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1 May, 19:22

Let A ⊆ B ⊆ C be rings. Suppose C is a finitely generated A-module. Does it follow that B is a finitely generated A-module?

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  1. 1 May, 19:27
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    Let A ⊆ B ⊆ C be rings. If C is a finitely generated A-module. Then B is a finitely generated A-module.

    Step-by-step explanation:

    Draw a ring and call it A, then draw another circle with a longer radius from the same centre of A and call it B, then draw another from the same centres of A and B, but with the longest radius and call it C.

    Then, when you say A ⊆ B ⊆ C, this means that A is a subset of and equal to B which is a is a subset of and equal to C.

    Meaning:

    1. A is in B and B is in C.

    2. The values in A are the only values in B. i. e

    If A = {2,4,6} then B = {2,4,6}

    3. The values of ring B are the only values in ring C. i. e. if B = {2,4,6} then C = {2,4,6}.

    4. There is no more values in B that is not in A.

    5. There are no more values in C that is not in B.

    Since they are subsets of each other defined by ⊆, which makes the subset exactly the same as the host set or superset.

    So the same rule that applies to C will apply to B

    A finitely generated module is a module that has a finite generating set. A finitely generated module over a ring A may also be called a finite A-module.
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