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@SunilPasupulati: Magma told me so (I did the computation conditional on GRH). However, the argument does not really depend on that. If the group was $C_3\times C_3$, the entire argument would still work: the extension would be split, and the subgroup of order $5$ in $I$ would still act trivially, so that you could find the Hilbert class field by looking at the sextic subfield of $\mathbb{Q}(\zeta_{31})$.
@FranzLemmermeyer: it is! How did you find that polynomial? Do feel free to edit the answer, or alternatively I could edit in your much nicer polynomial, but I would be very interested in how you arrived at it.
The head of a projective indecomposable module over an Artinian ring is simple, so there is exactly one isomorphism class of simples that $P_S$ can surject onto (the equality should really be an isomorphism, not an equality). I am sure this is in Alperin, but I don't have the book in front of me right now.
