Annihilator of a direct sum and product of modules

Question

Let be $R$ a ring and $\{M_i\}_{i \in I} $ a family of $R$-modules. I am trying to proof that $\text{Ann}_R(\prod_{i \in I} M_i)
= \text{Ann}_R(\oplus_{i \in I}M_i) $

Applying definitions :

$\prod_{i \in I} M_i = \{(x_i)_{i\in I}\hspace{2mm}|\hspace{2mm}x_i \in M_i \hspace{2mm} \forall i \in I\}$

$\text{Ann}_R(M)=\{r \in R \hspace{2mm}|\hspace{2mm} rx=0 \ \forall x \in M\}$

so:

$\text{Ann}_R(\prod_{i \in I} M_i)=\{r \in R \hspace{2mm}|\hspace{2mm} r x_i=0 \forall x_i \in M_i,i\in I\}$

and that is : $\cap_{i \in I}\text{Ann}_R(M_i)$

I arrive to the exact same set when applying the definition for direct sum. It almost seems trivial, so I want to know if there is any way to do a more "formal " proof.

Answer 1

Your approach is fine. In general though, if you want to show that two sets $A$ and $B$ are equal, I would suggest showing that $x \in A \implies x \in B$ and then that $x \in B \implies x \in A$. This leads to cleaner arguments, and reduces the chance of error. I'll show you what I mean in this case.

Let $r \in \operatorname{Ann}_R(\prod_{i\in I} M_i)$. Since $\bigoplus_{i\in I}M_i\subseteq \prod_{i\in I}M_i$, certainly $r$ kills every element of $\bigoplus_{i\in I}M_i$, so $r \in \operatorname{Ann}_R(\bigoplus_{i \in I}M_i)$.

Now let $r \in \operatorname{Ann}_R(\bigoplus_{i\in I}M_i)$. Let $\alpha \in \prod_{i\in I}M_i$. Then we may write $\alpha$ as $(\alpha_i)_{i\in I}$ where $\alpha_i \in M_i$ is the component of $\alpha$ in $M_i$. For each $j$, there is an element of $\bigoplus_{j\in I}M_j$ whose $M_i$-component is $\alpha_i$, and whose other components are $0$. This must be killed by $r$, so $r\cdot \alpha_i = 0$ for all $i$. Therefore $r\cdot \alpha = 0$. Since $\alpha$ was arbitrary, $r \in \operatorname{Ann}_R(\prod_{i\in I}M_i)$.

The reason I prefer this is that it argues in a linear fashion; each statement follows the previous ones. When - as you did in your proof - you rearrange definitions of sets to make them equal, it requires reading back and forth a few times, which I think makes for a less pleasing argument.