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Hello? I have a simple question.

Is $\mathbb{Z}_p$ flat $\mathbb{Z}_pG$-module for a finite $p$-group $G$? Here, $p$ is prime and $\mathbb{Z}_p$ means the integers localized at $(p)$. If not, is it false even for a finite abelian $p$-group $G$?

Please let me know.

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  • $\begingroup$ Can you compute $\operatorname{Tor}^{\mathbb Z_{(p)}G}_\bullet(\mathbb Z_{(p)},\mathbb Z_{(p)})$ ? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jun 27, 2011 at 5:42
  • $\begingroup$ (Try it with $G$ the cyclic group of order $p$ first) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jun 27, 2011 at 5:56
  • $\begingroup$ @Mariano Suárez-Alvarez : Isn't $\operatorname{Tor}_*^{\mathbb{Z}_{(p)}G}(\mathbb{Z}_{(p)},\mathbb{Z}_{(p)})=0$ for $*\neq 0$? when $G=\mathbb{Z}/p\mathbb{Z}$ $\endgroup$
    – qkqh
    Commented Jun 27, 2011 at 8:14
  • $\begingroup$ @Mariano Suárez-Alvarez : I used a (augmented) free resolution $\cdots \to \mathbb{Z}_{(p)}G \to \mathbb{Z}_{(p)}G \to \mathbb{Z}_{(p)}G (\to \mathbb{Z}_{(p)} \to 0)$ such that first map sends $g$ to $1$, second sends $g$ to $g-1$, (last one is the augmentation map) and the fore two maps are repeated. Then by tensoring $\mathbb{Z}_{(p)}$, $\cdots \to \mathbb{Z}_{(p)} \to \mathbb{Z}_{(p)} \to \mathbb{Z}_{(p)}$ is obtained wih identity map and zero map. And I guess it is similar when $G$ is another finite $p$-group. Does this have wrong parts?^^;; $\endgroup$
    – qkqh
    Commented Jun 27, 2011 at 8:36

3 Answers 3

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(This is answering a comment to the main question)

$\newcommand\ZZ{\mathbb Z}$ If $G$ is cyclic of order $p$, then there is a resolution of $\ZZ$ looking like $$\cdots\to\ZZ G\xrightarrow{d_{\mathrm{odd}}} \ZZ G\xrightarrow{d_{\mathrm{even}}}\cdots\to\ZZ G\xrightarrow{d_{\mathrm{even}}} \ZZ G\xrightarrow{d_{\mathrm{odd}}}\ZZ G\xrightarrow{\varepsilon}\ZZ$$ in which $\varepsilon$ is the usual augmentation, the odd differentials $d_{\mathrm{odd}}$ are given by multiplication by $g-1$, and the even ones $d_{\mathrm{even}}$ are given by multiplication by $1+g+\cdots+g^{p-1}$.

If we tensor over $\ZZ$ this complex with the flat $\ZZ$-module $\ZZ_{(p)}$, we get a new exact complex $$\cdots\to\ZZ_{(p)} G\xrightarrow{d_{\mathrm{odd}}} \ZZ_{(p)} G\xrightarrow{d_{\mathrm{even}}}\cdots\to\ZZ_{(p)} G\xrightarrow{d_{\mathrm{even}}} \ZZ_{(p)} G\xrightarrow{d_{\mathrm{odd}}}\ZZ_{(p)} G\xrightarrow{\varepsilon}\ZZ_{(p)}$$ which is clearly a free $\ZZ_{(p)}G$-resolution of $\ZZ_{(p)}$. Notice that this is not the complex you mentioned in your comment.

Drop the rightmost term, tensor it now with $\ZZ_{(p)}$ over $\ZZ_{(p)}G$, and we end up with a complex which looks like $$\cdots\to\ZZ_{(p)}\xrightarrow{0} \ZZ_{(p)}\xrightarrow{p}\cdots\to\ZZ_{(p)} \xrightarrow{p} \ZZ_{(p)} \xrightarrow{0}\ZZ_{(p)}$$ In particular, $Tor\_1^{\ZZ_{(p)}G}(\ZZ_{(p)},\ZZ_{(p)})$ is $\ZZ\_{(p)}/p\ZZ\_{(p)}=\ZZ/p\ZZ$.

