If $M=\langle v_1,...,v_n\rangle $ is an $R$-module, is $\langle v_1\rangle$ always a submodule of $M$?

Question

Note $R$ is a PID.

In my mind, as $\langle v_1\rangle$ is a subgroup of $M$, but maybe it's not closed under scalar multiplication. Could you give me an counterexample, if $\langle v_1\rangle $ is not always a submodule of $M$? Thank you!

Answer 1

Sure, it's a submodule. In fact $\langle v_1 \rangle$ is closed under addition and scalar multiplication essentially by definition: if $x,y \in \langle v_1 \rangle$ and $r \in R,$ then $x = sv_1$ and $y=tv_1$ for some $s,t \in R,$ and so the linear combination $x + ry = (s+rt)v_1 \in \langle v_1 \rangle$ as required.