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I teach a course of undergraduate algebraic geometry. I noticed that students have difficulty grasping the proof of Noether normalizsation Lemma (given in Reid's undergraduate algebraic geometry). Also, the proof given in Mumford red book is also non-elementary. Does anyone know of some easier proof of the result? Thank you

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The proof given in Mumford seems to be pretty easy and natural. –  Anton Fonarev Mar 27 '12 at 9:07
Is it easier than Reid's? –  user16974 Mar 27 '12 at 9:34
Why should any proof of Noether Normalization be non-elementary? You just make an induction and manipulate polynomials, look at their degrees etc. ... no deep and complicated ideas involved. Perhaps a better question would be to give a geometric motivation for the proof. –  Martin Brandenburg Mar 27 '12 at 10:30
@Ali the difference between the two proofs is just in the class of transformations used to obtain a monic polynomial. I personally think that the one from the Red Book is a bit more natural. –  Anton Fonarev Mar 27 '12 at 14:31
Dear Ali, this answer mathoverflow.net/questions/42275/… provides a geometric proof of Noether normalization. I don't know if it is easier in some technical sense, but it reflects how I think about it, and remember it. Regards, –  Emerton Mar 27 '12 at 19:01
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Here is the proof from my class notes last time I taught the course. I don't remember where I borrowed it from. Hope it helps. It is tedious but elementary in the sense of Martin's comment. It is the special case where the base field is infinite. It might become clearer if you give the various steps as exercises to your class.

Noether normalization lemma: Given any infinite field k, and any finitely generated k - algebra B, then we can decompose the extension k in B, into k in A in B where A is isomorphic to a polynomial algebra over k, and B is a module - finite extension of A. Proof: Proof by induction on the number of k algebra generators of B over k. First do the case of one generator. If B = k[c] and c is transcendental over k we are done. If c is algebraic over k then B ≈ k[T]/(f(T)) where f is the monic irreducible polynomial of least degree d satisfied by c over k. Then B = k[c] is generated over k by the elements 1, c,...,cd-1, and in particular B is a finite vector space over k. So take A = k, and we are done.

Now assume B = k[c1,...,cn] where n ≥ 2.  Consider the canonical surjective k algebra map k[T1,...,Tn]-->B taking Ti-->ci.  If this map is also injective we are done, since then we may take A = B.
If this map is not injective then some non constant polynomial f maps to zero.  If f, considered as a polynomial in Tn, is monic of degree  d ≥ 1, then k[c1,...,cn] is generated as a module over k[c1,...,cn-1] by the finite set of monomials of degree < d in cn.  In this case we are done, since by the inductive hypothesis k[c1,...,cn-1] is module finite over some subalgebra isomorphic to a polynomial algebra over k.  Then by transitivity of finiteness,  B = k[c1,...,cn] is also finite over that same polynomial subalgebra of B.

We claim we can “change coordinates” linearly in k[T1,...,Tn], until the polynomial f becomes monic in Tn.  That will mean there is another k algebra surjection k[T1,...,Tn]-->B = k[e1,...,en], taking Ti-->ei such that some element of the kernel is monic in Tn.  Then we are done by the same inductive argument.  So all we have to do is the change of variables argument, sometimes known as the Weierstrass preparation lemma.
Let f(T1,...,Tn) be any non constant polynomial, and consider the linear change of coordinates Ti --> (Ti+aiTn), 1≤i≤n-1, and Tn-->Tn, where the a’s are elements of k.  Write f as fd + lower degree terms, where d = degree f, and fd is homogeneous of degree d.  
Under this transformation, f --> f(T1+a1Tn,...,Tn-1+an-1Tn, Tn).  We claim it is possible to choose the coefficients ai so that this new polynomial has non zero coefficient of Tnd.  Just expand, and get f(T1+a1Tn,...,Tn-1+an-1Tn, Tn) = 

fd(T1+a1Tn,...,Tn-1+an-1Tn, Tn) + lower degree terms.

Then expand the top degree term and get
fd(T1+a1Tn,...,Tn-1+an-1Tn, Tn) = fd(a1,...,an-1, 1)Tnd + lower degree terms in Tn. Adding gives f(T1+a1Tn,...,Tn-1+an-1Tn, Tn) = fd(a1,...,an-1, 1)Tnd + lower degree terms in Tn. Thus we only have to choose the a’s so that fd(a1,...,an-1, 1) ≠ 0. Since fd is a non zero homogeneous polynomial of degree d ≥ 1, fd(a1,...,an-1, 1) is a non zero polynomial of degree ≤ d in the a’s. By induction on d, since our field k is infinite, it does not vanish for all choices of the a’s. QED.

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@roy smith: Thank you for your nice answer! –  user16974 Mar 28 '12 at 14:22
You are welcome. As I guess is obvious, this proof shows that all you need to notice for the induction is that, 1) finiteness is transitive, 2) an integral element over R generates a finite R module, 3) by changing variables, we can achieve integrality. Pardon me, but I always prefer three line arguments to 50 line ones. –  roy smith Mar 28 '12 at 17:37
Even more useful would be to relate this to the geometric argument linked above. –  roy smith Mar 28 '12 at 18:24
By the way, if I remember correctly, this is the proof written in Ried's book. –  Anton Fonarev Mar 28 '12 at 21:58
Thanks Anton. This is in his book on Undergraduate algebraic geometry, rather than his book on commutative algebra, which is why I missed it at first search. –  roy smith Mar 29 '12 at 5:28
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