In fact, it is not hard (using for example a change-of-rings argument for the localization map $\ZZ G\to\ZZ\_{(p)}G$) that for a general group one has $$Tor\_1^{\ZZ\_{(p)}G}(\ZZ\_{(p)},\ZZ\_{(p)})=Tor\_1^{\ZZ G}(\ZZ\_{(p)},\ZZ\_{(p)})=\bigr(Tor\_1^{\ZZ G}(\ZZ,\ZZ)\bigr)\_{(p)}=(G\_\mathrm{ab})\_{(p)},$$ and the last group is not zero if, say, $G$ is a finite $p$-group.

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  • $\begingroup$ @Mariano Suárez-Alvarez : Right, right. I had some mistakes on the map $d_{even}$, so I got a wrong $\operatorname{Tor}_*^{\mathbb{Z}_{(p)}}(\mathbb{Z}_{(p)},\mathbb{Z}_{(p)}). Thank you ^^. $\endgroup$
    – qkqh
    Commented Jul 5, 2011 at 7:15
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The question can be considered and answered in greater generality:

Let $R$ be a (not necessarily commutative) ring with unit and let $G$ be a finite group. Then the trivial $RG$-module $R$ is flat iff $|G|$ is invertible in $R$.

Proof: By a result of Benson, $R$ is flat iff it's projective (see Theorem 1.2), which is equivalent to the splitting of the augementation $\epsilon: RG \to R$ (over $RG$).

Assume $i: R \to RG$ is a spliting of $\epsilon$ and $i(1) = \alpha$. From $g \cdot i(1) = i(g\cdot 1) = i(1)\;\; (g \in G)$ it follows that $\alpha = r \cdot N_G$ for some $r \in R$ and the norm element $N_G = \sum_{g \in G}g$. Appyling $\epsilon$ yields $1=r\cdot |G|$, so $|G|$ is a unit in $R$.

Conversely, if $|G|$ is a unit in $R$, than $i:R \to RG, 1 \to |G|^{-1}\cdot N_G$ is easily seen to be a splitting of $\epsilon$.

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  • $\begingroup$ @Ralph : Oh, this is really general. Thank you^^ $\endgroup$
    – qkqh
    Commented Jul 5, 2011 at 8:19
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Consider the sequence of Z_p G-modules 0 -> I -> Z_p G -> Z_p -> 0

where I is the augmentation ideal. Tensor this over Z_p G with Z_p = Z_p G/ I and you get

0 -> I / I^2 - > Z_p -> Z_p -> 0

which is no longer exact at the left; thus the module is not flat. Indeed, Tor_1(Z_p, Z_p) is exactly I/I^2, if I haven't miscomputed.

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  • $\begingroup$ Why is $I \otimes \mathbb{Z}_{(p)}G/I \cong I/I^2$? This isn't at all a question of your correctness, I'm simply new to this material and trying to learn. $\endgroup$
    – David White
    Commented Jun 27, 2011 at 14:41
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    $\begingroup$ $M \otimes_R R/I \cong M/IM$ for all commutative (unital, associative) rings $R$, ideals $I$ of $R$, and modules $M$ over $R$. Basically this is because $m\otimes 0 = m \otimes i = mi \otimes 1$ (so the tensor product is a quotient of M/IM, and clearly there is a bilinear map from M x R/I to M/IM given by multiplication, so the quotient is in fact an isomorphism). $\endgroup$
    – Jack Schmidt
    Commented Jun 27, 2011 at 16:37
  • $\begingroup$ @Jack, the same is true without commutativity if you are consistent about sides :) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jun 27, 2011 at 17:38

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