ac.commutative algebra – Can an infinite abelian $p$-group be tall and thin?

  1. Does there exist an abelian $p$-group $A$ with countable Ulm invariants and uncountable height?

Here by height, I mean the minimal ordinal $rho$ such that $p^rho A$ is divisible (1). For an ordinal $alpha < rho$, the $alpha$th Ulm invariant of $A$ is $dim_{mathbb F_p} frac{(p^alpha A)(p)}{(p^{alpha+1} A)(p)}$ where $(p)$ denotes the kernel of multiplication by $p$. These are in some sense a measure of the “width” of $A$.

  1. More generally, let $lambda$ be an uncountable cardinal. Does there exist an abelian $p$-group with all Ulm invariants $<lambda$ and height $geq lambda$?

If we allow $lambda = omega$, then the answer is yes: $oplus_kmathbb Z / p^k$ has height $omega$ and Ulm invariants all equal to 1. More generally, by Ulm’s theorem there exist countable abelian $p$-groups with finite Ulm invariants of any countable height.

  1. Does there exist an abelian $p$-group with finite Ulm invariants and height $geq omega_1$?

(1) Recall that we inductively define $p^0 A = A$, $p^{alpha+1} A = p(p^alpha A)$, and $p^alpha A = cap_{beta < alpha} p^beta A$ for limit ordinals $alpha$. We have a descending chain $A supset pA supset dots supset p^rho A = p^{rho+1}A cdots$ which stabilizes only when we reach the height of $A$.