![commutative algebra - Every prime power ideal in a Noetherian Ring of dimension one can be written uniquely as a power of a prime. - Mathematics Stack Exchange commutative algebra - Every prime power ideal in a Noetherian Ring of dimension one can be written uniquely as a power of a prime. - Mathematics Stack Exchange](https://i.stack.imgur.com/XxRGa.png)
commutative algebra - Every prime power ideal in a Noetherian Ring of dimension one can be written uniquely as a power of a prime. - Mathematics Stack Exchange
![تويتر \ Constanza Rojas-Molina على تويتر: "Rings are sets with 2 binary operations *,+. Noetherian Rings, named after Emmy Noether, satisfy the Ascending Chain Condition: if ideals are contained in other ideals تويتر \ Constanza Rojas-Molina على تويتر: "Rings are sets with 2 binary operations *,+. Noetherian Rings, named after Emmy Noether, satisfy the Ascending Chain Condition: if ideals are contained in other ideals](https://pbs.twimg.com/media/EnfZHzDW4AARn9_.jpg:large)
تويتر \ Constanza Rojas-Molina على تويتر: "Rings are sets with 2 binary operations *,+. Noetherian Rings, named after Emmy Noether, satisfy the Ascending Chain Condition: if ideals are contained in other ideals
![Simple Noetherian Rings (Cambridge Tracts in Mathematics, Series Number 69): Cozzens, John, Faith, CArl: 9780521092999: Amazon.com: Books Simple Noetherian Rings (Cambridge Tracts in Mathematics, Series Number 69): Cozzens, John, Faith, CArl: 9780521092999: Amazon.com: Books](https://m.media-amazon.com/images/I/51E7rFTmgVL._AC_UF1000,1000_QL80_.jpg)
Simple Noetherian Rings (Cambridge Tracts in Mathematics, Series Number 69): Cozzens, John, Faith, CArl: 9780521092999: Amazon.com: Books
![SOLVED: Let R be a Noetherian ring; (a) If M, N, and T are R-modules, where N is a submodule of M, show that there is a bijection HomR(M/N,T) f ∈ HomR(M,T) SOLVED: Let R be a Noetherian ring; (a) If M, N, and T are R-modules, where N is a submodule of M, show that there is a bijection HomR(M/N,T) f ∈ HomR(M,T)](https://cdn.numerade.com/ask_images/5545e716c6114247b9c8372267d11ebc.jpg)
SOLVED: Let R be a Noetherian ring; (a) If M, N, and T are R-modules, where N is a submodule of M, show that there is a bijection HomR(M/N,T) f ∈ HomR(M,T)
![Constanza Rojas-Molina on Twitter: "#noethember day 20. A Noetherian ring is a ring with extra properties, in part. one that satisfies the ascending chain condition on left and right ideals. This means Constanza Rojas-Molina on Twitter: "#noethember day 20. A Noetherian ring is a ring with extra properties, in part. one that satisfies the ascending chain condition on left and right ideals. This means](https://pbs.twimg.com/media/DsemGRGX4AEbCxn.jpg:large)
Constanza Rojas-Molina on Twitter: "#noethember day 20. A Noetherian ring is a ring with extra properties, in part. one that satisfies the ascending chain condition on left and right ideals. This means
![abstract algebra - Examples of rings that are Noetherian but not finitely generated? - Mathematics Stack Exchange abstract algebra - Examples of rings that are Noetherian but not finitely generated? - Mathematics Stack Exchange](https://i.stack.imgur.com/fPNez.png)