Morita equivalence and connectivity

Question

Let $A, B$ be Morita equivalent $\mathbb{E}_1$-ring spectra. Fix an an $(A, B)$-bimodule $P$ and a $(B, A)$-bimodule $Q$ such that $P \otimes_B Q \cong A$ and $Q \otimes_A P \cong B$. If $A$ is bounded below then $B$ is bounded below, because $P, Q$ are perfect $A$-modules (hence bounded below) and \begin{align*} \operatorname{conn}(B) &= \operatorname{conn}(Q \otimes_A P) \\ &\geq \operatorname{conn}(Q \otimes P) \\ &\geq \operatorname{conn}(Q) + \operatorname{conn}(P) + 1 \\ &> -\infty. \end{align*}

Is it true that if $A$ is connective then $B$ is connective? Ie, can a connective and nonconnective ring spectrum be Morita equivalent? This is of course not possible for $\mathbb{E}_\infty$-ring spectra. (Edit: I wrote this thinking that Morita equivalence implied isomorphism like for ordinary commutative rings, but then realized the proof I had in mind doesn't work.)

One thing I was thinking about is that $\operatorname{End}_B(P) \cong A$ is connective, and if $\pi_*(P)$ is a faithful $\pi_*(B)$-module this would imply $B$ must be connective as well. It seems plausible for it to be faithful in some sense because it's invertible. But I'm not if the "some sense" would persist after taking homotopy groups.

Edit: If this is not the case, is there an intrinsically defined invariant of a compactly generated stable category that can tell us when it's the category of modules over a connective ring spectrum, or maybe more generally of $\mathsf{Sp}$-enriched presheaves on a category enriched over $\mathsf{Sp}^\mathrm{cn}$?

Answer 1

Suppose $A$ is any nontrivial connective ring spectrum, with right module $P = A \oplus A[1]$, and let $B = End_A(P)$.

The ring $B$ satisfies $$B \simeq A \oplus A[1] \oplus A[-1] \oplus A$$ and in particular it is not connective because $A[-1]$ is not.

However, $B$ is Morita equivalent to $A$, via the $B$-$A$ bimodule $P$ and the $A$-$B$ bimodule $Hom_A(P,A)$.

Roughly, this is the same as the Morita equivalence between $R$ and $M_n(R)$, except that instead of $R^n$ we are allowed to suspend factors.