Question

Given a field $k$ and a finitely generated $k$-algebra $R$ without zero divisors, one knows that there exist $x_1, \ldots, x_n$ algebraically independent such that $R$ is integral over $k[x_1, \ldots, x_n]$. How can one choose actually the $x_i$'s ?

More precisely, if $a\in R$ is transcendent over $k$, can one find $x_2, \ldots x_n$ such that $R$ is integral over $k[a,x_2 \ldots, x_n]$ ? If it is false, can one get the conclusion under stronger assumptions ? In this discussion, I am also interested by geometric explanations.

Answer 1

This is an amplification of Mohan's suggestion to consider the context of projective varieties. I will asume that $k$ is infinite.

Given your $k$-algebra $R$, we get Spec $R$ as a variety $V$ in $\mathbb A^d$ for some $d$. We may take the projective closure (i.e. Zariski closure) of $V$ in $\mathbb P^d$ to obtain a projective variety $\overline{V}$.

If $V = \mathbb A^d$ there is nothing to say: $R = k[x_1,\ldots,x_d]$ and witnesses its own Noether normalization. So assume that $V$ is a proper subvariety of $\mathbb A^d$, so that $\overline{V}$ is a proper subvariety of $\mathbb P^d$.

Because $k$ is infinite, we can (and do) choose a point $P$ (defined over $k$) lying at infinity, and not on $\overline{V}$. We can (and do) also choose a hyperplane $H$ (defined over $k$) which does not contain $P$. We may then apply the projection from $P$ to $H$ to obtain a morphism $\pi: \overline{V} \to H$. Properness of projective morphisms implies that this map has closed image in $H$, say $\overline{W}$.

Because $P$ was chosen to lie at infinity, this restricts to a morphism $V \to H \cap \mathbb A^d$, which again has closed image, say $W$. (As the notation suggests, $\overline{W}$ will then be the projective closure of the affine variety $W$.) This map is finite, in the sense that $V \to W$ corresponds to a map $R' \to R$ with $R$ finite over $R'$.

Now if $W = H\cap \mathbb A^d$, we have obtained our Noether normalization of $R$, since $R'$ is a polynomial ring in this case. If not, we proceed inductively, replacing $V$ by $W$ and $\mathbb A^d$ by $H\cap \mathbb A^d$ (an affine space of one dimension less). Eventually we will reach a stage where the projection to the hyperplane is surjective. (If $V$ has dimension $n$ then we have to perform $d -n$ projections altogether.)

So you see that you have a lot of flexibility in how to achieve the normalization (because at each stage there are a lot of choices of $P$ and $H$), but the choices are not completely arbitrary (because of the condition that $P$ not lie on $\overline{V}$, and that $H$ not contain $P$). For example, thinking this way, you will easily see what goes wrong in Mohan's counterexample. (Taking $a = x$ in his example corresponds to projecting from a point $P$ that does lies on the projective closure of the hyperbola.)

Answer 2

In general, if $n=1$, then you clearly can not choose $a$ as for example, if $R=k[x,x^{-1}]$ and $a=x$. So, already the choice of $a$ is not so arbitrary.

I am not sure what you mean by choosing the variables, but the proof of the Normalization does give you a `choice'. For simplicity, if you assume that $k$ is infinite and $R=k[x_1,\ldots, x_m]$ (some set of generators for $R$), if these are actually algebraically independent, you are done. Otherwise they satisfy an equation $f(x_1,\ldots, x_m)=0$ and then you can change the $x_i$'s linearly, so that $f$ is monic in $x_m$. This of course is slightly non-constructive. Then $R$ is integral over $S=k[x_1,\ldots, x_{m-1}]$ and you can continue.

May be you should look at the proof for projective varieties, where the successive choices are easier to make.