I am wondering if there is a simple example of a manifold such that, given a value for the scalar curvature $R$, I can find a manifold such that the Ricci tensor has all zero components except for one component which takes the value $R$.


I feel like this can be achieved using a warped product of two metrics to separate out one coordinate and then just solve the differential equations so that the first coefficient vanishes, but obviously the coefficient of the non-zero component needs to be $R$.



Example of a Manifold which has One Non-zero Component of Ric corresponding to Scalar Curvature
Your intuition is correct that warped product metrics will do the trick.


Maybe try to redo the computations yourself, but if I'm not mistaken the Ricci curvature of $g=dt^2+f(t)^2dx^2$ (where $x\in\mathbb{R}^n$ and $dx^2$ denotes the euclidean metric on $\mathbb{R}^n$) is given by :
$$\text{Ric}\_g=-(n-1)\frac{f''(t)}{f(t)}dt^2-\left(f''(t)f(t)+(n-2)f'(t)^2\right)dx^2.$$
So any positive non constant solution to $f''f+(n-2)(f')^2=0$ will give you a solution, these are given by : $f(t)=C\_2(C\_1+(n-1)t)^{\tfrac{1}{n-1}}$.
However note that the scalar curvature will not be constant in that example and that the singularity which happens when $f$ vanishes is genuine (the scalar curvature will blow up).


Maybe we can get something better (smooth for instance) using $\mathbb{R}\times \mathbb{S}^n$ or $\mathbb{R}\times \mathbb{H}^n$, this way the ODE becomes $f''f+(n-2)(f')^2=\pm 1$. The trick which solved the ODE in the first case doesn't work here though and a more careful analysis is required.


I don't know if one can find less symmetric examples, maybe studying the vector field given by the eigenvector of the Ricci tensor associated to the nonzero eigenvalue can give something.


