Let $k$ be a field. Is $k[x,x^{-1}]$ free over $k[x]$?

Question

I see that $\{x, x^{-1}\}$ do not form a basis over $k[x]$ since for example $x^2 \cdot x^{-1} +(-1)\cdot x =0$.

I'm not sure if this is enough to conclude that $k[x,x^{-1}]$ is not free over $k[x]$.

Not sure how to proceed further.

Answer 1

No. If we have two nonzero elements of $k[x,x^{-1}]$ we can write then as $x^{-r}f(x)$ and $x^{-s}g(x)$ where $f$ and $g$ lie in $k[x]$ and $r$, $s\ge0$. These are linearly independent over $k[x]$: $$x^rg(x)[x^{-r}f(x)]-x^sf(x)[x^{-s}g(x)]=0.$$ So if $k[x,x^{-1}]$ is free over $k[x]$ it must be free of rank $1$.

But a typical nonzero element of $k[x,x^{-1}]$ is $x^{-r}f(x)$ where $f(x)\in k[x]$ and then $$x^{-r-1}\notin x^{-r}f(x)k[x].$$

Answer 2

There are similarities between $k[x]$ and $\Bbb Z$, so the question is similar to asking whether the $\Bbb Z$-module $\Bbb Z[\frac12] = \{\frac a {2^k} \in \Bbb Q: a, k\in \Bbb Z\}$ is free over $\Bbb Z$.

One possible proof is as follows:

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Suppose $\Bbb Z[\frac12]$ is a free $\Bbb Z$-module. Let $(x_i)_i$ be a basis, and let $x_0$ be one of the $x_i$. We then consider the element $x_0/2$, which can be written as $$x_0/2 = \sum a_i x_i$$ for some integers $a_i$. Multiplying by $2$ gives $$x_0 = \sum (2a_i) x_i.$$ But on both sides we have linear combinations of basis elements, so for every basis element, the coefficients on both sides must be equal.

In particular, this means $2a_0 = 1$, which is impossible as $a_0$ is an integer.

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The same proof works for your question: just replace $\Bbb Z$ and $2$ with $k[x]$ and $x$, respectively.