## Torsion elements in the field of rational expressions [closed]

Suppose I consider the field of rational expressions over the rational numbers in the indeterminants $X_{1},\ldots, X_{t}$ (where $t > 1$). That is, I consider the field of fractions of the polynomial ring $\mathbb{Q}[X_{1},\ldots, X_{t}]$.

Here is my question: is the multiplicative group of this field torsion-free? Or are there polynomials $f,g \in \mathbb{Q}[X_{1},\ldots, X_{t}]$ such that $g$ is not the zero polynomial, $f/g\ne 1$, but $(f/g)^{k}=1$ for some positive integer $k$?

Thanks.

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The field contains -1, which has order 2. – S. Carnahan Jun 22 2010 at 6:14
In the absence of additional motivation or context in the question, I'm voting to close. Incidentally, if you replace $\mathbb{Q}$ with any field $k$, the torsion subgroup of the multiplicative group of a rational function field $k(x_1,\dots,x_n)$ is contained in $k^\times$. – S. Carnahan Jun 22 2010 at 6:46
To expand on Scott's statement, $k[x_1, \ldots, x_t]$ has unique factorization so, if $(f/g)^n=1$, then $f/g$ is in $k$. – David Speyer Jun 22 2010 at 7:38
Scott: I don't think that this question should be closed. – Martin Brandenburg Jun 22 2010 at 10:35

## closed as no longer relevant by Victor Protsak, Ben Webster♦Aug 18 2010 at 5:44

Right, I didn't mention that I'm not interested in elements in the subfield $\mathbb{Q}$. But unique factorisation in $\mathbb{Q}[X_{1},\ldots, X_{t}]$ eliminates the possibility of any others.