![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 X: "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 in an Constanza Rojas-Molina on X: "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 in an](https://pbs.twimg.com/media/EnfZHzDW4AARn9_.jpg)
Constanza Rojas-Molina on X: "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 in an
![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)
abstract algebra - Examples of rings that are Noetherian but not finitely generated? - Mathematics Stack Exchange
![If the Localization is Noetherian for All Prime Ideals, Is the Ring Noetherian? | Problems in Mathematics If the Localization is Noetherian for All Prime Ideals, Is the Ring Noetherian? | Problems in Mathematics](https://i0.wp.com/yutsumura.com/wp-content/uploads/2016/12/ring-theory-eye-catch-e1497227610548.jpg?resize=720%2C340&ssl=1)