# Does $u|x \sim N(0,\sigma^2)$ implies independence?

I'm a bit confused by the following statement found in "Introductory Econometrics, A Modern Approach" (Wooldridge, 2019):

The phrase seems to imply the assumption is about independence and Normality. But as far as I can tell, $$u \sim N(0,\sigma^2)$$ already implies complete independence between $$u$$ and the explanatory variables. Maybe he is just being explicit about the implications of Normality without going onto the details of what implies what, details that might confuse a novice reader (this might be supported by the title of the assumption, which is "Normality" rather than "Independence and Normality").

Still, my point is: $$u \sim N(0,\sigma^2)$$ implies independence not only for the first two moments of $$u$$ but for the whole distribution. Is this the case? Am I misunderstanding Wooldridge's point?

Stating the distribution of $$u$$ does not imply independence.
Consider $$v = 2u$$. Thus, $$v \sim \mathcal N (0, 4\sigma^2)$$, yet $$v$$ is not independent of $$u$$.
$$u\sim\mathcal N(0,\sigma^2)$$ does not imply independence, as @Firebug stated. But I would add that $$u|x_1,x_2,\cdots,x_k\sim\mathcal N(0,\sigma^2)$$ does imply independence, since it means that the conditional distribution of $$u$$ is always the same, regardless of the values of the covariates.