ & = -h_2(\underline{x};C_n) + h_1(\underline{x};M_n) -h_0(\underline{x};M_n) + h_0(\underline{x};\overline{M}_n) \\ &= -h_2(\underline{x};M_n) + h_1(\underline{x};M_n) -h_0(\underline{x};M_n) + h_0(\underline{x};\overline{M}_n) \\ & = \chi_1(\underline{x};M_n) - (h_0(\underline{x};M_n) - h_0(\underline{x};\overline{M}_n)). \end{align*} $$

Now we know that

$$ \begin{align*} \chi_1(\underline{x};M_n) \sim_{\mathcal{M}} 0, \end{align*} $$

and by (e) and (1) above, we have

$$ \begin{align*} 0 \leq h_0(\underline{x};M_n) - h_0(\underline{x};\overline{M}_n) \leq h_0(\underline{x};C_n) = \chi_1(\underline{x};C_n) \sim_{\mathcal{M}} 0. \end{align*} $$

Thus,

$$ \begin{align*} \lim_{n \to \infty} \frac{ h_1(\underline{x};\overline{M}_n)}{\nu_R(M_n)} = 0, \end{align*} $$

and the sequence $\{\overline {M}_n\}$ is lim Cohen–Macaulay.

In the case where $\{M_n\}$ is a weakly lim Ulrich sequence, it remains to check that

$$ \begin{align*} \lim_{n \to \infty} \frac{e_R(\overline{M}_n)}{\nu_R(\overline{M}_n)} = 1. \end{align*} $$

But $e_R(\overline {M}_n) = e_R(M_n)$ and $\nu _R(\overline {M}_n) \sim _{\mathcal {M}} \nu _R(M_n)$ , so the condition immediately follows, and thus, $\{\overline {M}_n\}$ is a lim Ulrich sequence of R.

Proof. Part (a) is clear by the choice of t. Part (c) follows immediately from parts (a) and (b). It remains to prove part (b). Define

$$ \begin{align*} \varphi: H_1(x^t,y^t;M) \to (M:_{M \otimes_R \mathcal{K}}(x^t,y^t))/M \end{align*} $$

to be the map

$$ \begin{align*} [(u, v)] \mapsto \Big[\frac{u}{y^t} \Big] = \Big[\frac{-v}{x^t} \Big], \end{align*} $$

where the equality follows from the relation $ux^t + vy^t = 0.$ This map is well-defined. If $[(u,v)]$ is trivial, then there exists $w \in M$ such that $[(u,v)] = [y^tw, -x^tw]$ . But $[y^tw, -x^tw]$ is mapped to $[(y^tw)/y^t] = [w/1] =0.$

For the map going the other direction, define

$$ \begin{align*} \psi: (M:_{M \otimes_R \mathcal{K}}(x^t,y^t))/M \to H_1(x^t,y^t;M) \end{align*} $$

to be the map

$$ \begin{align*} [f] \mapsto [(y^tf, -x^tf)]. \end{align*} $$

This is clearly well-defined. The maps $\varphi $ and $\psi $ are inverses, so we are done.

Proof. We first prove that

$$ \begin{align*} \nu_R(M_n) \sim_{\mathcal{M}} \nu_R(M_nS). \end{align*} $$

Let $Q_n = M_nS/M_n.$ Note that $Q_n$ has finite length because $S/R$ has finite length. The short exact sequence

$$ \begin{align*} 0 \to M_n \to M_nS \to Q_n \to 0 \end{align*} $$

yields the long exact sequence

$$ \begin{align*} \ldots \to \text{Tor}_1^R(Q_n,k) \to M_n \otimes_R k \to M_nS \otimes_R k \to Q_n\otimes_R k \to 0. \end{align*} $$

Then

$$ \begin{align*} \nu_R(M_n) \leq \nu_R(M_nS) + \ell(\text{Tor}_1^R(Q_n,k)) \leq \nu_R(M_n) + \nu_R(Q_n) + \ell(\text{Tor}_1^R(Q_n,k)), \end{align*} $$

and so it suffices to show that

$$ \begin{align*} \ell(\text{Tor}_1^R(Q_n,k))\sim_{\mathcal{M}}0 \hspace{1em}\text{ and }\hspace{1em} \nu_R(Q_n) \sim_{\mathcal{M}}0. \end{align*} $$

