ring of Laurent polynomial as a module over polynomial ring I believe that this question should be easy, but I failed to prove it on my own.
Assume that we are in the category of $\mathbb{Z}[x]$-modules. Then we can view the ring of Laurent polynomials $\mathbb{Z}[x,x^{-1}]$ as a $\mathbb{Z}[x]$-module. Now consider a short exact sequence of the following form:
$0\to \mathbb{Z}[x,x^{-1}]\to M \to \mathbb{Z}[x,x^{-1}]\to 0$
Can we say that the above sequence splits, so that $M\cong \mathbb{Z}[x,x^{-1}]\oplus \mathbb{Z}[x,x^{-1}]$ as a $\mathbb{Z}[x]$-module?
Of course this is true when $\mathbb{Z}[x,x^{-1}]$ is projective or injective, but it seems this is not true? My algebraic knowledge stops when it comes to infinitely generated modules like $\mathbb{Z}[x,x^{-1}]$ over $\mathbb{Z}[x]$, but $\mathbb{Z}[x,x^{-1}]$ is one of the simplest infinitely generated module, so I believe there should be an elementary answer for this question...
 A: First, note that multiplication by $x$ is a bijection on $M$.  For instance, this follows from the five lemma, since multiplication by $x$ is a bijection on $\mathbb{Z}[x,x^{-1}]$.
So, $M$ is a $\mathbb{Z}[x,x^{-1}]$-module, and we can consider the short exact sequence as a short exact sequence of $\mathbb{Z}[x,x^{-1}]$-modules.  Now we see that it splits, since $\mathbb{Z}[x,x^{-1}]$ is projective as a $\mathbb{Z}[x,x^{-1}]$-module.

Here is a more concrete argument that is essentially the same (just with all the relevant lemmas worked out explicitly in this case).  Let us label the maps in the short exact sequence as $i:\mathbb{Z}[x,x^{-1}]\to M$ and $p:M\to\mathbb{Z}[x,x^{-1}]$.  Choose an element $m_0\in M$ such that $p(m_0)=1$.  Now choose an element $m_1\in M$ such that $p(m_1)=x^{-1}$.  Then $xm_1-m_0$ is in the kernel of $p$, and thus the image of $i$; let $xm_1-m_0=i(a)$.  Now let $m_1'=m_1-i(x^{-1}a)$.  Note that $$xm_1'=xm_1-xi(x^{-1}a)=xm_1-i(a)=m_0.$$  Also, since $pi=0$, $p(m_1')=x^{-1}$.
So, replacing $m_1$ with $m_1'$, we can choose $m_1\in M$ such that $p(m_1)=x^{-1}$ and $xm_1=m_0$.  But then we can repeat this argument to choose $m_2\in M$ such that $p(m_2)=x^{-2}$ and $xm_2=m_1$, and so on.  We can then get a splitting of $p$ by mapping $x^{-n}$ to $m_n$ for each $n$.
