# 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 $$ and height $$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 $$A$$.

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