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I am trying to find out the $\ker(f\otimes g)$ where $f:M \rightarrow P$ and $g:N \rightarrow Q$ are $A$ linear maps where $A$ is not a field. So $(f\otimes g):M\otimes N\rightarrow P\otimes Q $ is $A$ linear map.The question is: can the following inclusion be proper?

$(\ker f\otimes N)\cup(M\otimes \ker g)\subset \ker(f\otimes g) $.

Is there any condition on the modules like flatness (other than vector space case) which forces the reverse containment ?

I am not getting anything,Help me.Thanks in advance.

Robert Cardona
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Via
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    Why do you expect a module to be the union of two submodules, not their sum? Recall that unions of submodules, not contained in each other, are never submodules. Or did you mean the union in the correct category, namely modules, and not what 99% of other mathematicians understand by union, namely the union of underlying sets? – Martin Brandenburg Nov 24 '14 at 21:22
  • That should not be union ..mistaken – Via Nov 24 '14 at 21:23

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$\ker(f) \otimes N$ doesn't have to be a submodule of $\ker(f \otimes g)$. But there is a canonical homomorphism $\ker(f) \otimes N \to \ker(f \otimes g)$. Similarly, we get a canonical homomorphism on $M \otimes \ker(g)$. Hence, we get a canonical homomorphism $\ker(f) \otimes N \oplus M \otimes \ker(g) \to \ker(f \otimes g)$. This turns out to be surjective when $f,g$ are surjective, because of the right exactness of the tensor product.

For more on this, see What is the kernel of the tensor product of two maps?

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Martin Brandenburg
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