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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.

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    Hint: Is $T$ invertible in $K[T]$? Is $X$ invertible in $K[X,Y]/(XY-1)$? What is $XY$ equal to? – 2017-02-22
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    $K[x,y]/(xy-1) \cong K[x,\frac{1}{x}] $ – 2017-02-22

2 Answers 2

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

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    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 $k=\mathbb Q$). – 2018-12-03
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Suppose $K[x,y]/(xy-1)$ is isomorphic to $K[t]$.

Then there is a surjective homomorphism $f:K[x,y] \to K[t]$ with $\text{ker}(f) = (xy-1)$.

Let's assume you've previously shown that the group of units of $K[t]$ is $K^*$ (if not, you need to show this).

Since $f$ is a homomorphism, $f$ must map units to units, hence $f(K^*) \subseteq K^*$. It follows that $f(K) \subseteq K$.

Let $p = f(x)$, $q = f(y)$.

Then $0 = f(xy - 1) = f(x)f(y) - f(1) = pq - 1$, hence $pq = 1$.

Thus, $p,q$ are units in $K[t]$, so $p,q \in K^*$.

Let $m \in K[x,y]$ be a monomial. Write $m = ax^iy^j$ where $a \in K$ and $i,j$ are nonnegative integers.

Then $f(m) = f(ax^iy^j) = f(a)f(x^i)f(x^j) = f(a)f(x)^if(y)^j = f(a)p^iq^j$, hence $f(m) \in K$.

Let $s \in K[x,y]$.

Then $s$ is a sum of finitely many monomials, hence since $f$ maps each monomial to an element of $K$, it follows that $f(s) \in K$.

Thus, $f$ maps every element of $K[x,y]$ to an element of $K$, so $f$ is not surjective, contradiction.

It follows that $K[x,y]$ is not isomorphic to $K[t]$.