# Tate module of elliptic curves; Commuting Hom functor and tensor product in the second coordinate

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.

• 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. – Pavel Čoupek Oct 20 at 7:05
• 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? – Qiaochu Yuan Oct 20 at 7:28
• @Yuan My L is the Iwasawa Algebra which is not finitely presented over $\mathbb{Z}_p$. – user100603 Oct 20 at 14:15
• @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? – user100603 Oct 20 at 14:27
• @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)? – Pavel Čoupek Oct 21 at 1:49