1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.