If $M,N$ are D-modules, where $D$ is a PID and $Ann(M)=\langle a\rangle, Ann(N)=\langle b\rangle$, $a\neq 0 \neq b$. Prove that $Ann(M \oplus N)=\langle c\rangle$ where $c=\gcd(a,b)$.
Does the statement makes sense? Every element in $Ann(M \oplus N)$ is of the form $(m,n)$, making it strange to claim that it is generated by $\gcd(a,b)$.
Or should I understand it as Ann set giving the $c$ elements such that $c(m,n)=0$? Because in the latter case $c(m,n)=(cm,cn)$ and since the annihilators of $M,N$ are generated by $a,b$ then $c=a^p, c=b^q$ with $p,q\in \mathbb{Z}$. But I can't conclude $c$ must be $\gcd(a,b)$.
In general, if $M$ and $N$ have annihilators $I$ and $J$ then $M\oplus N$ has annihilator $I\cap J$. To prove that note that elements of $I\cap J$ do annihilate $M\oplus N$, and if $a$ annihilates $M\oplus N$ is must annihilate $M\oplus\{0\}$ and $\{0\}\oplus N$ and so also $M$ and $N$.
In a PID, if $I=\left<a\right>$ and $J=\left<b\right>$ with $ab\ne0$ then $I\cap J=\left<c\right>$ where $c$ is a least common multiple of $a$ and $b$.