Gelfand-Kirillov dimension of a finite extension. In my commutative algebra course we learned about the Gelfand-Krillov dimension. This is the definition:
$\newcommand{\GK}{\mathsf{GKdim}}$
Let $A$ be a $K$-algebra of finite type and $V$ a $K$-vector subspace of $A$ of finite dimension such that $A=K[V]$. The Gelfand--Kirillov dimension of $A$ is defined as:
$$\GK(A)=\lim_{n\to\infty}\frac{\log_2(\text{dim}_K(K+V+\dots+V^n))}{\log_2(n)}$$
We also proved that this dimension coincides with the Krull dimension (because we are in the commutative case) and, as a corollary to this, we have the following trivial result:
Let $B\left/A\right.$ be a finite extension of finitely generated $K$-algebras. Then $\GK(A)=\GK(B)$.
That is indeed a trivial result using the fact that Gelfand--Kirillov and Krull dimension are equal. My question is whether there is a relatively easy proof of this corollary just using the definition. My guess is that this is possible, but since this is quite a difficult definition I don't know how to start.
This is what my intuition tell me we should do:
Of course $A$ is f.g. so we can find $V$ such that $A=K[V]$ and $B\left/A\right.$ is finite, so we should be able to find $W$ such that $B=K[V+W]$. From this, if we can prove that there is some $N(n)$ such that for every $n$:
$$K+V+\dots+V^n\subset K+(V+W)+\dots+(V+W)^n\subset K+V+\dots+V^{N(n)}$$
Then we are done.
Maybe none of this steps are correct and this has no sense, but this is the only idea I have.
Edit: I would also like if you could recommend me a good book covering the Gelfand--Kirillov dimension.
 A: Lets first reduce to a simpler case, since $B/A$ is finite, we can break it up into iterated extensions, each dominated by extensions of the type $A'\subset A'[x]/\langle f(x)\rangle$, for $f(x)$ monic of finite degree with coefficients in $A'$:
$$A\subset A[x_1] \subset A[x_1,x_2] \subset ... A[x_1,..x_n]=B$$
So it suffices to prove the claim in this case, since the dimension can only get smaller when we map onto these subrings. Now let this (monic) polynomial be $f(x)=\sum_{i=0}^k a_i x^i$ with $a_i\in A$. If $V$ is our subspace generating $A$, consider the generating subspace of $A$:
$$V'=V+\text{span}_K(a_i)_{i=0}^{k}$$
And the following subspaces of $B$:
$$S=\text{span}_K(x^i)_{i=0}^{k-1}$$
$$W=V'+S$$
Then note that we have $S^2\subset V'S$, since products of $x^i$ have coefficients in $V'$, so we have $$W^2=(V')^2+V'S$$ In general, we have $$W^n=(V'+S)^n=\sum_{i=0}^n (V')^{n-i}S^i= (V')^n+(V')^{n-1}S$$
By repeated applications of $S^2\subset V'S$.
But now our subspace $S$ had finite dimension $k$, and $1$ was in $V'$, so $(V')^{n-1}\subset (V')^{n}$, so we have the bound of: $$\dim W^n\leq (k+1)(\dim V')^n$$ Then, since the $GK$ dimension can be computed with respect to any generating subspace, we get our desired bound, since the $k+1$ vanishes in this limit.
If one wanted to avoid that corollary of the main theorem, that $GK$ dimension is independent of the generating subspace $V$ chosen, a similar method to the above can be used to show this directly, add elements to your subspace one at a time.
