When do we have $m\otimes n = 0$

Question

Let $M$ and $N$ be $R$-modules ($R$ a commutative ring with identity). Let $m \in M$ and $n \in N$. Is there any necessary and sufficient condition to have $m\otimes n = 0$ (as an equation in $M\otimes_RN)$.

For example when $ann(n)+ann(m)=R$. – user52045 Dec 23 '13 at 19:44 — user52045, Dec 23 '13 at 19:44
@user52045 your condition is only sufficient. – user117432 Dec 23 '13 at 19:47 — user117432, Dec 23 '13 at 19:47

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

Yes there is such a criterion (Bourbaki, Algèbre commutative, ch. I, §2, no. 11, lemma 10). See also the duplicate question math.SE/288431.

There is also a "trivial" but still useful criterion: For every bilinear map $\beta : M \times N \to T$ we have $\beta(m,n)=0$.

edited Apr 13 '17 at 12:20

Community

1

answered Dec 23 '13 at 20:05

Martin Brandenburg

163,620

Merry Xmas Martin. May I ask you a small question in number theory? It is not that big I post it in site? I want just a hint or good direction and don't want to make you any troubles. May I ask it here? :-) – Mikasa Dec 26 '13 at 18:27
Elsewhere. Press "Ask a question". – Martin Brandenburg Dec 26 '13 at 19:05

When do we have $m\otimes n = 0$

1 Answers1