Injection from $(Kotimes_{mathbb{Z}}hat{mathbb{Z}})^*to prod_p(Kotimes_{mathbb{Z}}mathbb{Z}_p)^*$?

In fact, more is true: the ring homomorphism $Kotimes_{Bbb Z}hat{Bbb Z} rightarrow prod_p(K otimes_Bbb Z Bbb Z_p)$ is injective.

To see this, we first look at the case $K = Bbb Q$. It is easy to see that every element of $hat{Bbb Q} = Bbb Qotimes_{Bbb Z} hat{Bbb Z}$ can be written as a pure tensor (because a "common denominator" can be found for any finite sum of pure tensors), thus you should be able to understand this special case.

For general $K$, we fix an isomorphism $K simeq Bbb Q^d$ as $Bbb Q$-vector spaces. We then have $$Kotimes_{Bbb Z}hat{Bbb Z} simeq Kotimes_{Bbb Q}Bbb Q otimes_{Bbb Z} hat{Bbb Z} simeq hat{Bbb Q}^d$$ and also $$Kotimes_{Bbb Z} Bbb Z_p simeq Kotimes_{Bbb Q}Bbb Q otimes_{Bbb Z}Bbb Z_p simeq Bbb Q_p^d.$$

Together with the projection maps $Kotimes_{Bbb Z} hat{Bbb Z}rightarrow Kotimes_{Bbb Z}Bbb Z_p$ and $hat{Bbb Q}^d rightarrow Bbb Q_p^d$, we get a commutative square. The injectivity then follows from the above special case.

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There is actually a more explicit description of $Kotimes_Z hat{Bbb Z}$. It is equal to the following set: $${(x_p)inprod_p (Kotimes_{Bbb Z}Bbb Z_p): x_pin mathcal O_Kotimes_{Bbb Z}Bbb Z_p, text{almost all } p},$$ where "almost all" means "all but finitely many". This is sometimes called a "restricted tensor product".

The group of units $(K otimes_{Bbb Z}hat{Bbb Z})^times$ can be described similarly, by adding $^times$ everywhere.

You may read the wiki page on adele rings or many books on the subject for more information.