Is a vector of independent Brownian motions a multivariate Brownian motion?

Given a filtered probability space $(\Omega, \mathcal{F}, \mathcal{F}_{t\geq 0}, P)$:

If $B_1, B_2, \dots, B_m $ are all real $\mathcal{F}_t$ Brownian motions, jointly independent. Is the resulting vector $B = (B_1, B_2, \dots, B_m )$ an $\mathcal{F}_t$ $m$-dimensional Brownian motion or just a natural one?

To me it seems that I can't conclude that $B_t - B_s$ is independent of $\mathcal{F}_s$ even though all components are.

In fact $P(\{B_{1,t}-B_{1,s} \in \Lambda_1\}, \dots,\{B_{m,t}-B_{m,s} \in \Lambda_m\}, A)$ for $\Lambda_i$ a borel set, and $A \in \mathcal{F}_s$ can't be factorized, because all events are only pairwise independent with $A$.

Am I missing something or am I correct?

To summarise, if $X$ and $Y$ are independent, and also pairwise independent of a sigma-algebra $\mathcal{D}$, I need that $\sigma(X,Y)$ independent of $\mathcal{D}$ to factorize the probabilities. Or in other words a random vector whose components are independent of a sigma algebra is not necessarily independent itself.

Take $m=2$. Let $\mathcal F_t$ be the natural filtration and consider a variable $X$ such that $$F_{X,B_{1,1},B_{2,1}}(x,y,z) = 2F(x)F(y)F(z) \text{ if } xyz>0, 0 \text{ otherwise}$$ where $F$ stands for the standard normal cdf, and this definition is extended by "independence" (i.e. you build the rest of your Brownian motions by drawing mutually independent normal variables).

Consider the filtration $\mathcal G_t = \sigma(\mathcal F_t \cup \sigma_X)$.

Then you can show that $\mathcal G_t$ is a filtration for $B_1$ and $B_2$, essentially because $X$ is independent with $B_{1,1}$ and with $B_{2,1}$, however it is not a filtration for $(B_1, B_2)$, because the sign of $X$ determines the sign of the product $B_{1,1}.B_{2,1}$, so $(B_{1,1}, B_{2,1})$ is not independent of $\mathcal G_0$.