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?

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

**(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.)**

*Exercises 1 - 3** Exercise 5*: Can you generalize

**to an arbitrary field in place of the complex field $\mathbb{C}$?**

*Exercises 1 - 4*I hope this helps!

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.