Completion at a prime

Proof. We have to show that \(\hom (\S /p^{\infty }[-1],X)\) is \(p\)-complete and that the map \(X \to \hom (\S /p^{\infty }[-1],X)\) is an \(\S /p\)-equivalence. For the latter, we may equivalently show that its fiber is \(\S /p\)-acyclic, or equivalently that its fiber is \(\S [\frac {1}{p}]\)-local by Observation 7.3.2. Because of the exact sequence \(\S /p^{\infty }[-1] \to \S \to \S [\frac {1}{p}]\), this fiber is isomorphic to \(\hom (\S [\frac {1}{p}],X)\). But this is clearly \(\S [\frac {1}{p}]\)-local: for any \(\S [\frac {1}{p}]\)-acyclic spectrum \(Y\) we have \(\hom (Y,\hom (\S [\frac {1}{p}],X)) \simeq \hom (Y \otimes \S [\frac {1}{p}],X) \simeq \hom (0,X) = 0\).

We will now show that \(\hom (\S /p^{\infty }[-1],X)\) is \(p\)-complete, i.e. \(\S /p\)-local. To this end, let \(A\) be an \(\S /p\)-acyclic spectrum, i.e. the map \(p\colon A \to A\) is an isomorphism. We need to show that \(0 = \hom (A,\hom (\S /p^{\infty }[-1],X)) \simeq \hom (A \otimes \S /p^{\infty }[-1],X)\). For this, it will suffice to show that \(A \otimes \S /p^{\infty } = 0\). Since we may write \(\S /p^{\infty }\) as a colimit of \(\S /p^n\), we may similarly write \[ A \otimes \S /p^{\infty } \simeq \colim _n(A \otimes \S /p^n) \simeq \colim _n(A/p^n). \] But as \(A\) is \(\S /p\)-acyclic, the map \(p\colon A\to A\) and hence also \(p^n\colon A \to A\) is an isomorphism, and so \(A/p^n = 0\). This shows that \(A \otimes \S /p^{\infty }\simeq 0\), as desired. □

Proof. The exact sequence \(\S /p^n[-1] \to \S \xrightarrow {\cdot p^n} \S \) induces an exact sequence \[ \hom (\S ,X) \xrightarrow {\cdot p^n} \hom (\S ,X) \to \hom (\S /p^n[-1], X). \] Since \(\hom (\S ,X) \simeq X\), it follows that \(\hom (\S /p^n[-1], X) \simeq X/p^n\) for all \(n\). Furthermore, under these isomorphisms the maps \(\S /p^n \xrightarrow {\cdot p} \S /p^{n+1}\) induce the canonical maps \(X/p^{n+1} \to X/p^n\), as these are the ones induced on vertical cofibers in the following commutative diagram:

Using Theorem 7.3.5, we may then compute that \begin {align*} X^{\wedge }_p \simeq {} & \hom (\S /p^{\infty }[-1],X) \simeq \hom (\colim _n \S /p^n[-1],X) \\ \simeq & \lim _n \hom (\S /p^n[-1],X) \simeq \lim _n X/p^n, \end {align*}

as desired. □

Proof. Since \(X\) is \(p\)-complete, the localization map \(X \to X^{\wedge }_p\) is an \(\S /p\)-equivalence between \(\S /p\)-local spectra, hence is an isomorphism by Lemma 7.1.6. By Corollary 7.3.6, we may therefore write \(X \cong \lim _n(X/p^n)\). Each \(X/p^n\) is \(p\)-local by Proposition 7.2.12. By Lemma 7.2.8, \(p\)-local spectra are \(\S [P^{-1}]\)-local for \(P\) the set of all primes different from \(p\), so they are closed under limits by Exercise 7.1.4. □

Proof. The \(q\)-completeness of \(X\) means that \(X\) is \(\S /q\)-local. The \(p\)-completeness of \(X\) implies, via Corollary 7.3.7, that the map \(q\colon X \to X\) is an isomorphism, hence that \(X\) is \(\S /q\)-acyclic. It then follows from Lemma 7.1.6 that \(X = 0\). □

Proof. Write \(\Z /p^{\infty } \cong \colim _k \Z /p^k\) for the filtered colimit along the multiplication-by-\(p\) injections \(\Z /p^k \xrightarrow {\ \cdot p\ } \Z /p^{k+1}\). Since filtered colimits are exact in \(\Ab \), the colimit in \(\D (\Z )\) of a filtered diagram of abelian groups concentrated in degree zero is again concentrated in degree zero, with degree-zero homology given by the colimit in \(\Ab \). Thus the displayed colimit also computes the colimit in \(\D (\Z )\). Mapping out of it gives \[ \hom _{\D (\Z )}(\Z /p^{\infty },A) \simeq \lim _k \hom _{\D (\Z )}(\Z /p^k,A). \] For every \(k \geq 1\), the standard resolution of \(\Z /p^k\) identifies \[ \Hom _{\Z }(\Z /p^k,A) \cong A[p^k] \qquad \text {and}\qquad \Ext ^1_{\Z }(\Z /p^k,A) \cong A/p^k A. \] Under these identifications, the transition maps are multiplication by \(p\) on the first tower and the quotient maps on the second. The two claims now follow by applying the Milnor sequence of Corollary 4.4.37 in degrees \(0\) and \(-1\).

