Want to prove that some $\mathbb R[x]$-Module has no basis So here is my question,
Consider the $\mathbb R[X]$-module $\mathbb R[X,X^{-1}]$ i.e the $\mathbb R[x]$-module of all Laurent-Polynomials. I want to show that is module is not free i.e it has no basis.
I already feel I see the problem, since if one multiplies an element $p(x)\in\mathbb R[X^{-1}]$ with some element of the ring $\mathbb R[X]$ it is only possible to increase the degree of $p(x)$ but it is not possible to decrease it.
Moreover, I think if we I assume for contradiction that there exists a basis 0f $\mathbb R[X,X^{-1}]$ than it contains at least some reel constant $c$ as basis of $\mathbb R[X]\subset\mathbb R[X,X^{-1}]$. Then every basis has to be of the form $\{c,p_1,p_2,...\}$ where $p_i\in\mathbb R[X,X^{-1}]$ - $\mathbb R[X$] i.e is some polynomial of degree strictly less the zero. And this polynomials can not be linear independent since one can always increase the degree until it becomes again positv...?
 A: Let $K$ be any field (nothing special about the real numbers here). Suppose $K[X,X^{-1}]$ were a free $K[X]$-module on some basis with cardinality $\gamma$ (finite or infinite). You already argued that $\gamma=1$ is not possible.
Now this would imply $K[X,X^{-1}]\otimes_{K[X]}K(X)$ is a $K(X)$-vector space of dimension$~\gamma$. But is is actually a $K(X)$-vector space of dimension$~1$, since any two elements of $K[X,X^{-1}]$ are obviously linearly dependent over$~K(X)$. Contradiction.
A: Suppose $\mathcal{B}$ is a set of generators for $\mathbb{R}[X, X^{-1}]$ as an $\mathbb{R}[X]$-module. There must be terms of arbitrarily low lowest term, since - as you've correctly noted - multiplying elements of $\mathbb{R}[X, X^{-1}]$ by elements of $\mathbb{R}[X]$ can only increase the degree of the lowest term. So in particular, $|\mathcal{B}| = \infty$
But $1$ can be written as a finite $\mathbb{R}[X]$-linear sum of elements of $\mathcal{B}$, say $1 = f_1b_1 + ... + f_rb_r$. Let $b \in \mathcal{B}$ be such that $b \neq b_i$ for all $i$, and assume the lowest term of $b$ is degree $r < 0$. Then $bX^r \in \mathbb{R}[X]$ (and $bX^r \neq 0$).
But now $(f_1bX^r)b_1 + ... + (f_rbX^r) - bX^r$ is a non-trivial $\mathbb{R}[X]$-linear relation between elements of $\mathcal{B}$, so $\mathcal{B}$ cannot be a basis.
A: Observation: Any two nonzero $\Bbb R[x]$ submodules of $\Bbb R[x,x^{-1}]$ intersect. If $M$ and $N$ are two such submodules and $a\in M$, $b\in N$ are nonzero elements. You can find powers of $x$ such that $x^ia\in \Bbb R[x]$ and $x^kb\in \Bbb R[x]$, and then you can find their gcd in $\Bbb R[x]$. That is, there exist $p,q$ such that $px^ia=qx^kb\in M\cap N$, showing the intersection is nonzero.
Using that observation, if $\Bbb R[x,x^{-1}]$ were free, it could not have more than one copy of $\Bbb R[x]$ in its decomposition. That would mean $\Bbb R[x,x^{-1}]$ is a cyclic $\Bbb R[x]$ module. But as you observed, no single element would suffice to generate all of $\Bbb R[x,x^{-1}]$, since the set of things a single element can generate has a definite lower bound on its degree.
A: Here is a base-free argumentation. Let us assume that $\mathbb{R}[X^{\pm 1}]$ is a free $\mathbb{R}[X]$-module then it is also projective. Let $g : \mathbb{R}[X,Y] \to \mathbb{R}[X^{\pm 1}]$ be the surjective $\mathbb{R}[X]$-algebra-homomorphism mapping $Y \mapsto X^{-1}$. Now by projectivity there is a $\mathbb{R}[X]$-module-homomorphism $f : \mathbb{R}[X^{\pm 1}] \to \mathbb{R}[X,Y]$ such that $g \circ f = id$. This leads easily to a contradiction: because of $X^n \cdot f(X^{-n}) = f(1) \in \mathbb{R}[X,Y]$ we see that $X^n$ divides $f(1)$ for all positive integers $n$ which forces $f(1)=0$. But then $X = gf(X) = X \cdot gf(1) = 0$.
Of course this argument still works if we replace $\mathbb{R}$ by any other (nonzero) commutative ring.
