"Derived algebraic geometry" usually means the study of geometry locally modeled on "$Spec R$" where $R$ is a *connective* $E_\infty$ ring spectrum (perhaps with further restrictions). Why "connective", though?

In my (limited) understanding, approaches to the subject like that of Toen and Vezzosi are motivated as an approach to studying things like intersection theory in ordinary algebraic geometry. The picture is that a connective $E_\infty$ ring $R$ (incarnated as a simplicial commutative ring, usually) is an "infinitesimal thickening" of the ordinary ring $\pi_0 R$. This picture breaks down if $R$ is not connective, motivating one to restrict attention to the connective case (moreover, I don't know of a way to model nonconnective ring spectra analogous to simplicial rings).

But another motivation comes from homotopy theory, $TMF$, and the moduli stack of elliptic curves, which is a nonconnective derived Deligne-Mumford stack. When the basic motivating objects are nonconnective, it leaves me puzzled that Lurie continues to focus primarily on the connective case in *Spectral Algebraic Geometry*.

I see basically two mutually exclusive possible reasons for this:

The theory of nonconnective derived algebraic geometry is wild / ill-behaved / hard to understand, so one restricts attention to the connective cases which is more tractable.

The theory of nonconnective derived algebraic geometry is a straightforward extension of the theory of connective derived algebraic geometry; it is easy to study nonconnective objects in terms of connective covers, but the results are most naturally phrased in terms of the connective objects, so that's the way the theory is expressed.

**Question A.** Which of (1) / (2) is closer to the truth?

Maybe as an illustrative test case, here are two statements pulled at random from SAG. Let $R$ be a connective $E_\infty$ ring, let $Mod_R$ denote its $\infty$-category of modules, and $Mod_R^{cn}$ its $\infty$-category of connective modules (both of which are symmetric monoidal), and let $M \in Mod_R$.

$M$ is perfect(=compact in $Mod_R$) iff $M$ is dualizable in $Mod_R$.

$M$ is locally free (= retract of some $R^n$) iff $M$ is connective and moreover dualizable in $Mod_R^{cn}$.

**Question B.** Do these statements have analogs when $R$ is nonconnective? If so, are they straightforward extensions of these statements from the connective case?

For Question B, feel free to substitute a better example of a statement if you like.