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3 votes

Characterize rings $R$, such that the countable product $P=R^N$ has the property that every finitely generated submodule of $P$ is free

I think your condition is equivalent to being a semifir (but see the remark below). See Chapter 2 of Paul Cohn, "Free ideal rings and localization in general rings". A commutative semifir is ...
Dave Benson's user avatar
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2 votes

Characterize rings $R$, such that the countable product $P=R^N$ has the property that every finitely generated submodule of $P$ is free

Assuming rings are commutative with unity, these are exactly the Bézout domains (and the zero ring). Let $R$ be a non-zero ring. The following are equivalent: $R$ is a Bézout domain (a domain in ...
Alex Kruckman's user avatar
5 votes
Accepted

A commutative ring with unity which does not have relatively pseudo-injective ideals with zero intersection

Can we find a commutative ring $𝑅$ with unity such that there exist two ideals $A, B\subseteq R$ such that $A\cap B=0$ and $𝐴_𝑅$ is NOT pseudo-$𝐵_𝑅$-injective? Let $R=\mathbb Z[x,y]/(x^2,xy,y^2)$,...
Keith Kearnes's user avatar
4 votes

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

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}}...
Dave Benson's user avatar
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9 votes

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

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 ...
Achim Krause's user avatar
  • 9,139

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