# $K[X,Y]/(XY-1)$ not isomorphic to $K[T]$ [duplicate]

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.

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

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

• 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 $k=\mathbb Q$). – user26857 Dec 3 '18 at 19:49

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 $$K[t]$$.