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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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