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?

  • 1
    $\begingroup$ Yes, your second paragraph is the point. There is not much sense in talking about the emptyset as a linearly independent set of elements, although you could if you wanted to. $\endgroup$
    – rschwieb
    Nov 18, 2014 at 3:55
  • $\begingroup$ @rschwieb Thanks, I fixed the typo. Then the proof is ok? $\endgroup$
    – Cure
    Nov 18, 2014 at 4:07
  • 1
    $\begingroup$ Yes, as I said, that's the idea. It's been proven. $\endgroup$
    – rschwieb
    Nov 18, 2014 at 4:17
  • 3
    $\begingroup$ Indeed, $\operatorname{card}(B) \leq 1$ would be the correct statement. Be careful with your proof: I don't think it works well for $n \geq 2$. Use the linear dependency $b_1 \cdot b_2 + b_2 \cdot \left(-b_1\right) + b_3 \cdot 0 + b_4 \cdot 0 + \cdots + b_n \cdot 0 = 0$ instead (at least one of $b_2$ and $-b_1$ is nonzero if $B$ is really a set; otherwise it's obvious anyway). $\endgroup$ Nov 18, 2014 at 5:11
  • 3
    $\begingroup$ Subsets of linearly independent subsets are linearly independent. Therefore it suffices to consider the case $n=2$. $\endgroup$ Nov 18, 2014 at 6:59


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