Let $\Lambda$ be the Iwasawa Algebra of the Galois group of the cyclotomic $\mathbb{Z}_p$-extension $\mathbb{Q}_{cyc}$ of $\mathbb{Q}$. Let $\widehat{\Lambda}$ be its Pontryagin dual (i.e the dualizing module is $\mathbb{Q}_p/\mathbb{Z}_p$). Let $\tilde{\kappa}$ be the "universal $p$-adic cyclotomic character" $G_\mathbb{Q} \rightarrow Gal(\mathbb{Q}_{cyc}/Q) \rightarrow \Lambda^\times$.

Let $\rho: G_\mathbb{Q}\rightarrow Aut_{\mathbb{Z}_p}(T_p(E))$ be the Galois representation attached to the Tate module of an elliptic curve over $\mathbb{Q}$. Let $\tilde{\rho}: G_\mathbb{Q} \rightarrow Aut_{\Lambda}(\tilde{T}_p(E))$ be the cyclotomic deformation, where $\tilde{T}_p(E)= T_p(E) \otimes_{\mathbb{Z}_p} \Lambda(\tilde{\kappa})$. Here $\Lambda(\tilde{\kappa})$ is the $\Lambda$ but the Galois action is given by $\tilde{\kappa}$.

Let $\widehat{\Lambda}(\tilde{\kappa})$ denote $\widehat{\Lambda}$ where $g \in G_\mathbb{Q}$ acts by $\tilde{\kappa}(g)$. Then we know (by Greenberg) that $$\widehat{\Lambda}(\tilde{\kappa})=Hom(\widehat{\Lambda}(\tilde{\kappa}^{-1}), \mathbb{Q_p}/\mathbb{Z}_p).$$

What I am trying to understand is the following.

Greenberg says that $$T_p(E) \otimes_{\mathbb{Z}_p}Hom(\widehat{\Lambda}(\tilde{\kappa}^{-1}), \mathbb{Q_p}/\mathbb{Z}_p) \cong Hom(\widehat{\Lambda}(\tilde{\kappa}^{-1}), T_p(E) \otimes_{\mathbb{Z}_p}\mathbb{Q_p}/\mathbb{Z}_p) $$

I know from this math overflow post Does module Hom commute with tensor product in the second variable? that to have $Hom_A(L,M) \otimes_A N \cong Hom_A(L, M \otimes_A N)$ we need $L$ to be finitely presented and $N$ to be flat.

But in Greenberg's situation, $\widehat{\Lambda}(\tilde{\kappa}^{-1})$ does not seem to be finitely generated $\mathbb{Z}_p$-module. This is the problem I am facing. Any help is appreciated.

Greenberg's statement is attached. It is an extract of his paper "Iwasawa theory and p-adic deformation of motives" (page 214). See the red highlighted part. enter image description here

  • 2
    $\begingroup$ For a finite free $R$-module $F\simeq R^n$, one always has $F\otimes_R\mathrm{Hom}_R(M,N)\simeq \mathrm{Hom}_R(M,N)^n \simeq \mathrm{Hom}_R(M,N^n)\simeq \mathrm{Hom}_R(M,F\otimes_R N)$. While this looks like cheating, it is not, at least in the sense that all the isomorphisms above are natural. $\endgroup$ – Pavel Čoupek Oct 20 at 7:05
  • $\begingroup$ As I mentioned in my answer to the linked question, the desired commutativity also holds if $L$ is finitely presented projective, with no hypotheses on $M$ or $N$. And the Tate module is free of rank $2$, yes? $\endgroup$ – Qiaochu Yuan Oct 20 at 7:28
  • $\begingroup$ @Yuan My L is the Iwasawa Algebra which is not finitely presented over $\mathbb{Z}_p$. $\endgroup$ – user100603 Oct 20 at 14:15
  • $\begingroup$ @Pavel. Thanks. It makes sense. More generally, if I have a ring extension $A\rightarrow B$ two rings. $F$ is a free $B$-module of finite rank. Lets assume that $N$ is a $B $-module and $M$ is an $A$-module. Then we also have $F \otimes_B Hom_A(M,N) \cong Hom_A(M,F \otimes_B N)$ as $A$-modules. Is this correct? $\endgroup$ – user100603 Oct 20 at 14:27
  • $\begingroup$ @user100603 I don't think so, I assumed, perhaps incorrectly, that the $\mathrm{Hom}$ stands for homomorphism of $\mathbb{Z}_p$-modules. Is it not the case? What is the precise meaning of the $\mathrm{Hom},$ is it meant as continuous homomorphisms of Abelian topological groups (as would be the case with Pontrjagin duality, usually)? $\endgroup$ – Pavel Čoupek Oct 21 at 1:49

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