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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.

7 votes

Kernel of evaluation map into field of quotients

First, the answer to Question 2 is YES and it is not hard to show. For $q=\frac{a}{b}$, by assumption $(a,b)=R$, therefore by standard argument $(a^k,b)=R$ for any positive integer $k$. For any $f\in\ …
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