Let $(M,g)=(N,\ddot{g})\times f(B,\bar{g})$ be an Einstein warped-product manifold (i.e. $Ric=\lambda g$) where $f:N \rightarrow (0, \infty)$ (positive scalar function) and with $g= \ddot{g}+f^2 \bar{g}$.
If $(B, \bar{g})$ is Ricci flat, being $(M, g)$ an Einstein manifold, this means that $(M, g)$ must be only Ricci flat or not? Or $(M, g)$ can be Einstein (not Ricci flat) even though $(B, \bar{g})$ is Ricci flat?
$(M,g)$ need not be Ricci flat.
In the half space description $\mathbb{R}_+\times\mathbb{R}^{n-1}$ of the hyperbolic space, the hyperbolic metric on $\mathbb{H}^n$ can be described as : $g=\tfrac{1}{t^2}(dt^2+g_0)$ where $g_0$ is the flat metric on $\mathbb{R}^{n-1}$. Setting $s=\ln t$, we get $g=ds^2+e^{-2s}g_0$ which is Einstein and not Ricci flat although its fiber is flat.
When we have an Einstein warped-product manifold where the base is a Riemannian manifold, independently of dimension, and the fiber is a Ricci-flat, we have: $|\nabla f|^2+[\frac{\lambda (m-n)+ R}{m(m-1)}]f^2=0$ (with $n$ and $m$ the dimension of the base and the fiber, respectively and $R$ is the scalar curvature of the base). Then, either $R$ $\leq$ $\lambda (n − m)$ or $f$ is trivial. This means that the Einstein warped-product manifold will not necessarily be flat even though the fiber is Ricci-flat (for details see, for example, https://www.researchgate.net/publication/319151263_On_the_structure_of_Einstein_warped_product_semi-Riemannian_manifolds).