Let $\underline {x} = x_1, x_2$ be a system of parameters for R. By Lemma 5.6, we know that $\ell (Q_n) \leq h_1(x_1^t, x_2^t;M_n)$ for some fixed t. But $\mathcal {M}$ is a lim Cohen-Macaulay sequence, so $h_1(x_1^t, x_2^t;M_n) \sim _{\mathcal {M}} 0$ and

$$ \begin{align*} \ell(Q_n) \sim_{\mathcal{M}} 0. \end{align*} $$

Then

$$ \begin{align*} \nu_R(Q_n) = \ell(Q_n/ \mathfrak{m} Q_n) \sim_{\mathcal{M}}0. \end{align*} $$

Next, by taking a prime cyclic filtration of $Q_n$ , one can observe that

$$ \begin{align*} \ell(\text{Tor}^R_1(Q_n,k)) \leq \ell(Q_n)\ell(\text{Tor}^R_1(k,k)). \end{align*} $$

Then it immediately follows that

$$ \begin{align*} \ell(\text{Tor}_1^R(Q_n,k))\sim_{\mathcal{M}}0. \end{align*} $$

We now show that $\mathcal {N} = \{M_nS\}$ is a lim Cohen–Macaulay sequence of R-modules. It is enough to show that

$$ \begin{align*} h_1(\underline{x};M_nS) \sim_{\mathcal{M}} 0. \end{align*} $$

Because $M_n$ and $M_nS$ are torsion-free over R, we have the long exact sequence

(2) $$ \begin{align} \begin{aligned} 0 &\to H_2(\underline{x};Q_n) \to H_1(\underline{x};M_n) \to H_1(\underline{x};M_nS) \to H_1(\underline{x}; Q_n)\\ &\to H_0(\underline{x};M_n) \to H_0(\underline{x};M_nS) \to H_0(\underline{x};Q_n) \to 0. \end{aligned} \end{align} $$

Next we show that for all $i \geq 0$ ,

$$ \begin{align*} h_i(\underline{x}; Q_n) \sim_{\mathcal{M}} 0. \end{align*} $$

We see that

$$ \begin{align*} h_2(\underline{x}; Q_n) \sim_{\mathcal{M}} 0 \end{align*} $$

because $H_2(\underline {x};Q_n)$ injects into $H_1(\underline {x};M_n)$ . We already proved that $\ell (Q_n) \sim _{\mathcal {M}} 0$ . It immediately follows that

$$ \begin{align*} h_0(\underline{x}; Q_n) = \ell(Q_n/ \underline{x} Q_n) \sim_{\mathcal{M}} 0. \end{align*} $$

Then it follows from $\chi (\underline {x}; Q_n) = 0$ that

$$ \begin{align*} h_1(\underline{x}; Q_n) \sim_{\mathcal{M}} 0. \end{align*} $$

From the long exact sequence (2), we have

$$ \begin{align*} h_1(\underline{x};M_nS) \leq h_1(\underline{x};M_n) + h_1(\underline{x};Q_n). \end{align*} $$

But $h_1(\underline {x};M_n) \sim _{\mathcal {M}} 0$ and $h_1(\underline {x};Q_n)\sim _{\mathcal {M}} 0.$ Therefore, $\mathcal {N} = \{M_nS\}$ is a lim Cohen–Macaulay sequence over R.

If $\mathcal {M}$ is lim Ulrich, it immediately follows that $\mathcal {N}$ is lim Ulrich because $e_R(M_n) = e_R(M_nS)$ and $\nu _R(M_n) \sim _{\mathcal {M}} \nu _R(M_nS).$ It remains to check that $\mathcal {N} = \{M_nS\}$ is a lim Cohen–Macaulay sequence for $S.$

For any i, the Koszul homology $H_i(\underline {x};M_nS)$ does not change whether we think of $M_nS$ as an R-module or an S-module. We also have

$$ \begin{align*} \nu_R(M_nS) \leq \nu_R(S)\nu_S(M_nS), \end{align*} $$

which yields

$$ \begin{align*} \frac{\nu_R(M_nS)}{\nu_R(S)} \leq \nu_S(M_nS). \end{align*} $$

Then

$$ \begin{align*} \lim_{n\to \infty} \frac{h_1^S(\underline{x};M_nS)}{\nu_S(M_nS)} \leq \lim_{n\to \infty} \frac{\nu_R(S)h_1^R(\underline{x};M_nS)}{\nu_R(M_nS)} =\nu_R(S) \lim_{n\to \infty} \frac{h_1^R(\underline{x};M_nS)}{\nu_R(M_nS)} = 0. \end{align*} $$

Thus, $\mathcal {N} = \{M_nS\}$ is a lim Cohen-Macaulay sequence over S.

