Uniqueness of the rank of the core of a lattice - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:50:27Z http://mathoverflow.net/feeds/question/103338 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103338/uniqueness-of-the-rank-of-the-core-of-a-lattice Uniqueness of the rank of the core of a lattice Jeff 2012-07-27T18:50:13Z 2012-07-28T13:23:59Z <p>In the paper </p> <p>P.J. Webb: Bounding the ranks of ZG-lattices by their restrictions to elementary abelian groups. J. Pure Appl. Algebra 23 (3) (1982), 311-318. </p> <p>the author writes in the introduction without giving any reference or explanation ($G$ denotes a finite group): </p> <blockquote> <p>If $M$ is any $\mathbb{Z}G$-lattice, that is a $\mathbb{Z}G$-module which is a free abelian group of finite rank, we may write $M=\text{core}(M)\oplus \text{proj}(M)$ where $\text{proj}(M)$ is a projective module and $\text{core}(M)$ is a submodule of $M$ with no projective summands. In general the submodule $\text{core}(M)$ is not uniquely determined by these requirements, but all the possible choices for $\text{core}(M)$ will have the same ranks as an abelian group. </p> </blockquote> <p>I managed to show that each lattice can be decomposed in the described way, but I'm not able to prove that all $\text{core}(M)$ have the same rank. </p> <p><strong>Question:</strong> Why is the rank of $\text{core}(M)$ uniquely determined ? </p> http://mathoverflow.net/questions/103338/uniqueness-of-the-rank-of-the-core-of-a-lattice/103378#103378 Answer by Florian Eisele for Uniqueness of the rank of the core of a lattice Florian Eisele 2012-07-28T11:58:32Z 2012-07-28T13:23:59Z <p>Even more is true: Two possible cores of $M$ lie in the same genus, i.e. they become isomorphic in the localization at any prime of $\mathbb Z$. <del>However, I couldn't find an explicit proof of this anywhere in literature</del> <strong>Edit:</strong> It's Proposition 5.4 in Gruenberg: "Relation Modules of Finite Groups" (though unfortunately there's apparently no online version of this book besides google books). </p> <p>Assume you have two decompositions $M=C_1\oplus P_1 = C_2 \oplus P_2$ with $P_1$ and $P_2$ projective. Assume furthermore that one cannot split off any further non-zero projective summands from $C_1$ and $C_2$. </p> <p>By a theorem of Swan (see <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183523366" rel="nofollow">Swan: Projective modules over finite groups</a>) $\mathbb Q\otimes P_1$ and $\mathbb Q\otimes P_2$ are free, and therefore so is $\mathbb Z_{(p)} \otimes P_1$ and $\mathbb Z_{(p)}\otimes P_2$ ($\mathbb Z_{(p)}$ denotes the localization of $\mathbb Z$ at $p$; this simply uses the fact that since the columns of the decomposition matrix of a finite group are linearly independent, the isomorphism type of the $\mathbb Q$-span of a projective $\mathbb Z_{(p)}$-module determines the isomorphism type of the projective module; alternatively see Theorem (77.14) on p. 548 in Curtis, Reiner: Representation theory of finite groups and associative algebras). </p> <p>So now apply cancellation for $\mathbb Z_{(p)}G$-lattices (see <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183524252" rel="nofollow">Corollary 1 in this paper of Reiner</a>) to find that unless $P_1$ and $P_2$ have the same $\mathbb Z$-rank (in which case we would be done), either $C_1$ or $C_2$ (say, w.l.o.g. $C_1$) has a non-zero free summand in every localization. So if this is the case, then take for each prime $p_1,\ldots,p_n$ that divides $G$ an epimorphism $$\phi_i: \mathbb Z_{(p_i)} \otimes C_1 \twoheadrightarrow \mathbb Z_{(p_i)}G \subset \mathbb QG$$ By multiplying each $\phi_i$ with a sufficiently large integer relatively prime to $p_i$ we may assume $\phi_i(C_1) \subseteq \mathbb Z G$ for each $i$. Now we get a map $$\phi := \sum_{i=1}^n \left(\prod_{j\neq i} p_j \right)\cdot \phi_i|_{C_1}$$ $\phi$ is a homomorphism from $C_1$ to $\mathbb Z G$, which is an epimorphism in the localization at each $p_i$ (by construction: the factor $\prod_{j\neq i} p_j$ is a unit in $\mathbb Z_{(p_i)}$ and in the radical of each $\mathbb Z_{(p_j)}$. Hence the image of $\phi$ in the localization at $p_i$ is the same as the image of $\phi_i$). The image of $\phi$ in the localization at any other prime is projective anyways, since if $p$ doesn't divide the order of $G$, then any $\mathbb Z_{(p)}G$-lattice is projective. So now we have a homomorphism from $C_1$ onto the image of $\phi$, and the image of $\phi$ is projective in each localization. But then by Theorem (78.1) on p. 550 in Curtis&amp;Reiner the image of $\phi$ is projective, and therefore the image of $\phi$ is a projective summand of $C_1$. This contradicts the assumption that we couldn't split off any further projective summands from $C_1$.</p>