# Adding some more rigour to a simple exercise in Hartshorne [duplicate]

I have just begun working through Hartshorne's algebraic geometry.

In part (b) of exercise 1.1, we need to show that $A(Z)$ is not isomorphic to a polynomial ring, where $Z$ is the plane curve $xy =1$.

Since $k[x,y]/\langle xy - 1 \rangle\cong k[x, \frac{1}{x}]$, it makes sense that $A(Z)$ is not isomorphic to a polynomial ring. But how do I rigorously prove that $$k[x, \frac{1}{x}] \not \cong k[x,x^n]$$ for some $n$?

• May I ask what k[x,x^n] means? – CJD Jun 14 '17 at 4:07

If an integral domain $R$ is a polynomial ring, $S[t]$ say, then every unit of $R$ is contained in $S$. Here $R=k[x,1/x]$. The units of $R$ include $x$, $1/x$ and the nonzero elements of $k$. These all have to lie in $S$, but they generate $R$ as a ring. Therefore $R=S$ and $R=S[t]$ which is impossible.