Geometric interpretation of Noether normalisation and finding a transcendence basis 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 module-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
 A: 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 $x-z=0$):

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