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For modules $M$ and $N$ over a commutative ring, why does $\operatorname{Supp}(M\oplus N)=\operatorname{Supp}(M)\cup\operatorname{Supp}(N)$?

I tried justifying it with the following, but I'm not fully confident, since I think I only consider $M+N$, not $M\oplus N$.

Denote by $L=M\oplus N$. Suppose $L_\mathfrak{p}=0$ and that $m/s\in M_\mathfrak{p}$, $n/s\in N_\mathfrak{p}$. Since $m/s$, $n/s$ are zero in $L_\mathfrak{p}$, there exists $x\notin\mathfrak{p}$ such that $xm=0$ and $xn=0$. But then $m/s$ is zero in $M_\mathfrak{p}$, and $n/s$ is zero in $N_\mathfrak{p}$. That is, $M_\mathfrak{p},N_\mathfrak{p}=0$.

Conversely, suppose $M_\mathfrak{p},N_\mathfrak{p}=0$. If $(m+n)/s\in L_\mathfrak{p}$, then there exist $x,y\notin\mathfrak{p}$ such that $xm=0$ and $yn=0$, so $(xy)(m+n)=0$, and thus $L_\mathfrak{p}=0$. Hence $L_\mathfrak{p}=0$ if and only if $M_\mathfrak{p},N_\mathfrak{p}=0$. This implies $\operatorname{Supp}(M\oplus N)=\operatorname{Supp}(M)\cup\operatorname{Supp}(N)$.

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1 Answer 1

up vote 4 down vote accepted

There is no computation to do - note that $(M\oplus N)_\mathfrak{p} \cong M_\mathfrak{p} \oplus N_\mathfrak{p}$, and that the direct sum of two modules is zero if and only if both are zero.

That being said, your computations are just fine!

(Also, it doesn't make much sense to write $M+N$ unless $M$ and $N$ are given to you as submodules of some common module.)

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Thanks a lot, Bruno. –  Jakucha Feb 25 '12 at 21:51
    
You are most welcome, @Jakucha! –  Bruno Joyal Feb 25 '12 at 23:38

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