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Let $k$ be an algebraically closed field and $A\neq 0$ be a finitely generated $k$-algebra; we know that $A\cong k[x_1,\dots,x_n]/\mathfrak a$, for a proper ideal $\mathfrak a\subset k[x_1,\dots,x_n]/\mathfrak a$. Call $\bar x_1,\dots,\bar x_n$ the classes of equivalence in $A$ of the variables $x_1,\dots,x_n$: they are a set of generators for $A$, so the Noether normalization lemma guarantees that exist algebraically indepent elements $\bar y_1,\dots,\bar y_r\in A$ such that $A$ is integral over $B:=k[\bar y_1,\dots,\bar y_r]$. Now choose $y_1,\dots, y_r\in k[x_1,\dots,x_n]$ whose classes of equivalence in $A$ are respectively $\bar y_1,\dots,\bar y_r$, and denote with $\mathfrak b$ the contraction $B\cap \mathfrak a$. We have a commutative diagram:

$$\require{AMScd} \begin{CD} k[y_1,\dots,y_r] @>i >> k[x_1,\dots,x_n] \\ @VqVV @VpVV \\ k[y_1,\dots,y_r]/\mathfrak b @>j>>k[x_1,\dots,x_n]/\mathfrak a \end{CD} $$ where the horizontal homomorphisms are injective and the vertical ones are surjective. Since the $\bar y_1,\dots,\bar y_r$ are algebraically independent, also the $y_1,\dots,y_r$ are, so $B\cong k[y_1,\dots,y_r]$. This gives rise to another commutative diagram: $$\require{AMScd} \begin{CD} k^r@<i^*<<k^n \\ @AIdAA @A\iota AA \\ k^r@<j^*<<\operatorname{Spec}A \end{CD}$$ where $Id$ is the identity, $\iota$ is an injection and $j^*$ is surjective, being $A$ integral over $B$. This means that for an affine algebraic variety $X\subseteq k^n$ with coordinate ring $A$, there is a continuous (respect to Zariski) surjection $X\to k^r$, that extends to a continuous function (obviously surjective) $k^n\to k^r$.

My problem is that I should prove not that those functions are continuous, but that $i^*$ (and so it's restriction $j^*$) is $k$-linear. I probably must use that each $\bar y_h$, $h\in \{1,\dots ,r\}$, can be chosen as a linear combination of the $\bar x_1,\dots,\bar x_n$; however I don't understand how. The hint in the text of the exercise says that I should use the fact that if $R\subseteq S$ is an integral extension, any homomorphism $A\to k$ extends to a homomorphism $B\to k$, for $k$ an algebraically closed field, but again, I don't see how to use it.

Dr. Scotti
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    It is not the case that $i^$ must be linear: consider $A=\Bbb C[x_1,x_2]/(x_1x_2-1)$, $r=1$, $i(y_1)=x_0^2+x_1^2$. I suspect you probably meant to ask for a proof that you can choose things so $i^$ is linear, in which case this is a duplicate of Find a finite morphism $F:X\to\mathbb{P}^k$: the answer there shows how to construct a noether normalization as a linear projection when the base field is infinite. – KReiser Jan 04 '22 at 22:22
  • I don't think I meant to ask that; I was trying to do the final part of the exercise 5.16 of Atiyah-Macdonald: if $k$ is algebraically closed and $X$ is an affine algebraic variety in $k^n$ with coordinate ring $A\neq 0$, then there exists a linear subspace $L$ of dimension $r$ in $k^n$ and a linear mapping of $k^n$ onto $L$ which maps $X$ onto $L$. – Dr. Scotti Jan 04 '22 at 22:59
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    That is equivalent to what I said and what the procedure in the linked duplicate shows. – KReiser Jan 04 '22 at 23:17

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