I'm reading these notes, trying to understand why Regev encryption is circularly secure. They say (at the very end) that the following pair of values is indistinguishable from uniform:
$$\left(a' - \lfloor{\frac{q}{2}} \rfloor u_i, <a', s> + e\right). \text{ (1)}$$
where $q$ is an integer, $a'$ and $s$ are uniformly selected vectors, and $u_i$ is the $i^{th}$ standard basis vector. $e$ is a small error term selected from a special distribution so that the pair $(a', <a', s> + e)$ looks like it was selected from the uniform distribution - this is the learning with errors (LWE) assumption.
I can see why (1) would look uniform, however the original value they wanted to show was uniformly random was:
$$\left(a, <a, s> + e + s_i \lfloor{\frac{q}{2}} \rfloor \right) . \text{ (2)}$$
Where $s_i$ is an element of $s$, the secret key, and $a$ is another uniformly selected vector. In the notes they chose a' so that (2) = (1) and the claim follows. My question is, why was this necessary at all? Isn't $<a, s> + e$ independent from $s_i$ because of a? (Unless $s = 0$). Can't one just argue directly that both sides of (2) look uniformly random and independent?
Many thanks for any help.