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.


2 Answers 2


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.)

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    $\begingroup$ I think that a nice follow-up reference would be Lemma 13.2 and Theorem 13.3 of Eisenbud's commutative algebra book. That discussion has a very similar flavor as Emerton's answer, though it is phrased more algebraically. It also explains how to choose a Noether normalization in the case where $k$ is a finite field. $\endgroup$ Oct 18, 2010 at 19:14
  • $\begingroup$ @Emerton: Sir, can you please look at my question in this link math.stackexchange.com/questions/1752850/… and suggest some answer? The question is in the same spirit as your answer above, but it is not clear to me if the answer to my question follows from your above argument. Thanks in advance! $\endgroup$ Apr 27, 2016 at 17:41

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.


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