  Consider the many shared resource scheduling problem where jobs have to be
scheduled on identical parallel machines with the goal of minimizing the
makespan. However, each job needs exactly one additional shared resource in
order to be executed and hence prevents the execution of jobs that need the
same resource while being processed. Previously a $(2m/(m+1))$-approximation
was the best known result for this problem. Furthermore, a $6/5$-approximation
for the case with only two machines was known as well as a PTAS for the case
with a constant number of machines. We present a simple and fast
5/3-approximation and a much more involved but still reasonable
1.5-approximation. Furthermore, we provide a PTAS for the case with only a
constant number of machines, which is arguably simpler and faster than the
previously known one, as well as a PTAS with resource augmentation for the
general case. The approximation schemes make use of the N-fold integer
programming machinery, which has found more and more applications in the field
of scheduling recently. It is plausible that the latter results can be improved
and extended to more general cases. Lastly, we give a $5/4 - \varepsilon$
inapproximability result for the natural problem extension where each job may
need up to a constant number (in particular $3$) of different resources.
