# Is $B$ a finitely generated $R$-module?

Proposition 4.29 of Rotman's Introduction to Homological Algebra states that if $$R$$ is a left semihereditary ring, then every finitely generated submodule $$A$$ of a free $$R$$-module is a direct sum of a finite number of finitely generated left ideals.

In his proof, Rotman argues in the first paragraph that we may assume that $$F$$ is a finitely generated free left $$R$$-module with a basis $$\{x_1, \dots, x_n \}.$$ Next, he proceeds by induction. If $$n > 1,$$ then we define $$B$$ as the intersection of $$A$$ and $$R x_1 + \cdots + R x_{n - 1};$$ Rotman now says that by the inductive hypothesis, $$B$$ is a direct sum of a finite number of finitely generated left ideals.

My question is how are we sure that $$B$$ is finitely generated, as there are cases where submodules of finitely generated modules are not finitely generated?

I might be overcautious here, as on page 163 of this book, it states that if $$R$$ is a domain that is not Noetherian, then $$R$$ has an ideal $$I$$ that is not finitely generated. Also, if $$B$$ is an $$R$$-module that can be generated by $$n$$ elements and $$C$$ is a finitely generated $$R$$-submodule of $$B,$$ then $$C$$ may require more than $$n$$ generators. Thank you very much!

I don't think the suggested post answers my question since in the second to last paragraph of that proof, it states that the intersection of B and A is f.g., which seems to be exactly my question.

Every element $$b$$ of $$B$$ can be written as $$b = r_1 x_1 + \cdots + r_{n - 1} x_{n - 1}$$ for some elements $$r_i$$ of $$R.$$
Explicitly, given an element $$b$$ of $$B,$$ we have that $$b = s_1 a_1 + \cdots + s_m a_m$$ for some elements $$s_i$$ of $$R$$ by hypothesis that $$B$$ is in $$A = R \langle a_1, \dots, a_m \rangle.$$ Observe that each $$a_i$$ is an element of the free $$R$$-module $$F,$$ hence for each $$a_i,$$ we have that $$a_i = t_{1i} x_1 + \cdots + t_{ni} x_n$$ for some element $$t_{ji}$$ of $$R.$$ We can therefore write $$b = s_1(t_{11} x_1 + \cdots + t_{n1} x_n) + \cdots + s_n(t_{1m} x_1 + \cdots + t_{nm} x_n).$$ Combining like terms, we find that $$b = (s_1 t_{11} + \cdots + s_n t_{1m})x_1 + \cdots + (s_1 t_{n1} + \cdots + s_n t_{nm})x_n.$$ But by hypothesis, we have also that $$B$$ is in $$Rx_1 \oplus \cdots \oplus Rx_{n - 1},$$ hence this expression of $$b$$ as a linear combination of $$x_i$$ is unique, and the coefficient on $$x_n$$ must be 0. At any rate, the claim is established.
• But you have an explicit form for the $R$-submodule $B.$ Every element of $B$ is an element of $A$ and an element of $R x_1 + \cdots + R x_{n - 1}.$ What do the elements of the latter look like? – Carlo Jul 2 '20 at 2:51