For finite group G and field k of char=p, if P,P′ are projective k[G]-modules with [P]=[P′], is it true that P=P′ ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:42:44Z http://mathoverflow.net/feeds/question/115940 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115940/for-finite-group-g-and-field-k-of-charp-if-p-p-are-projective-kg-modules-wit For finite group G and field k of char=p, if P,P′ are projective k[G]-modules with [P]=[P′], is it true that P=P′ ? George 2012-12-10T01:27:16Z 2012-12-11T20:59:34Z <p>That is: is it true that if projective k[G]-modules have same composition factors then they are isomorphic?</p> <p>This is easy to see for char(k)=0, or if G is a composition of a p-group and a p′-group. Serre in "Linear Representations of Finite Groups" (a remark in 16.2 after Corr.2) states this as a well-known fact: "Indeed we know that the equality [P]=[P′] ... is equivalent to P=P′". But unfortunately no references.</p> http://mathoverflow.net/questions/115940/for-finite-group-g-and-field-k-of-charp-if-p-p-are-projective-kg-modules-wit/115959#115959 Answer by Geoff Robinson for For finite group G and field k of char=p, if P,P′ are projective k[G]-modules with [P]=[P′], is it true that P=P′ ? Geoff Robinson 2012-12-10T08:42:04Z 2012-12-11T20:59:34Z <p>This is true for finite groups, and it is a consequence of the non-singularity of the Cartan matrix (whose determinant is a power of \$p\$) in the algebraically closed case. The Cartan invariant \$c_{ij}\$ gives the multiplicity of the \$j\$-th simple module as a composition factor of the \$i\$-th projective indecomposable. If there were two non-isomorphic projectives with the same composition factors, the Cartan matrix would certainly be non-singular. I believe the result may have been stated by R. Swan.</p> http://mathoverflow.net/questions/115940/for-finite-group-g-and-field-k-of-charp-if-p-p-are-projective-kg-modules-wit/115975#115975 Answer by Jim Humphreys for For finite group G and field k of char=p, if P,P′ are projective k[G]-modules with [P]=[P′], is it true that P=P′ ? Jim Humphreys 2012-12-10T13:18:23Z 2012-12-10T13:18:23Z <p>Some clarifications to the question are needed. First, you are referring to Section 16.1 of Serre's book (not 16.2), where he is formulating the main results in Brauer theory. These were originally derived (as in the 1962 Curtis-Reiner text) more concretely in terms of Brauer characters, but then recast in the language of Grothendieck groups. The result you are asking about is formally stated by Serre as Corollary 2 of Theorem 35 in that same section. </p> <p>Serre's parenthetic remark after Corollary 2 of Theorem 34 in 16.1 refers back to the earlier and more elementary step in his Corollary 2 of Proposition 42 in 14.4. </p> <p>There are two essential steps here, based always on the fact that you have a triple <code>\$(K,A,k)\$</code> involving a residue field <code>\$k\$</code> of characteristic <code>\$p&gt;0\$</code> coming from a suitable ring <code>\$A\$</code> whose field of fractions <code>\$K\$</code> has characteristic 0. (The deeper results require some hypotheses on completeness, splitting fields, etc.) First you compare the Grothendieck groups of projective modules for <code>\$KG\$</code> and <code>\$AG\$</code> (as in Serre's Section 14). Then you build the <code>\$cde\$</code>-triangle as in Section 16 and obtain there the injectivity of the map <code>\$c\$</code> as Geoff indicates. (The later book <em>Methods of Representation theory I</em> by Curtis-Reiner has a more detailed version of all this theory.) </p> <p>Let me mention that working with a finite group is essential here, since a partial parallel exists for certain finite group schemes (in the guise of restricted Lie algebras) for which the matrix of Cartan invariants has determinant 0 and the behavior of projectives is more complicated. </p>