# Crafting Suspension Spectra

There is a theorem by Hopkins and Smith which states that for every $n > 0$ there is an ideal $I_n = (v_0^{k_0}, \dots, v_n^{k_n})$ such that there exist a spectrum $X_n$ with the following homology: $$BP_*(X_n)=BP_*/I_n.$$

I was wondering if it was possible to have a similar but more detailed result.

Is there, for every $n$, an ideal $I_n = (v_0^{k_{0,n}}, \dots, v_n^{k_{n,n}})$ such that:

1) For every $i$, the limit of the $k_{i,n}$ goes to infinity when $n$ goes to infinity,

2) There exist a suspension spectrum $X_n$ with $BP_*(X_n)=\Sigma^{d_n}BP_*/I_n$, for a natural number $d_n$

3) The $d_n$ don't go to infinity when $n$ goes to infinity.

if it's not possible to satisfy 1,2,3, what about just 1,2? Thanks!!!

For large $i_0,\cdots,i_n$ we can realize $BP_\ast/(v_0^{i_0},\cdots,v_n^{i_n})$ as the $BP$-homology of a finite spectrum $S/(v_0^{i_0},\cdots,v_n^{i_n})$. This follows from Devinatz-Hopkins-Smith. Suspending a finite spectrum enough times gives a suspension spectrum. This means that there is some space $X_n$ such that $\Sigma^{k_n} S/(v_0^{i_0},\cdots,v_n^{i_n}) \simeq \Sigma^\infty X_n$ for $k_n\gg 0$. I'm not sure how to address the question about the limit of the $k_n$'s.
• But how would the $BP$-homology of $\Sigma X_n$ look like? Do we have any tool to find out? Thanks – Alfred Dec 27 '17 at 15:52
• @Alfred Since $\Sigma^{k_n} S/(v_0^{i_0},\cdots,v_n^{i_n}) \simeq \Sigma^\infty X_n$ the $BP$-homology of $\Sigma^\infty X_n$ will just be $BP_\ast/(v_0^{i_0},\cdots,v_n^{i_n})$ shifted by $k_n$. – skd Dec 27 '17 at 16:16