Proof. Let $x, y$ be a regular system of parameters for S. We have

$$ \begin{align*} a_n = \nu_S(N_n) = h_0(x,y; N_n) \end{align*} $$

and

$$ \begin{align*} b_n = h_1(x,y; N_n). \end{align*} $$

Then $\{N_n\}$ is lim Cohen–Macaulay over S if and only if

$$ \begin{align*} \lim_{n \to \infty} \frac{h_1(x,y; N_n)}{\nu_S(N_n)} = \lim_{n \to \infty} \frac{b_n}{a_n} = 0. \end{align*} $$

Moreover, we have

$$ \begin{align*} \lim_{n \to \infty} \frac{e_S(N_n)}{\nu_S(N_n)} = \lim_{n \to \infty} \frac{a_n - b_n}{a_n} = 1. \end{align*} $$

Note that $e_S(N_n) = a_n-b_n$ by [Reference SerreS, Theorem 1 on pg. 57]. Thus, $\{N_n\}$ is a lim Ulrich sequence for $S.$

Proof. First, (c) and (d) are equivalent by definition. Next (b) and (c) are equivalent by Theorem 4.2. It is clear that (b) implies (a). It remains to show that (a) implies (b).

Suppose R has a weakly lim Ulrich sequence. Then by Theorems 5.3 and 5.7, there exists a lim Ulrich sequence $\mathcal {M} = \{M_n\}$ of torsion-free R modules that are also S modules. Consider the minimal free resolution

$$ \begin{align*} 0 \to S^{b_n} \to S^{a_n} \to M_n \to 0, \end{align*} $$

where $a_n = \nu _S(M_n).$ Now

$$ \begin{align*} \nu_R(S^{b_n}) = b_n\nu_R(S) \end{align*} $$

and

$$ \begin{align*} e_R(S^{b_n}) = b_n e_R(S). \end{align*} $$

Then Theorem 5.8 yields

$$ \begin{align*} \lim_{n \to \infty} \frac{\nu_R(S^{b_n})}{\nu_R(M_n)} = \lim_{n \to \infty} \frac{\nu_R(S)b_n}{\nu_R(M_n)} \leq \lim_{n \to \infty} \frac{\nu_R(S)b_n}{\nu_S(M_n)} = \lim_{n \to \infty} \frac{\nu_R(S)b_n}{a_n} = 0, \end{align*} $$

and

$$ \begin{align*} \lim_{n \to \infty} \frac{e_R(S^{b_n})}{\nu_R(M_n)} = \lim_{n \to \infty} \frac{e_R(S)b_n}{\nu_R(M_n)} \leq \lim_{n \to \infty} \frac{e_R(S)b_n}{\nu_S(M_n)} = \lim_{n \to \infty} \frac{e_R(S)b_n}{a_n} = 0. \end{align*} $$

Consequently, by the minimal free resolution above, we have

(3) $$ \begin{align} \nu_R(M_n) \sim_{\mathcal{M}} \nu_R(S^{a_n}) = \nu_R(S)a_n, \end{align} $$

and

(4) $$ \begin{align} e_R(M_n) \sim_{\mathcal{M}} e_R(S^{a_n}) = e_R(S)a_n. \end{align} $$

Combining equivalences 3 and 4, we have

$$ \begin{align*} \frac{e_R(S)}{v_R(S)} = \lim_{n \to \infty} \frac{e_R(S)}{v_R(S)} = \lim_{n \to \infty} \frac{e_R(M_n)}{v_R(M_n)} = 1. \end{align*} $$

Thus, S is an Ulrich module of R.

Proof. This is immediate by Theorem 4.3 by taking $d = 2$ and applying Theorem 6.1.

Proof. This is immediate by taking k to be perfect and $\text {char}(k)>0$ in Counterexample 4.8 and applying Theorem 6.1.