First of all it is good to differentiate what can be curved. One way to deal with <PERSON> equations is to choose a space-time foliation (global when possible). In the <PERSON> models one has a set isometries which define the homogeneus and isotropic spatial hypersurfaces, it can be shown that these hypersurfaces have a normal vector field $n^\mu$ which is geodesic $(n^\nu\nabla_\nu{}n^\mu=0)$ where $\nabla_\nu$ is the connection compatible with the <PERSON> metric $g_{\mu\nu}$ (signature $(-1,1,1,1)$), i.e., $\nabla_\alpha{}g_{\mu\nu}=0$.
Given this choice of hypersurfaces we have the projector $\gamma_{\mu\nu}\equiv{}g_{\mu\nu}+n_\mu{}n_\nu$ which acts as a metric in the spatial sections. 578
This metric we have a unique covariant derivative (in the hypersurfaces) $D_\alpha\gamma_{\mu\nu} = 0$, and this derivative in turn defines a <PERSON> tensor on the spatial section $$(D_\mu{}D_\nu-D_\nu{}D_\mu)v_\alpha = \mathcal{R}{\mu\nu\alpha}{}^\beta{}v\alpha,$$ where $\gamma_\alpha{}^\beta{}v_\beta=v_\alpha$. For the <PERSON> metric we have that $$\nabla_\mu{}n_\nu = \Theta_{\mu\nu} = \frac{\Theta}{3}\gamma_{\mu\nu},$$ where $\Theta_{\mu\nu}$ is the extrinsec curvature and $\Theta$ its trace.
In short, the notion of a foliation induces a concept of curvature and a extrinsec curvature in the hypersurfaces in a way that the four dimensional Riemann tensor $$R_{\mu\nu\alpha}{}^\beta{}v_\beta = (\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu)v_\alpha,$$ is split in $\mathcal{R}{\mu\nu\alpha}{}^\beta{}v\alpha$ plus combinations of $\Theta_{\mu\nu}$ and $D_\mu$, i.e., $$R_{\mu\nu\alpha\beta} = \mathcal{R}{\mu\nu\alpha\beta} + 2\Theta{\mu[\alpha}\Theta_{\beta]\nu} - 4(D_{[\mu}\Theta_{\nu][\alpha}n_{\beta]} + D_{[\alpha}\Theta_{\beta][\mu}n_{\nu]}) + 4(n^\sigma\nabla_\sigma(n_{\mu}\Theta_{\nu][\alpha}n_{\beta]}) + n_{[\mu}\Theta_{\nu]}{}^\sigma\Theta_{\sigma[\alpha}n_{\beta]}),$$ where $T_{[\mu\nu]} = (T_{\mu\nu}-T_{\nu\mu})/2$.
All this said, we can see that it is possible to have a zero spatial curvature (Ricci tensor) $\mathcal{R}{\mu\alpha} = 0$, but a non-zero four dimensional $R{\mu\alpha}$. For a <PERSON> metric we have $D_\mu\Theta=0$, $\mathcal{R}{\mu\nu} = 2K\gamma{\mu\nu}$ and $D_\mu K=0$. 578