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Look at the abelian group $G=\mathbb{Z}1\oplus\mathbb{Z}\sqrt{2}\subset\mathbb{R}$. As a $\mathbb{Z}$-module it is isomorphic to $\mathbb{Z}^2$ so if we calculate $G\otimes_\mathbb{Z} \mathbb{R}$ it should give us $\mathbb{R}^2$ but clearly it is $\mathbb{R}$. What is going on here? Something must be wrong, I just don't know what. Shouldn't $R^2\otimes_R N=N^2$ be true for all rings $R$ and $R$-modules $N$?

Somge
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  • $\mathbb{R}$ and $\mathbb{R^2}$ are isomorphic as abelian groups. ($\mathbb{Z}$-modules) Actually, they are even isomorphic as $\mathbb{Q}$-vector spaces, because they have the same dimension. – Mark Oct 13 '23 at 23:49
  • Since @Mark's comment is non-obvious, here's an explanation for the OP https://math.stackexchange.com/questions/1744066/are-bbbr-and-bbbr2-isomorphic-what-about-bbbz-and-bbbz2 – Chickenmancer Oct 13 '23 at 23:58
  • @Chickenmancer It's always good to see another answer (so no need to remove the link), but the answer in the link is pretty much the same as what I wrote, at least in this part. The explanation there is that they have the same dimension over $\mathbb{Q}$, without any more details. – Mark Oct 14 '23 at 00:19
  • @Mark: this is irrelevant. The tensor product is an $\mathbb{R}$-module. – Qiaochu Yuan Oct 14 '23 at 08:36
  • @QiaochuYuan Yeah, I didn't check it to be honest. I saw a part of the question where OP was clearly wrong and decided to write a comment. – Mark Oct 14 '23 at 10:04
  • @Mark for what it's worth I was sharing that link to give more context to your comments for the OP to get more context about their error. The additional commentary in distinguishing $\Bbb{Z}$ and $\Bbb{Z}^2$ I thought might be helpful to Somge. – Chickenmancer Oct 14 '23 at 16:37

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clearly it is $\mathbb{R}$.

This is incorrect. The tensor product is $\mathbb{R}^2$, not only as an $\mathbb{R}$-vector space but even as an $\mathbb{R}$-algebra. Explicitly, we have

$$\mathbb{Z}[\sqrt{2}] \otimes \mathbb{R} \cong \mathbb{R}[x]/(x^2 - 2) \cong \mathbb{R}[x]/((x - \sqrt{2})(x + \sqrt{2})) \cong \mathbb{R}^2$$

by the Chinese remainder theorem, where the two induced maps $\mathbb{Z}[\sqrt{2}] \to \mathbb{R}$ are the two real embeddings, sending $\sqrt{2}$ to $\sqrt{2}$ and to $-\sqrt{2}$.

Qiaochu Yuan
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