Proof
The following proof is due to Nagata and is taken from Mumford's red book. A proof in the geometric flavor is also given in the page 127 of the red book and this mathoverflow thread.
The ring A in the lede is generated as k-algebra by elements, say, such that are algebraically independent over k and the rest are algebraic over . We shall induct on m. If, then the assertion is trivial. Assume now . It is enough to show that there is a subring S of A that is generated by elements and is such that A is finite over S, for, by inductive hypothesis, we can find algebraically independent elements of S such that S is finite over . Since, there is a nonzero polynomial f in m variables over k such that
- .
Given an integer r which is determined later, set
Then the preceding reads:
- .
Now, the highest term in of looks
Thus, if r is larger than any exponent appearing in f, then the highest term of in also has the form as above. In other words, is integral over . Since are also integral over that ring, A is integral over S. It follows A is finite over S.
Read more about this topic: Noether Normalization Lemma
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