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?

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.

the answer is:

mean: 6

mad: 2.14

explanation:

remove parenthesis:

=

apply rule

=

multiply fractions:

=

=

divide:

=

mordancy.