I'm having a confusion about doing algebra with spectra. My question occured while reading p. 10 of Sanath Devalapurkar's Chromatic Homotopy Theory where we are in the course of proving the thick subcategory theorem.
Let $R$ be a ring spectrum and suppose that the Hurewicz image of $x \in \pi_{\bullet}R$ inside $K(n)_{\bullet}R$ is zero for all $0 \leq n \leq \infty$. We define $$ P(n) = BP/(p, v_1, \cdots , v_{n-1}).$$ Sanath then claims that by assumption there is some finite $N$ for which $P(N) \wedge x^{-1} R$ is contractible.
I'm not too experienced with such computations yet, so I'm confused about how this follows. It seems like that if one tries to translate this into ordinary algebra, then the $K(n)$-conditions amount to $x^{-1}R \otimes \mathbb{F}_p[v_n, v_n^{-1}] = 0$ (for an ordinary ring $R$ and $x \in R$) for all $n$ and the desired condition is $x^{-1}R \otimes \mathbb{F}_p[v_N^{-1}, v_N, v_{N+1}, v_{N+2}, \cdots] = 0$ for some $N \in \mathbb{N}$.
It sounds like one can already choose $N = 1$? Either I'm messing up the translation or I'm confused about ordinary algebra.
Let me start by commenting on the algebraic analogue. I don't quite see why the desired condition in the algebraic analogue is $x^{-1}R\otimes \mathbb{F}_p[v_{N}^{-1}, v_N, v_{N+1}, v_{N+2},\ldots]=0$, and not only $x^{-1}R\otimes \mathbb{F}_p[v_N, v_{N+1}, v_{N+2},\ldots]=0$. The absence of this inversion of $v_{N}$ makes that you could not take $N=1$ and be done with it. Maybe this is where there is a mistake in you algebraic analogue?
The condition that we have to use to show that $P(N)\otimes x^{-1}R$ is contractible for some $N$ is the fact that the image of $x$ in $K(n)_*R$ is $0$ for $n=\infty$ specifically. To summarize the argument: we know that $K(\infty)\simeq \mathbb{F}_p\simeq P(\infty)$, which will give us that $P(\infty)\otimes x^{-1}R\simeq 0$. After that we use the analogue of the following algebraic fact: if $A$ is a commutative ring and $I=(a_1,a_2,\ldots)$ is an ideal of $A$, then if an element $x\in A$ is contained in $I$ it is also contained in some ideal $I_N=(a_1,\ldots,a_N)$, and as such, if the image of $x$ in $A/I$ is $0$, so is its image in $A/I_{N}$ for some finite $N$. In the case of spectra, this will precisely tell us that $P(N)\otimes x^{-1}R\simeq 0$ for some finite $N$.
Let us start being precise. For notational simplicity, I will assume that all elements $v_i$ and $x$ itself lie in degree $0$, because I will mess up degrees in the formulas otherwise. The argument in the general case is conceptually the same.
Let $A\in\mathrm{Sp}$ and suppose given elements $a_1,a_2,\ldots\in\pi_0(A)$. We can inductively define $A/(a_1,\ldots,a_n)$ as cofiber of $-\cdot a_n\colon A/(a_1,\ldots a_{n-1})\to A/(a_1,\ldots a_{n-1})$. Define $I_n:\simeq\mathrm{fib}(A\to A/(a_1,\ldots,a_n))$. Of course, we want to think about $I_n$ as ''being the ideal $(a_1,\ldots,a_n)$''. (Note that $A/I_n\simeq A/(a_1,\ldots a_n)$ holds by definition.) Let $$ B:\simeq\mathrm{colim}(A\to A/I_1\to A/I_2\to\ldots) $$ and put $I:\simeq\mathrm{fib}(A\to B)$. We want to think about $I$ as the ideal $(a_1,a_2,\ldots)$, and therefore want to show that $I\simeq\mathrm{colim}_n\,I_n$. For this, consider the diagram $$ \require{AMScd} \begin{CD} A @>{\mathrm{id}}>> A @>{\mathrm{id}}>>A @>{\mathrm{id}}>>\ldots\\ @VVV @VVV@VVV\\ A @>>>A/I_1@>>>A/I_2@>>>\ldots \end{CD} $$ We know that colimits commute with finite limits in $\mathrm{Sp}$ (this holds in any stable $\infty$-category, because finite products are finite coproducts, and pullback squares are also pushout squares), so if we take fibers of the vertical maps and then a sequential colimit (giving us $\mathrm{colim}_n\,I_n$), we get the same result as first taking sequential colimits and then a fiber (giving us $I$), and therefore $I\simeq\mathrm{colim}_n\,I_n$. In particular, if $x\in\pi_0(A)$ is such that the composite $\mathbb{S}\xrightarrow{x} A \to A/I$ is $0$, then $x\colon\mathbb{S}\to A$ factors as $\mathbb{S}\to I \to A$ by the universal property of the fiber. But $\mathbb{S}$ is compact, and therefore there exists an $N$ such that $\mathbb{S}\to I$ factors as $\mathbb{S}\to I_N \to I$. In particular, there exists an $N$ such that the composite $\mathbb{S}\xrightarrow{x} A\to A/I_N$ is already $0$.
Now to the problem at hand. To show that some spectrum $P(n)\otimes x^{-1}R$ is contractible, it suffices to show that the map $x\colon\mathbb{S}\to P(n)\otimes x^{-1}R$ is equivalent to the zero map, because $x$ is also a unit of this spectrum. Since the image of $x$ in $K(\infty)_*R$ is $0$ and $K(\infty)\simeq\mathbb{F}_p\simeq P(\infty)$, we know that $x\colon\mathbb{S}\to P(\infty)\otimes x^{-1}R$ is the zero map. But by definition, $P(\infty)\simeq BP/I'$ where $I':\simeq(p,v_1,v_2,\ldots)$ is an ideal in the sense above. Let $I'_n:\simeq (p,v_1\ldots,v_{n-1})$, and write $I$ and $I_n$ for the corresponding ideals of $A:\simeq BP\otimes x^{-1}R$. Since the tensor product is cocontinuous in each variable, we can write $P(n)\otimes x^{-1}R$ as $A/I_n$ for $1\leq n\leq \infty$ (where $I_\infty:\simeq I$). In the beginning of this paragraph, we have established that $x\colon\mathbb{S}\to A/I$ is the zero map, and by the previous paragraph this implies that there exists a finite $N$ such that $x\colon\mathbb{S}\to A/I_N$ is the zero map as well. As we previously established, this is equivalent to $P(N)\otimes x^{-1}R$ being contractible, and we are done.