I'm studying Dummit and Foote's Abstract Algebra, and on page $352$ there are some examples: Let $R$ be a ring with $1$ and let $M$ be the (left) $R$-module $R$ itself. Note that $R$ is a finitely generated, in fact cyclic, $R$-module because $R = R1$ (i.e., we can take $A = \{1\}$).

Submodules of a finitely generated module need not be finitely generated: take $M$ to be the cyclic $R$-module $R$ itself where $R$ is the polynomial ring in infinitely many variables $x_1, x_2, x_3, \dots$ with coefficients in some field $\mathbb{F}$. The submodule (i.e., $2$-sided ideal) generated by $\{x_1, x_2, \dots\}$ cannot be generated by any finite set.

I don't understand why the argument in the first part cannot be applied to the polynomial ring, where we say $R=R(1)$. Then the generating set is just $\{1\}$. I know that the polynomial ring doesn't satisfy the conditions in the first part, specifically there is no multiplicative identity. But still, for any polynomial $p(x)$, $\ p(x)=1\cdot p(x)$, so $R=R(1)$.

I would greatly appreciate your help.

  • $\begingroup$ I'm terribly sorry-I don't know how to use Latex. Are there any tutorials on this website on how to use it? $\endgroup$
    – 010110111
    Nov 10, 2013 at 15:42
  • $\begingroup$ You can check out the MathJaX basic tutorial :) $\endgroup$
    – Maethor
    Nov 10, 2013 at 15:53
  • $\begingroup$ I know that the polynomial ring doesn't satisfy the conditions, but I don't understand why its not finitely generated, with the set being ${1}$. Since $F$ is a field, it has a multiplicative identity ${1}$, so $\ p(x)=1\cdot p(x)$. Then $R=R1$ which means the polynomial ring is finitely generated. Obviously there is something wrong with this argument, and I'm asking what that is. $\endgroup$
    – 010110111
    Nov 10, 2013 at 16:55

1 Answer 1


Everything is correct. Argument from first paragraph applies to polynomial ring. $R:=\mathbb F[x_1,x_2,...]$ is cyclic module over itself because $R$ has unit. The example says that submodule of $R$ generated by $x_1,x_2,...$ is not cyclic. Now you asking if ideal $(x_1,x_2,...)$ is generated by one element and it's not because every finite set of polynomials contains only finitely many of variables. You can't use argument with unite to ideals because ideals by definition do not contain unite.


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