Tensor product over $\mathbb{Z}$ and p-adic integer ring $\mathbb{Z}_p$

Question

Thanks for your reading. Suppose we have two $\mathbb{Z}_p$-modules $A,B$. Do we always have $A \otimes_{\mathbb{Z}} B \simeq A \otimes_{\mathbb{Z}_p} B$, as abelian groups or $\mathbb{Z}_p$-modules? Here $\mathbb{Z}_p$-module could be given by multiplication on left or right item of the tensor. If not, in which cases is it true?

The motivation is when I read many arithmetic geometry papers, I see many authors will just write tensor without subscript, which I think means tensor over $\mathbb{Z}$, though according to background of text it should be tensor over $\mathbb{Z}_p$.

I get some hints from @Geoffrey Trang. For a commutative ring $R$ and two $R$-modules $A$ and $B$, where every $\mathbb{Z}$-linear map between $R$-modules is automatically $R$ -linear, we can have the two tensor isomorphic as abelian groups. It is true for $R=Z/nZ$, but also for $R=\mathbb{Q}$, and more generally, whenever $\mathbb{Z} \rightarrow R$ is an epimorphism in the category of commutative rings. So, I think since $\mathbb{Z}→\mathbb{Z}_p$ is an epimorphism in the category (and by completion under $p$-adic topology and definition of epimorphism), we have the two tensor over $\mathbb{Z}$ and $\mathbb{Z}_p$ are isomorphic as abelian groups.

Besides, the two $\mathbb{Z}_p$-module structure on $A \otimes_{\mathbb{Z}} B$ are the same also due to $\mathbb{Z} \rightarrow \mathbb{Z}_p$ is epimorphism. And it will be true for any $\mathbb{Z} \rightarrow R$ epimorphism. This point is obtained also by @Geoffrey Trang's hint.

Are the results listed above right? If right, why the equivalence of the two kinds of linear maps, $\mathbb{Z},R$ gives the equivalence of the two tensor?( I guess by some hom-tensor adjunction?)

And, why epimorphism condition gives us the equivalence of the two kinds of linear maps, and the equivalence of the two kinds of module? Moreover, is the converse of the question true? Like if the two kinds of linear maps are equivalent, do we have $\mathbb{Z} \rightarrow R$ epimorphism?

Answer 1

No, $\mathbb{Z}_p\otimes_{\mathbb{Z}}\mathbb{Z}_p$ is much larger than $\mathbb{Z}_p\otimes_{\mathbb{Z}_p}\mathbb{Z}_p=\mathbb{Z}_p$. To see this, observe that $\mathbb{Z}_p/\mathbb{Z}$ is $p$-torsion free and $\mathbb{Z}_p$ is $p$-local, so $\mathrm{Tor}_1(\mathbb{Z}_p, \mathbb{Z}_p/\mathbb{Z})=0$, and we have an exact sequence $$ 0 \to \mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z} \to \mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_p \to \mathbb{Z}_p\otimes_{\mathbb{Z}} (\mathbb{Z}_p/\mathbb{Z})\to 0. $$ Next, the term on the right is in fact a $\mathbb{Q}$-vector space: multiplication with $p$ is invertible on $\mathbb{Z}_p/\mathbb{Z}$, and multiplication with numbers coprime to $p$ is invertible on $\mathbb{Z}_p$, so every integer acts invertibly on $\mathbb{Z}_p\otimes_{\mathbb{Z}} (\mathbb{Z}_p/\mathbb{Z})$. Tensoring a $\mathbb{Q}$ vector space with $\mathbb{Q}$ doesn't change its value, so $$ \mathbb{Z}_p\otimes_{\mathbb{Z}} (\mathbb{Z}_p/\mathbb{Z}) \cong (\mathbb{Z}_p\otimes_{\mathbb{Z}} (\mathbb{Z}_p/\mathbb{Z}) )\otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}_p \otimes_{\mathbb{Q}} (\mathbb{Q}_p/\mathbb{Q}). $$ As $\mathbb{Q}_p$ is an uncountable $\mathbb{Q}$-vector space, so is this tensor product. So the above exact sequence, which is split by the multiplication map $\mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_p \to \mathbb{Z}_p$, proves that $$ \mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_p \cong \mathbb{Z}_p \oplus \text{uncountable $\mathbb{Q}$-vector space} $$ Nonetheless, the category of $p$-complete abelian groups carries a symmetric-monoidal structure given by the $p$-completed tensor product, i.e. tensor product over $\mathbb{Z}$ followed by $p$-completion, and sometimes this might also just be denoted by the symbol $\otimes$.

Answer 2

To add to Achim's answer, Bousfield and Kan, "The core of a ring" JPAA 1972 and (correction) 1973 define a ring $R$ to be a "solid" if the multiplication map $R\otimes_{\mathbb{Z}}R\to R$ is an isomorphism. They then go on to characterise solid rings. There is also a related discussion on page 44 of their book, "Homotopy limits, completions and localizations".

If a ring $R$ is solid, then for any right $R$-module $A$ and left $R$-module $B$, we have $$A\otimes_{\mathbb{Z}} B \cong A\otimes_R R\otimes_{\mathbb{Z}} R \otimes_R B \cong A\otimes_R R \otimes_R B\cong A\otimes_R B.$$ So your question is asking whether the ring of $p$-adic numbers $\mathbb{Z}_p$ is solid.

The ring of $\mathbb{Z}_p$ is not solid in the sense of Bousfield and Kan. This follows from their paper, or from the answer of Achim above.