Existence of non-zero pseudo-null submodules Let $p$ be a rational prime, and let $\Lambda_d$ be the Iwasawa algebra in $d$ variables, i.e. $\Lambda_d = \mathbb{Z}_p[[T_1, \ldots, T_d]]$. Let $A$ be a finitely generated and torsion $\Lambda_d$-module. We will denote by $A^\circ$ the maximal pseudo-null submodule of $A$, where a finitely generated $\Lambda_d$-module $X$ is called pseudo-null iff it is annihilated by two relatively prime elements of $\Lambda_d$.
In case $d = 1$, it is known that $A^\circ = \{0\}$ if and only if the projective dimension of $A$ is at most 1 (see Neukirch, Schmidt, Wingberg, Cohomology of Number Fields, Proposition (5.3.19)).
Is the analogous statement true in general? In other words, suppose that $A$ is a finitely generated and torsion $\Lambda_d$-module, $d \ge 1$ arbitrary. Is $A^\circ$ non-trivial if and only if $\text{pd}_{\Lambda_d}(A) > 1$?
 A: The `$\Rightarrow$'-part is not true anymore if $d > 1$. To give a counterexample, consider the maximal ideal $\mathfrak{m} = (p, T_1, \dots, T_d)$ of $\Lambda_d$. Clearly, $\mathfrak{m}^\circ = 0$ since $\Lambda^\circ = 0$. I however claim that the projective dimension of $\mathfrak{m}$ as a $\Lambda_d$-module is $d$.
Note that $\Lambda_d$ is a regular local ring and therefore, by the Auslander-Buchsbaum formula, it is enough to show that $\mathfrak{m}$ has depth 1. Now, any non-zero element $x$ of $\mathfrak{m}$ is not a zero divisor on $\mathfrak{m}$ but any element $y \in \mathfrak{m}$ is a zero divisor on $\mathfrak{m} / x \mathfrak{m}$. This shows that $\{ x \}$ is a maximal regular sequence for $\mathfrak{m}$, as required.
It is maybe worth pointing out that the `$\Leftarrow$'-part does hold true more generally.
Lemma: If $A$ is a finitely generated $\Lambda_d$-module of projective dimension at most one, then $A^ \circ = 0$.
Proof: Assume there is a non-trivial pseudo-null submodule $M \subseteq A$. This means that the support $\mathrm{Supp} (M)$ of $M$ contains no prime of height $\leq 1$. It follows that the set of associated primes $\mathrm{Ass} (M) \subseteq \mathrm{Supp} (M)$ contains a prime $\mathfrak{p}$ of height $\geq 2$. Since $\mathrm{Ass} (M)  \subseteq \mathrm{Ass} (A)$ we see that $\mathfrak{p}$ is also an associated prime of $A$.
This implies that
$$
\mathrm{depth} (A) \leq \dim (\Lambda_d / \mathfrak{p}) = (d + 1) - \mathrm{ht} (\mathfrak{p}) \leq d - 1.
$$
By the Auslander-Buchsbaum formula we therefore have $\mathrm{pd} (A) > 1$, contradiction. $\square$
