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Take the polynomial $f = x_1 x_2 + x_2 x_3 + x_3 x_1 \in k[x_1,x_2,x_3]$, and define the set $V = Z(f) = \{(x_1,x_2,x_3) \ | \ f(x_1,x_2,x_3) = 0 \} \subset \mathbb A ^3$. Consider the coordinate ring of $V$, given by $k[x_1,x_2,x_3]/(f) \cong k[a_1,a_2,a_3]$, where $a_i = x_i \ \mathrm{mod} \ (f)$.

Noether normalisation says there exists $ \{y_1, \ldots, y_m | \ m \leq 3\} \subset k[a_1,a_2,a_3]$ such that the $y_i$ are algebraically independent over $k$ and that $k[a_1,a_2,a_3]$ is a finite $k[y_1, \ldots y_m]$-algebra. If I can find such $y_i$, then I can explicitly construct a morphism $\phi : V \to \mathbb A^m$ that is surjective and has finite fibres (this is the geometric interpretation of Noether normalisation).

My question is: how do I go about finding such $y_i$?

Thanks

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In your example, $m \le 2$ for dimension reasons. I think any sufficiently general choice of (linear) polynomials will provide such $y_i$. The geometric idea, after all, is to find a projection from your variety down to affine space, and it is reasonable to believe that for some (but not necessarily any) choice of directions, the projection has only finite fibres. –  Zhen Lin Mar 5 '12 at 21:36
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@Zhen, it isn't sufficient for the projection to have finite fibers for the corresponding extension to be integral. For instance, the projection of $xy=1$ on the $x$-axis has finite fibers, but $k[x,y]/(xy-1)$ isn't integral over $k[x]$. –  Bruno Joyal Mar 5 '12 at 23:40
    
@Bruno: Oops! I meant non-empty finite fibres. –  Zhen Lin Mar 6 '12 at 0:34
    
@Zhen Even then I don't think that is sufficient... I'm quite certain that it's possible for an extension to have the going-up property, finite fibers and yet not be integral. I can't think of a counter-example now, maybe you have an idea? –  Bruno Joyal Mar 6 '12 at 0:40
    
Apparently, it is true that proper + quasi-finite implies finite, at least when the codomain is locally noetherian. I'm not sure how to justify that a generic projection is proper, but it feels like something which ought to be true... –  Zhen Lin Mar 6 '12 at 7:49

1 Answer 1

up vote 5 down vote accepted

If you study the proof of Noether's normalization lemma, you'll find that it's actually a constructive proof - it provides an algorithm to solve your problem. I'm going to do it for you; see if you can figure out what's going on. (I'm not going to describe the algorithm in detail, but if you pay attention you should be able to understand why I'm doing what I'm doing.)

First, note that we should expect $m=2$; indeed, the variety defined by a single irreducible polynomial has codimension 1. So it will be a "surface" in $k^3$. We should therefore expect the coordinate ring to have transcendence degree $2$ over $k$.

Consider $f$ as a polynomial in $x_3$. The leading coefficient is $x_1+x_2$. Find a point where it doesn't vanish identically, say $x_1=1,x_2=0$. Then put $u_1=x_1+x_3$, $u_2=x_2$. Then we can rewrite $f$ as $$f=x_3(u_1-x_3+u_2)+(u_1-x_3)u_2 = -x_3^2+x_3u_1 + u_1u_2.$$

Magic! This is a monic polynomial in $x_3$, with coefficients in $k[u_1,u_2]$. Thus $R=k[x_1,x_2,x_3]/(f)$ is integral (of degree $2$) over $k[x_1+x_3,x_2]$.

Geometrically, this corresponds to projecting the surface $f=0$ on the plane spanned by the vectors $(1,0,1)$ and $(0,1,0)$ (i.e. the plane $x-z=0$):

enter image description here

As you can see, the projection is surjective (a consequence of the integrality of the extension), and the fibers have two points almost everywhere.

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Hi Bruno, i am studying the proof of Noether normalization theorem in Eisenbud p. 286-289 and i still don't a see a connection with what you are doing. Do you have another proof in mind or a hint on how to connect Eisenbud's proof with your construction? –  Manos Jun 16 at 18:37

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