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I'm stuck trying to prove that for two $R$-modules $M,N$ ($R$ commutative with a 1), then $$Ann(M+N)=Ann(M) \cap Ann (N)$$

I was trying to do double inclusion, and I can prove the RHS is contained in the LHS, but im stuck in the other direction since if I take $x \in Ann(M+N)$, then for any $m,n \in M,N$ (resp.) $x(m+n)=0$, but I dont see why this means $xm=0=xn$.

So I was wondering if I could get some hints on how to prove this

Thank you

Chris Birkbeck
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1 Answers1

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Hint without words:

$$x\in\text{Ann}(M+N)\implies x(m+n)=0\;,\;\;\forall\,m\in M\,,\,n\in N\implies$$

$$\implies x(m+0)=x(0+n)=0\;,\;\;\forall\,m\in M\,,\,n\in N\;\;\ldots$$

DonAntonio
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