From Kunen "Set Theory", Chapter II, Exercise 36:
If $T, T'$ are $\kappa$-trees, the product, $ T \times T' $ is the $\kappa$-tree whose $\alpha$-th level is $\mbox{Lev}_\alpha(T) \times \mbox{Lev}_\alpha(T')$, with order defined by $\langle{x,x' }\rangle < \langle{y,y' }\rangle$ iff $ x < y $ and $ x' < y' $.
Show that if $ T $ is a $ \kappa $-Suslin tree, then $ T \times T $ is not a $ \kappa $-Suslin tree.
I think I managed to prove this if $ \kappa $ is regular. I can't see how to deal with the case that $ \kappa $ is singular.
I was wondering if the exercise is actually true for singular $ \kappa $, and if it is, I am willing for a hint (not a solution). If not, I'd be happy for a counter-example.
Take $T = \{ ({k, \alpha }) \mid k \in \omega, \alpha \in \omega_k \}$ with $ ({k, \alpha)} > ({l, \beta)} $ iff $ k = l $ and $ \alpha > \beta $.
Then the product is $T \times T = \{ ( ({k, \alpha }), ({l, \alpha } ) ) \mid k,l \in \omega, \alpha \in \omega_k\cap\omega_l \}$ with $ ( ({k, \alpha }), ({l, \alpha } ) ) > ( ({k, \beta }), ({l, \beta } ) ) $ iff $ k = k, l = l, \alpha > \beta $.
This shows that every anti chain has at most $ \aleph_0 $ elements (since there are only $ \aleph_0 $ options for $ (k,l) $).