Consider the algebra homomorphism $\phi:\mathbb{Q}[x^{\pm 1},y^{\pm 1},z^{\pm 1}] \rightarrow \mathbb{Q}$ defined by $\phi(x)=\phi(y)=\phi(z)=1$, where $\mathbb{Q}[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$ denotes the ring of laurent polynomials in $x,y,z$. I would like to compute the kernel of $\phi$. My guess is that it is the ideal $I$ generated by $x^{\pm 1}-1,y^{\pm 1}-1,z^{\pm 1}-1$. Clearly $I\subset Ker(\phi)$, however to prove the reverse inclusion I would need some way to write an arbitrary laurent polynomial as a linear combination of the $x^{\pm 1}-1,y^{\pm 1}-1,z^{\pm 1}-1$ plus a remainder. Ultimately, I'm trying to show that remainder is 0.

I would like to know if such a division algorithm exists for laurent polynomials. I know there is a multivariable division algorithm in the ring $\mathbb{Q}[x,y,z]$. Does this algorithm extend?

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    $\begingroup$ You don't need the $\pm 1$'s in the exponents. Note that $x-1$ and $x^{-1}-1$ differ from each other just by the multiplicative factor $-x^{-1}$ (or $-x$), which is a unit in the Laurent polynomial ring. So you merely need the generators $x-1, y-1, z-1$. In order to see that $I$ is indeed generated by these three generators, you first show that they lie in $I$ (obvious) and then you show that every Laurent polynomial can be transformed into a constant by adding multiples of these three generators (indeed, you can transform each monomial into a constant this way, step by step). $\endgroup$ – darij grinberg Oct 30 '18 at 0:20

I don't know if there is such a division algorithm. There probably is, but you can also reduce this to finding an ideal in an ordinary polynomial ring. Namely, $$\mathbb Q[x_1,x_2,y_1, y_2,z_1,z_2]$$ The map $x_i, y_i, z_i\mapsto 1$ factors through $x_1x_2\mapsto 1$, $y_1y_2\mapsto 1$, $z_1z_2\mapsto 1$, and this middle map gives you the Laurent polynomial ring. You can use this to easily prove your conjecture, since the ideal in this larger ring is obvious.


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