Today we will certain properties of modules over commutative (i.e., $\mathbb{E}_ \infty$) rings. For some parts of the talk, we will need to require that $R$ is connective.
Projective modules#
Here, we need all $R$ to be connective.
Definition 1. Let $R$ be a connective ring. We say that $P \in \mathsf{LMod}_ R^\mathrm{cn}$ (the superscript means that it is connective) is projective if the map $$ \Hom(P, -) \colon \mathsf{LMod}_ R^\mathrm{cn} \to \mathcal{S} $$ preserves geometric realizations.
Remark 2. In the classical setting, geometric realization is a coequalizer of two maps $f, g$ that have a common section. We need to restrict to connective modules, because $\mathsf{LMod}_ R$ has no nonzero projective objects objects.
Proposition 3. The following are equivalent.
- The $R$-module $P$ is projective.
- For all $Q \in \mathsf{LMod}_ R^\mathrm{cn}$ and $i \gt 0$, the groups $\Ext^i(P, Q)$ are zero.
- Given a fiber sequence $N^\prime \to N \to N^{\prime\prime}$ all connective, we have $\Ext^0(P, N) \to \Ext^0(P, N^{\prime\prime})$ surjective.
Here, $\Ext^i(P, Q)$ is defined as $\pi^0 \Hom(P, Q[i])$ (shift convention is $\pi_k Q[1] = \pi_{i-1} Q$). The equivalence between (2) and (3) is formal. The really interesting thing is the equivalence between (1) and (2).
Proposition 4. The following are equivalent for an $R$-module $P$.
- $P$ is projective.
- There exists a free $R$-module such that $P$ is a retract of $M$.
Proof.
For the forward direction, first find a map $R^{\oplus I} \to P$ that is surjective on $\pi_0$. Then we can look at the kernel $$ N \to R^{\oplus I} \to P, $$ and because the map was chosen to be surjective on $\pi_0$, we see that $N$ is connective. Then take $\Hom(P, -)$ and use (3) of the previous proposition to lift $\id_P$ to a map $P \to R^{\oplus I}$.
For the reverse direction, the claim is that projectivity is preserved under retracts. This is because if $P$ is a retract of $S$ then $\Ext^i(P, Q)$ is a retract of $\Ext^i(S, Q)$, and we can use (2) of the previous proposition. Now we can check that free modules are projective.
Flat modules#
Here we don’t need to assume that $R$ is connective. We also only need $R$ to be $\mathbb{E}_ 1$.
Definition 5. We say that $M$ is flat over $R$ if $\pi_0 M$ is flat over $\pi_0 R$ and $$ \pi_m(R) \otimes_{\pi_0(R)} \pi_0(M) \to \pi_m(M) $$ are isomorphisms for all integers $m$.
Remark 6. Flatness is closed under coproducts, retracts, and filtered colimits. If $R$ is connective, projective implies flat.
Proposition 7. The following are equivalent, where $R$ and $N$ are connective.
- $N$ is a flat $R$-module.
- If $M$ is a discrete $R$-module then $M \otimes_R N$ is also discrete.
Proof.
There is a spectral sequence $$ E_{p,q}^2 = \Tor_p^{\pi_\ast R}(\pi_\ast M, \pi_\ast N)_ q \Longrightarrow \pi_{p+q}(M \otimes_R N). $$ Here, this $\Tor$ is computed in the category of graded modules over a graded ring.
Now for (1) to (2), we note that there is only one column $p = 0$. Then we can compute $$ \pi_n(M \otimes_R N) = (\pi_\ast M \otimes_{\pi_\ast R} \pi_\ast N)_ n = (\pi_\ast M \otimes_{\pi_0 R} \pi_0 N)_ n. $$ This means that $M$ discrete implies $M \otimes_R N$ discrete.
In the other direction, we are assuming that $- \otimes_R N$ restricts to $$ \mathsf{RMod}_ R^\heartsuit \simeq \mathsf{Mod}_ {\pi_0 R} \to \mathsf{Set}. $$ This implies that this functor is the same as $- \otimes_{\pi_0 R} \pi_0 N$, and this is necessarily exact because it is $- \otimes_R N$ is exact. This shows that $\pi_0 N$ is flat over $\pi_0 R$. Showing the other part is a bit trickier, but we can use the spectral sequence again.
Lemma 8. Let $f \colon M \to N$ be a map of flat $R$-modules. Then $f$ is an equivalence if and only if it induces a bijection $\pi_0 M \cong \pi_0 N$.
Proposition 9. Let $R$ be connective and let $M / R$ be flat. Then $M$ is projective if and only if $\pi_0 M$ is projective over $\pi_0 R$.
Proof.
Let’s only prove a weaker version. Assume $\pi_0 M$ is free over $\pi_0 R$. Then we can find a map $R^{\oplus I} \to M$ where on $\pi_0$ it is an isomorphism. Now we can apply the above lemma to conclude that $M$ is free.
Localizations#
Here we really assume $R$ is an $\mathbb{E}_ \infty$-ring. This implies that $\pi_\ast R$ is graded commutative. Let $S \subseteq \pi_\ast R$ be a set of homogeneous elements, closed under multiplication and containing $1$.
Definition 10. Let $M$ be an $R$-module.
- We say that $M$ is $S$-nilpotent if all $x \in \pi_\ast M$ is killed by some $s \in S$.
- We say that $M$ is $S$-local if for all $s \in S$, multiplication by $s$ gives an isomorphism $\pi_\ast M \to \pi_\ast M$.
This gives two subcategories $$ \mathsf{LMod}_ R^{S\mathrm{-nil}}, \mathsf{LMod}_ R^{S\mathrm{-loc}} \subseteq \mathsf{LMod}_ R. $$
Proposition 11. 1. The subcategory $\mathsf{LMod}_ R^{S\mathrm{-nil}}$ is stable and closed under all small colimits. Moreover, it is generated under colimit by $R/Rs[n]$, where for $s \in \pi_d R$ this $R/Rs$ is defined as the cofiber of $R[d] \xrightarrow{s} R$. 2. A module $M$ is $S$-local if and only if for all $s \in S$ and $n \in \mathbb{Z}$, $$ \Hom(R / Rs[n], N) $$ is contractible, and this is true if and only if for all $M \in \mathsf{LMod}_ R^{S\mathrm{-nil}}$, the set $\Hom(M, N)$ is contractible.
From (1) and some formal nonsense, we get a right adjoint $$ G \colon \mathsf{LMod}_ R \to \mathsf{LMod}_ R^{S\mathrm{-nil}} $$ to the inclusion. Then for any module $M$, we can define localization as the cofiber $$ G(M) \to M \to S^{-1}M. $$ Now if $N$ is nilpotent, we get a fiber sequence $$ \Hom(N, G(M)) \cong \Hom(N,M) \to \Hom(N, S^{-1} M) $$ and so the right term is contractible. This shows that $S^{-1} M$ is always $S$-local.
Proposition 12. The subcategories $(\mathsf{LMod}_ R^{S\mathrm{-nil}}, \mathsf{LMod}_ R^{S\mathrm{-loc}})$ gives a t-structure on $\mathsf{LMod}_ R$ with trivial heart.
Remark 13. We have $\pi_\ast (S^{-1} M) = S^{-1} \pi_\ast M$.