If \(A\) is finitely generated, its \(p\)-primary torsion is bounded, so the transition maps into every fixed stage of the first tower are eventually zero. It follows directly from the description in Definition 4.4.36 that both its limit and its first derived limit vanish. Finally, \(\lim _k A/p^k A \cong A \otimes _{\Z } \Z _p\) for every finitely generated abelian group \(A\). □

Proof. Let \(F := \bigoplus _{k \geq 0}\S \), with summand inclusions \(\iota _k\colon \S \to F\). Splicing the telescope sequence for \(\S [\frac {1}{p}]\) from Lemma 4.4.34 with the defining exact sequence \(\S \to \S [\frac {1}{p}] \to \S /p^{\infty }\), and then reindexing the source, gives an exact sequence \[ F \xrightarrow {\ d\ } F \longrightarrow \S /p^{\infty }, \] where \(d \circ \iota _0 = \iota _0\) and \(d \circ \iota _{k+1} = \iota _k-p\iota _{k+1}\) for \(k \geq 0\). On \(\pi _0\) this realizes the free resolution \[ 0 \longrightarrow \bigoplus _{k \geq 0}\Z \xrightarrow {\ d_*\ } \bigoplus _{k \geq 0}\Z \longrightarrow \Z /p^{\infty } \longrightarrow 0, \] in which the \(k\)-th basis vector of the middle group maps to \(p^{-k}+\Z \). Thus the exact sequence of spectra is obtained by realizing this free resolution by direct sums of spheres.

Apply \(\hom (-,X)\) to the exact sequence above. For every \(m \in \Z \), the universal property of the coproduct identifies \[ \pi _m\hom (F,X) \cong \Hom _{\Z }\Big (\bigoplus _{k \geq 0}\Z ,\pi _m X\Big ), \] and the map induced by \(d\) agrees with precomposition by \(d_*\). By Theorem 7.3.5, we have \(X_p^{\wedge } \simeq \hom (\S /p^{\infty },X)[1]\), so that \(\pi _n(X_p^{\wedge }) \cong \pi _{n-1}\hom (\S /p^{\infty },X)\). Write \(d_m^*\) for the endomorphism induced by \(d^*\) on \(\pi _m\hom (F,X)\). The long exact sequence associated to the exact sequence \(\hom (\S /p^{\infty },X) \to \hom (F,X) \xrightarrow {d^*} \hom (F,X)\) then gives a short exact sequence \[ 0 \to \coker (d_n^*) \to \pi _n(X_p^{\wedge }) \to \ker (d_{n-1}^*) \to 0. \] By Proposition 6.4.14, its outer terms are \[ \Ext ^1_{\Z }(\Z /p^{\infty },\pi _n X) = L_0^p(\pi _n X) \qquad \text {and}\qquad \Hom _{\Z }(\Z /p^{\infty },\pi _{n-1}X) = L_1^p(\pi _{n-1}X), \] respectively. This gives the asserted short exact sequence. □

Proof. By Lemma 7.3.9, the right-hand term of the short exact sequence of Theorem 7.3.10 vanishes, while its left-hand term is \(\pi _n(X) \otimes _{\Z } \Z _p\). □

Proof. For every \(p\in P\), the condition \(X/p=0\) means that multiplication by \(p\) on \(X\) is an isomorphism. Thus \(X\) is \(P\)-divisible, hence \(\S [P^{-1}]\)-local by Lemma 7.2.8. On the other hand, the isomorphism \(X[P^{-1}] \cong X\otimes \S [P^{-1}]\) shows that \(X\) is \(\S [P^{-1}]\)-acyclic. It follows from Lemma 7.1.6 that \(X = 0\). □

Proof. We have to show that the map \(X \to X[\frac {1}{p}] \times _{X^{\wedge }_p[\frac {1}{p}]} X^{\wedge }_p\) is an isomorphism, or equivalently that its fiber \(F\) is the zero spectrum. Note that \(F[\frac {1}{p}] = 0\), since the square

is a pullback square (as the two vertical maps are isomorphisms). We claim that also \(F/p = 0\), so that \(F = 0\) by Lemma 7.3.14, applied with \(P=\{p\}\). To see this, we need to show that the commutative square

is a pullback square. The two objects at the bottom are zero, since the maps \(p\colon X[\frac {1}{p}] \to X[\frac {1}{p}]\) and \(p\colon X^{\wedge }_p[\frac {1}{p}] \to X^{\wedge }_p[\frac {1}{p}]\) are isomorphisms by Corollary 7.2.5. In particular the bottom map is an isomorphism. But the top map is also an isomorphism, since the map \(X \to X^{\wedge }_p\) is by definition an \(\S /p\)-equivalence. This finishes the proof. □