Tuesday, 27 January 2015

ac.commutative algebra - Converse to Hilbert basis theorem?

Dear Zev,



There are some sources which give the converse. See e.g. pp. 64-65 of



http://math.uga.edu/~pete/integral.pdf



For that matter, see also pp. 32-33 of loc. cit. for the Chinese Remainder Theorem and its converse. (And I am not the only one to do this...)



Note that in both cases the converse is left as an exercise. I think (evidently) that this is the right way to go: it may not be so easy for the journeyman mathematician to come up with the statement of the converse, but having seen the statement it is a very valuable exercise to come up with the proof. (In particular, I believe that in a math text or course at the advanced undergraduate level and beyond, most exercises should indeed be things that one could find useful later on, and not just things which are challenging to prove but deservedly forgettable.)



Finally, by coincidence, just yesterday in my graduate course on local fields I got to the proof of the "tensor product theorem" on the classification of norms in a finite-dimensional field extension (which came up in a previous MO answer). The key idea of the proof -- which I found somewhat challenging to write; I certainly admit to the possibility of improvements in the exposition -- seems to be suspiciously close to the valuation-theoretic analogue of the converse of the Chinese Remainder Theorem! See pp. 18-19 of



http://math.uga.edu/~pete/8410Chapter2.pdf



if you're interested.

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