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Jun 14 at 0:18 comment added KConrad @Rellw well sure, when there is only one ring $R$ it's reasonable to assume that. But I can't promise you will never see a counterexample. Just pay attention to what is happening.
Jun 13 at 23:05 comment added Rellw @KConrad Thanks for your comment. If there is no special interpretation on context, where I can just find R-modules M and N, should I assume the tensor without subscript means tensor for R-modules?
Jun 13 at 23:01 comment added Rellw @Denis T Thanks very much for pointing my mistake. I will amend it later.
Jun 1 at 6:01 comment added KConrad "I see many authors will just write tensor without subscript, which I think means tensor over $\mathbb Z$." There is no such convention. When a ring is not indicated in the tensor product, you're just expected to know by context which ring is meant. It seems the whole motivation behind your post is due to this misunderstanding.
May 30 at 20:04 comment added Denis T Also it's easy to show that the essential image of restriction of scalars along an epimorphism $R \to S$ is closed under taking subobjects in the ambient category of R-modules. (Actually, those images are characterised as subcategories which are closed under subobjects, quotients, sums and products. This is Gabriel-de la Peña theorem.) $\Bbb Z \subset \Bbb Z_p$ clearly does not admit a structure of a $\Bbb Z_p$-module.
May 30 at 19:48 comment added Denis T Inclusion of integers into p-adics is NOT an epimorphism. Cardinality of the target of a ring epimorphism never excess the cardinality of its domain. A ring morphism $R \to S$ is epi iff the multiplication map of $R$-bimodules $S \otimes_R S \to S$ is an isomorphism; or, equivalently, when restriction of scalars is a fully faithful embedding.
May 30 at 17:52 answer added Dave Benson timeline score: 4
May 30 at 17:43 answer added Achim Krause timeline score: 9
May 30 at 17:25 history asked Rellw CC BY-SA 4.0