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sniffnoyBecause the rational root theorem is essentially that same sort of divisibility argument fundamentally, no? And how do you show stuff are algebraic integers in the first place in the usual proof? Schur's Lemma!
...except that in fact looking at it again I see the usual proof does use the fact that algebraic integers form a ring, which really is introducing some machinery. So I guess John & Gene's proof is more elementary after all.
-Harry
...except that in fact looking at it again I see the usual proof does use the fact that algebraic integers form a ring, which really is introducing some machinery. So I guess John & Gene's proof is more elementary after all.
-Harry
