Here is a problem I'm working on:
Show that the set of rational functions $P = \{\frac{1}{x-a} : a \in \mathbb{Q}\}$ is a $\mathbb{Q}$-linearly independent set in the field of rational functions $\mathbb{Q}(x)$.
First, the wording is throwing me off. Does the above amount to showing that $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$ $\Rightarrow$ $a_1 = a_2 = ... = a_n = 0$ for $v_1,v_2,...,v_n \in P$ and $a_1,a_2,...,a_n \in \mathbb{Q}$ ?
In other words, we need to show that **$a_1\frac{1}{x-q_1} + a_2\frac{1}{x-q_2} + ... + a_n\frac{1}{x-q_n} = 0$ $\Rightarrow$ $a_1 = a_2 = ... = a_n$ for some $a_1, a_2, ... , a_n \in \mathbb{Q}$ and rational numbers $q_1, q_2, ... , q_n$, right?
If that's the case, I suppose I haven't been able to see the trick yet as to how equation ** forces the scalars $a_i$ to be all zero. How can I show this?
Thanks!
You seem to have understood what the question is asking you to do.
As for how to solve it, consider the following: Start with a linear combination $$ a_1\frac1{x-q_1} + a_2\frac1{x-q_2} + \cdots a_n\frac1{x-q_n} = 0 $$ for rational $a_i$ and rational $q_i$, with all the $q_i$ are distinct (if any of them are equal, we can just put them together in a single term, adding together their respective $a_i$'s). From here, we can multiply on the left and on the right by $(x-q_1)(x-q_2)\cdots(x-q_n)$. This gives us $$ a_1(x-q_2)(x-q_3)\cdots(x-q_n) + a_2(x-q_1)(x-q_3)\cdots(x-q_n)+\\ \cdots + a_n(x-q_1)(x-q_2)\cdots(x-q_{n-1}) = 0 $$ Now insert $x = q_1$ in the above expression. What do you get (remember our assumption that all the $q_i$ are distinct)? What about if you insert $x = q_2$? Keep going, inserting all the different $q_i$ for $x$, and this problem is solved.