# Why does $\operatorname{Supp}(M\oplus N)=\operatorname{Supp}(M)\cup\operatorname{Supp}(N)$?

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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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.
(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.)