Refinement of "every finite dimensional normed vector space is isometrically isomorphic to $\mathbb{R}^n$". My textbook states that: "every finite-dimensional normed vector space is isometrically isomorphic to $\mathbb{R}^n$".
Is this statement strictly true? The set $\mathbb{Q}^n$ has dimension $n$, and can be turned into a normed vector space over the field $\mathbb{Q}$. Yet, it is obviously not isomorphic to $\mathbb{R}^n$. Does this theorem need to specify that the vector spaces need to be over a common field? In this case $\mathbb{R}$? 
Moreover, is the isometry part true? It seems to me that you could change norms and break the existence of an isometry.
 A: It would seem that for this statement the author assumes the vector spaces to be over $\Bbb R$ (and not even, for instance, over $\Bbb C$). It is certainly possible to make a normed-space-like notion for vector spaces over some field $k\ne \Bbb R,\Bbb C$, but it is uncommon at best. For instance, I would assume that, for an ordered field $(\Bbb F,\le)$, a field $k$ and a multiplicative, subadditive and non-negative map $\lvert \bullet\rvert:k\to \Bbb F$ such that $\lvert x\rvert=0$ if and only if $x=0$, it should be possible to define a concept of "norm" on $k$-vector spaces, and consequently a topology. Such topology should not be metrizable in general.
It should be pointed out that the statement itself is quite imprecise. For one thing, in order to claim a vector space to be linearly isometric to $\Bbb R^n$, one should specify a norm on $\Bbb R^n$. However, once a norm has been specified, the statement becomes false, because, for all $n\ge2$, it is quite easy to devise non-isometric norms on $\Bbb R^n$: for instance $\lVert x\rVert_\infty=\sup\{\lvert x_i\rvert\,:\, 1\le i\le n\}$ and $\lVert x\rVert_2=\sqrt{\sum_{k=1}^n \lvert x_k\rvert^2}$ are not linearly isometric for any $n\ge2$, because the closed $2$-balls are strictly convex and the closed $\infty$-balls aren't.
What is true is that every normed $\Bbb R$-vector space $V$ of finite dimension is linearly homeomorphic to $\Bbb R^{\dim V}$ (as the topology of $(\Bbb R^n,\lVert\bullet\rVert)$ is independent of the norm). Equivalently, any two normed $\Bbb R$-vector spaces of the same finite dimension are isomorphic by a bi-Lipschitz linear isomorphism.
