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
Famous quotes containing the word proof:
“War is a beastly business, it is true, but one proof we are human is our ability to learn, even from it, how better to exist.”
—M.F.K. Fisher (1908–1992)
“The insatiable thirst for everything which lies beyond, and which life reveals, is the most living proof of our immortality.”
—Charles Baudelaire (1821–1867)
“Right and proof are two crutches for everything bent and crooked that limps along.”
—Franz Grillparzer (1791–1872)