![abstract algebra - $a$ is integral over $I$ iff $R[a]$ is finitely generated $R-$module and $a\in\sqrt{IR[a]}$ - Mathematics Stack Exchange](https://i.stack.imgur.com/wPOrk.png "abstract algebra - $a$ is integral over $I$ iff $R[a]$ is finitely generated $R-$module and $a\in\sqrt{IR[a]}$ - Mathematics Stack Exchange")
abstract algebra - $a$ is integral over $I$ iff $R[a]$ is finitely generated $R-$module and $a\in\sqrt{IR[a]}$ - Mathematics Stack Exchange
Sam Walters ☕️ on Twitter: "One of my favorite #Algebra theorems. The classification theorem for finitely generated Abelian groups. It has a nice extension to modules over principal ideal domains (PIDs) in
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algebraic geometry - Invariant ring is a finitely generated $K$-algebra, confusion regarding the proof - Mathematics Stack Exchange
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abstract algebra - if R is a commutative ring in which all the prime ideals are finitely generated then R is Noetherian - Mathematics Stack Exchange
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![PDF] When is the Intersection of Two Finitely Generated Subalgebras of a Polynomial Ring Also Finitely Generated? | Semantic Scholar](https://d3i71xaburhd42.cloudfront.net/e846bf34979c105102ec9d9c89e1bf048d1ddd53/7-Figure2-1.png "PDF] When is the Intersection of Two Finitely Generated Subalgebras of a Polynomial Ring Also Finitely Generated? | Semantic Scholar")
PDF] When is the Intersection of Two Finitely Generated Subalgebras of a Polynomial Ring Also Finitely Generated? | Semantic Scholar
abstract algebra - if R is a commutative ring in which all the prime ideals are finitely generated then R is Noetherian - Mathematics Stack Exchange
Solved Some details in solution would be helpful. Thank you | Chegg.com
On Semiprime Finitely Generated Right Alternative Weakly M-Rings
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