What can be said about the group $\prod_\omega \mathbb{Z}\otimes_{\mathbb{Z}} \prod_\omega\mathbb{Z}$? Is it again $\prod_\omega \mathbb{Z}$? If yes, by which isomorphism, if no what is it then? Thanks in advance!
(The Tensor product is taken in abelian groups.)
Let me write $P=\prod_\omega\mathbb{Z}$ and $S=\bigoplus_\omega\mathbb{Z}$. There is a pairing $P\otimes S\to\mathbb{Z}$ given by $(a_n)\otimes(b_n)\mapsto \sum a_nb_n$, and I take it as known that this is a perfect pairing, i.e. it induces isomorphisms $P\to\operatorname{Hom}(S,\mathbb{Z})$ and $S\to\operatorname{Hom}(P,\mathbb{Z})$. (The first map being an isomorphism is easy and the second map being an isomorphism is a theorem of Specker.) In particular, the canonical map $P\to \operatorname{Hom}(\operatorname{Hom}(P,\mathbb{Z}),\mathbb{Z})$ is an isomorphism.
To prove that $P\otimes P\not\cong P$, then, it suffices to show that the canonical map $P\otimes P\to \operatorname{Hom}(\operatorname{Hom}(P\otimes P,\mathbb{Z}),\mathbb{Z})$ is not an isomorphism. First, note that $\operatorname{Hom}(P\otimes P,\mathbb{Z})\cong \operatorname{Hom}(P,\operatorname{Hom}(P,\mathbb{Z}))\cong \operatorname{Hom}(P,S)$. Note that $\operatorname{Hom}(P,\mathbb{Z})\cong S$ is countable so the image of any homomorphism $P\to S$ must have finite rank (since any infinite rank subgroup of $S$ would have uncountably many homomorphisms to $\mathbb{Z}$). It follows that the canonical map $\bigoplus_\omega \operatorname{Hom}(P,\mathbb{Z})\to \operatorname{Hom}(P,\bigoplus_\omega\mathbb{Z})=\operatorname{Hom}(P,S)$ is an isomorphism.
So, $\operatorname{Hom}(P,S)\cong \bigoplus_\omega S \cong \bigoplus_{\omega\times\omega}\mathbb{Z}$. Chasing through how all these canonical isomorphisms are defined, this isomorphism sends an element $(c_{mn})\in\bigoplus_{\omega\times\omega}\mathbb{Z}$ to the homomorphism $P\to S$ given by $(a_m)_{m\in\omega}\mapsto (\sum_m c_{mn}a_m)_{n\in\omega}$. (In other words, thinking of elements of $P$ and $S$ as infinite vectors, homomorphisms $P\to S$ are just infinite matrices acting in the usual way, where the matrices can have only finitely many nonzero entries.) The canonical map $P\otimes P\to \operatorname{Hom}(\operatorname{Hom}(P\otimes P,\mathbb{Z}),\mathbb{Z}) \cong \operatorname{Hom}(\bigoplus_{\omega\times\omega}\mathbb{Z},\mathbb{Z})\cong \prod_{\omega\times\omega}\mathbb{Z}$ then sends $(a_m)\otimes(b_n)$ to $(a_mb_n)_{m,n\in\omega}$. However, this map is not surjective; see here for instance.
