4
$\begingroup$

Why is $K[X,Y]/(XY-1)$ not isomorphic to $K[T]?$

As a hint I should think about units. But I don't get a helpful idea.

$\endgroup$
  • $\begingroup$ Hint: Is $T$ invertible in $K[T]$? Is $X$ invertible in $K[X,Y]/(XY-1)$? What is $XY$ equal to? $\endgroup$ – Michael Burr Feb 22 '17 at 18:21
  • 2
    $\begingroup$ $K[x,y]/(xy-1) \cong K[x,\frac{1}{x}] $ $\endgroup$ – Mustafa Feb 22 '17 at 18:33
-5
$\begingroup$

The group units of $k[x]$ is $k^*$, the units of $k[x,y]/(xy-1)$ is $k^*\times \{x^n\mid n\in \mathbb{Z}\}$.

$\endgroup$
  • 6
    $\begingroup$ I don't know why this answer got 4 upvotes. If $k$ is not algebraically closed, then this argument fails, that is, the groups $k^*$ and $k^*\times\mathbb Z$ can be isomorphic (for instance, when $k=\mathbb Q$). $\endgroup$ – user26857 Dec 3 '18 at 19:49
6
$\begingroup$

Let $K$ be a field, and let $R=K[x,y]/I$, where $I=(xy-1)$.

For $f\in K[x,y]$, let $\bar{f}$ denote the corresponding element of $R$.

Since $xy\equiv 1\;(\text{mod}\;I)$, it follows that $\bar{x}$ and $\bar{y}$ are units of $R$.

Also, if $a\in K^*$, then $\bar{a}$ is a unit of $R$.

Hence if $m\in K[x,y]$ is a nonzero monomial, then $\bar{m}$ is a unit of $R$.

Since every element of $K[x,y]$ is a finite sum of monomials, it follows that every element of $R$ is a finite sum of units.

But the units of $K[t]$ are just the elements of $K^*$, so the only elements of $K[t]$ which are finite sums of units are the elements of $K$.

In particular, the element $t$ in $K[t]$ is not a finite sum of units.

Therefore $R$ is not isomorphic to $K[t]$.

$\endgroup$