Let $R$ be a commutative ring, and consider $R$ as an $R$-module with the action given by the product of $R$. Prove that if $B\subset R$ is linearly independent, then $\operatorname{card}(B)=1.$
I would prove this by absurd: Suppose the cardinal of $B$ is 2, with $B=\{b_1,b_2\}$, then this set is dependent: Taking $b_1\cdot b_2+b_2\cdot(-b_1)=0$; in general if $B=\{b_1,\dots,b_n\}$ with $n\geq 2$ then the set is dependent: $b_1\cdot b_2\cdot \dots\cdot b_{n-1}\cdot b_n+b_n\cdot(-b_1\cdot\dots\cdot b_{n-1})=0$ then with $n$ elements is dependent, applying the same procedure to $n-1,n-2,\dots$, we get that any linearly independent set can't have cardinal bigger than 1.
Now something that wasn't explicit in the question $(a)$ was that $B$ must be non-empty, because the empty set would be linearly independent as well, right?. If $B$ non-empty, then its cardinal is non zero, and since it can't be bigger than $1$ follows that must be exactly $1$.
Is this what I'm supposed to do to prove it?