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Is there a term for an $A$-module $M$ such that $M \otimes_A -$ takes nonzero modules to nonzero modules?

Motivation: It is a standard theorem that if $B$ is faithfully flat over $A$, then $\hbox{Spec } B \to \hbox{Spec } A$ is surjective. However, looking at the proof, this only really requires the property above--you need to know that every fiber is nonempty, i.e., that the rings $B \otimes_A \Bbbk (x)$ are nonzero for $x \in \hbox{Spec } A$.

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It's equivalent to $M \otimes_A-$ being a faithful functor, i.e. being injective on Hom-sets. –  Graham Leuschke Nov 29 '10 at 22:25
    
I have seen the word 'faithful' used to describe a module; particularly a faithfully flat one. –  Greg Muller Nov 29 '10 at 22:50
    
According to mathworld.wolfram.com/FaithfulModule.html a faithful module is something different. In particular, $\mathbb{Q}$ is faithful over $\mathbb{Z}$. –  Charles Staats Nov 30 '10 at 0:35
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2 Answers 2

For flat modules $M$, the condition you cite is equivalent to $M$ being faithfully flat. But some modules that are not flat also satisfy your condition; for example, let $k$ be a field, let $A = k[[ x_1, \ldots, x_n ]]$ where $n>1$, and let $M$ be the maximal ideal of $A$.

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In fact, for any local ring $(A, \mathfrak{m})$, $\mathfrak{m} \otimes_A X =0$ implies that $\mathrm{Tor}_1^A(A/\mathfrak{m},X)=\mathfrak{m}X=0$. This says both that $X$ is free and a vector space [which is rare], so as long as $\mathfrak{m} \neq 0$, it is faithful in this sense. –  Graham Leuschke Nov 30 '10 at 14:14
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I've heard the term "conservative" used to describe functors which take nonzero modules to nonzero modules; see e.g. problem 4 from this problem set from a course on schemes, which is the converse of the situation you're looking at. I don't know if this terminology is standard, though; some googling reveals that "conservative" has several (possibly unrelated) meanings.

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