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Let $K$ be a quadratic field, $O$ an order of $K$ (say with basis, $\{1,\tau$}). Is it true that the map $x\mapsto x\otimes 1$ induces an inclusion $O\to O\otimes_{\mathbb Z}\hat{\mathbb Z}$ ? (where $\hat{\mathbb Z}=\prod_p\mathbb Z_p$)

I tried to do it by hand but I got confused because I should use the construction of the tensor product, which is very complicated. All I get is that if $x=z_1+\tau z_2$ is in the kernel of this map then $1\otimes z_1+\tau\otimes z_2=0$ and I don't know how to continue.

This also suggested to me the following question: having a tensor product of modules/rings/algebras $A\otimes B$, under which assumptions do we have an inclusion of $A$ or $B$ in the tensor product?

Blacksmith
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1 Answers1

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This is true. It follows abstractly from the fact that $\mathcal{O}$, being torsion-free, is a flat module (over $\mathbb{Z}$), and so tensoring with it preserves injections (here, the injection of $\mathbb{Z}$ into $\widehat{\mathbb{Z}}$). Explicitly, $\mathcal{O} \cong \mathbb{Z}^2$, so the inclusion of $\mathcal{O}$ into $\mathcal{O} \otimes \widehat{\mathbb{Z}}$ is just the inclusion $\mathbb{Z}^2 \to \widehat{\mathbb{Z}}^2$ on underlying abelian groups.

Generally, if $R$ is a commutative ring and $A$ and $B$ are $R$-algebras, then $A$ injects into $A \otimes_R B$ if both the unit map $R \to B$ is injective and $A$ is flat over $R$; these conditions are automatic if $R$ is a field but not otherwise.

Qiaochu Yuan
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  • Thank you. One should note also that tensoring the injection of $\mathbb Z$ into $\hat{\mathbb Z}$ gives an injection not only as modules but also as rings right? Your final general claim is not an if and only if, right? – Blacksmith Jan 15 '18 at 00:40
  • @Blacksmith: 1) injections are injections either way; that is, the forgetful functors from $k$-algebras to $k$-modules to sets both preserve and reflect monomorphisms. 2) Yes, that's why I wrote "if" and not "if and only if." – Qiaochu Yuan Jan 15 '18 at 00:47
  • Maybe you may be interested in this still open question https://math.stackexchange.com/questions/2604172/tensor-product-commutes-with-infinite-products/2604265?noredirect=1#comment5381776_2604265 – Blacksmith Jan 15 '18 at 14:30
  • If $k$ is a field, then $A$ is a vector space, so shouldn't it be flat automatically? – Chris Feb 09 '25 at 07:03
  • @Chris: the tensor product here is over $\mathbb{Z}$, so by "flat module" I mean flat over $\mathbb{Z}$. – Qiaochu Yuan Feb 09 '25 at 15:09
  • The tensor product in your last sentence is over $k$ though? – Chris Feb 09 '25 at 20:10
  • In the last sentence $k$ is an arbitrary commutative ring. I admit the notation is slightly confusing, I'll edit. – Qiaochu Yuan Feb 09 '25 at 20:50
  • @QiaochuYuan my apologies for being pedantic, but I think my point was more so that both conditions are automatic if $R$ is a field but as written it seems to imply only one is? – Chris Feb 10 '25 at 04:34
  • Sure, I guess that's worth clarifying, but if that was your point you should've said so. – Qiaochu Yuan Feb 10 '25 at 04:35