I don't believe it's true that all submodules of a finitely generated free module are free just based on results from google. Is there a canonical simple example of a finitely generated free module with a submodule that is not free?

  • $\begingroup$ Haven't you considered accepting an answer? $\endgroup$
    – user422112
    Oct 6, 2018 at 5:10

4 Answers 4


If $R$ is a ring which is not a principal ideal ring, then usually its ideals are not free. But they are submodules of $R$, which is free and finitely generated.

For example, the ideal $(x,y)$ in the ring $\mathbb C[x,y]$ is not free.


The following result is relevant to your question:

A submodule of a finitely generated free module over a principal ideal domain is free. (For example, we can use the structure theorem for finitely generated modules over principal ideal domains.)

In particular, we need to think about commutative rings that are not principal ideal domains in order to answer your question. The polynomial ring in two variables over a field is a good example of such a ring.

Let me expand on Mariano's nice example. I shall do this by introducing a sequence of exercises:

Exercise 1: Prove that $\{x,y\}$ is not a $\mathbb{C}[x,y]$-basis for the $\mathbb{C}[x,y]$-module $(x,y)$.

However, it is a little trickier to prove that $(x,y)$ is not a free $\mathbb{C}[x,y]$-module. We need to show, in this context, that there does not exist a $\mathbb{C}[x,y]$-basis for $(x,y)$, which is more difficult than showing that a particular tuple of elements of $(x,y)$ is not a $\mathbb{C}[x,y]$-basis for $(x,y)$.

Exercise 2: Prove that $(x,y)$ is not a free $\mathbb{C}[x,y]$-module. (Hint: tensor over $\mathbb{C}[x,y]$ with the fraction field $\mathbb{C}(x,y)$ of $\mathbb{C}[x,y]$. What is the dimension of $(x,y)\otimes_{\mathbb{C}[x,y]}\mathbb{C}(x,y)$ as a $\mathbb{C}(x,y)$-vector space?)

Finally, if you have seen the (important) notion of flatness, then the following exercise might serve as good practice with this notion:

Exercise 3: Can you prove that $(x,y)$ is not even a flat $\mathbb{C}[x,y]$-module? (Hint: do you remember the equational criterion for flatness (cf. the first theorem in chapter 2 of Hideyuki Matsumura's Commutative Algebra)?)

Commutative algebra and algebraic geometry are closely connected branches of mathematics and thus it is always a good idea to interpret results in commutative algebra in the geometric context:

Exercise 4: If you know some algebraic geometry (e.g., Hilbert's nullstellensatz), then interpret Exercises 1 - 3 (and their answers) geometrically. (Hint: flatness is a little difficult to interpret geometrically but is very natural in the geometric context; if you are not aware of the geometric interpretation of flatness, then please look it up.)

Exercise 5: Can you generalize Exercises 1 - 4 to an arbitrary field in place of the complex field $\mathbb{C}$?

I hope this helps!

  • $\begingroup$ Dear Amitesh, what do you have in mind with exercise 4 and what has Hilbert's Nullstellensatz got to do with it, especially since the result has absolutely no relation to algebraic closedness of the base field? And what interpretation of flatness do you want the OP to look up? (Let me add that, these remarks notwithstanding, I like your answer and upvoted it.) $\endgroup$ Oct 23, 2011 at 9:02
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    $\begingroup$ Sorry to revive, but to show that (x,y) is not free don't we only need to consider cases (?): (i) Our generating set, $S$, cannot consist of any element in $\mathbb{C}$ (ii) $S$ must contain at least, $kx$ and $k'y$, $k,k' \not=0 \in \mathbb{C}$ to generate $(x,y)$. (iii) But then we obtain non independence. $\endgroup$
    – Bryan Shih
    Mar 31, 2017 at 23:58

A very simple example is $\mathbb{Z}_6$ as a $\mathbb{Z}_6$-module is free (generated by $[1]$) and its ideal (regarded as a submodule) $2 \mathbb{Z}_6=\{[0],[2],[4]\}$ has no basis since $[2][3]=[4][3]=[0]$


A simple example is consider $\mathbb{Z}_2^{\mathbb{N}}$ as ring and as a free module over itself and $\mathbb{Z}_2$, and a simple argument of cardinality do the job, a free module is direct sum of copies of the ring.

  • $\begingroup$ How do you proof that $ \mathbb{Z}_{2}^{\mathbb{N}} $ is free ? $\endgroup$
    – user347910
    Nov 24, 2019 at 13:10
  • $\begingroup$ The ring is free as module over itself. $\endgroup$
    – Math.mx
    Nov 28, 2019 at 19:01

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