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.

  • $\begingroup$ thank you Carlo for editing the question. I was doing it and it shows that you've edited already! Thank you! $\endgroup$ – gerrard Jul 2 '20 at 2:37
  • $\begingroup$ 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 it has an ideal I that is not finitely generated. Also, if B as an R-mod can be generated by n elements and C is a f.g.submodule, C still may require more than n generators. $\endgroup$ – gerrard Jul 2 '20 at 2:44
  • $\begingroup$ 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? $\endgroup$ – Carlo Jul 2 '20 at 2:51
  • $\begingroup$ thanks Carlo! But each element of C can still be expressed by n generators of B, correct? and C requires more than n generators... $\endgroup$ – gerrard Jul 2 '20 at 2:58

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