Morphism $\Gamma :\mathbb{A}^n\rightarrow \mathbb{A}^d$ coresp. to $\gamma: k[y_1,\ldots,y_d]\rightarrow k[x_1,\ldots,x_n]$ is a projection Let $X\subset \mathbb{A}^n$ be an affine variety with $A(X)=k[x_1,\ldots,x_n]/I(X)$. Write $A=k[a_1,\ldots,a_n]$ for $A(X)$ (where $a_i$ is the image of $x_i$). Let $B=k[y_1,\ldots,y_d]$ be a Noether normalization of $A$.
I've previously shown that for $i=1,\ldots,d$ there exists linear polynomials $l_i=l_i(x_1,\ldots,x_n)$ and $k$-algebra homomorphism $\gamma: B\rightarrow k[x_1,\ldots,x_n]$ mapping $y_i\mapsto l_i$, st. $B\subset A$ factors as $\gamma$ composed with a surjection $k[x_1,\ldots,x_n]\rightarrow A$.
Let $\Gamma :\mathbb{A}^n\rightarrow \mathbb{A}^d$ be the morphism corresponding to $\gamma$. I want to show, that after a suitable linear change of coordinates on $\mathbb{A}^n$, then we may see $\Gamma$ as the projection onto the first $d$ coordinates.
From the correspondance, the I know that $\gamma(f)=f(\Gamma)$ for $f\in B$. So in fact $l_i=\gamma(y_i)=y_i(\Gamma)$. I've tried played around with this, but it doens't get me much further. I'm kinda stuck now, and don't see how I should make a linear change of the coordinates on $\mathbb{A}^n$ (as far as I understand, then I want to make variable change to the $x_i$'s st. that $\Gamma$ becomes the projection).
 A: Your question boils down to the following statement: given an injective map $f:k[y_1,\cdots,y_d]\to k[x_1,\cdots,x_n]$ so that $f(y_i)=c_i + \sum \lambda_{ij}x_j$, find an automorphism of $k[x_1,\cdots,x_n]$ so that after applying the automorphism, the map can be written as $f(y_i)=x_i$.
This turns out to just be a linear algebra problem. As $f$ is injective, the elements $\sum \lambda_{ij} x_j$ are linearly independent and thus span a subspace of dimension $d$ inside the vector space $k[x_i]_1$ of homogeneous polynomials of degree one. By completing this list to a basis and using the fact that for any two ordered bases of a vector space we can find an invertible linear transformation sending one to the other, we can find an automorphism of $k[x_i]_1$ which sends each $ \sum \lambda_{ij} x_j$ to $x_i$. This extends to an automorphism of $k[x_1,\cdots,x_n]$ (the inverse is given by the inverse matrix) so that after applying this automorphism, $f(y_i) = x_i+c_i$. Now the automorphism $x_i\mapsto x_i-c_i$ gives the desired result: we've rewritten $f$ as the map which takes each $y_i$ exactly to $x_i$, and this is the projection map on to the first $d$ coordinates.
