An elementary proof that $k[x,y]/(xy-1)\cong k[x]_x$, where $k$ is a field

Question

Letting $\phi:k[x,y]\to k[x]_x$, $\phi(x)=x$, $\phi(y)=\frac{1}{x}$, we see that $\ker \phi$ is prime, and $(1-xy)\subseteq\ker\phi$. Now, given that $k[x,y]$ has Krull dimension 2, $\ker\phi\neq (1-xy)$ would imply that $0\subsetneq (1-xy)\subsetneq\ker\phi$, and therefore $\ker\phi$ is a maximal ideal, so $k[x]_x$ is a field, which is easily checked to be false, and therefore $k[x,y]/(1-xy)\cong k[x]_x$. However, I was wondering if there was some way of proving this using only elementary methods.

Edit:

Claim: $k[x]_x$ is not a field.

Proof: Suppose $x-1\in k[x]_x$ is invertible. Then let $\frac{1}{x-1}=\frac{f(x)}{x^n}$, therefore $x^n= f(x)(x-1)$ in $k[x]$, therefore $1^n=1=0$, a clear contradiction.

Answer 1

Let $R$ be a $k$-algebra and $f\colon k[x]\to R$ be a $k$-algebra homomorphism where $f(x)=r$ is invertible. We want to see that there is a unique homomorphism $\hat{f}\colon k[x,y]/I\to R$, $I=(xy-1)$, such that $\hat{f}\circ p=f$, where $$ p\colon k[x]\to k[x,y]/I \qquad p(x)=x+(xy-1) $$ Define $g\colon k[x,y]\to R$ by $g(x)=r$ and $g(y)=r^{-1}$. Then $$ g(xy-1)=0 $$ proving that $\ker g\supseteq I$. Thus $g$ induces a $k$-algebra homomorphism as required.

Uniqueness of $\hat{f}$ is obvious, because $\hat{f}(x+I)=r$ and $\hat{f}(y+I)=r^{-1}$, because $\hat{f}\bigl((x+I)(y+I)\bigr)=\hat{f}(1+I)=1$; $\hat{f}$ is completely determined by the action on generators.

Since $k[x,y]/I$ satisfies the universal property of the ring of fractions with respect to the multiplicative set $\{x^n:n\ge0\}$, it is the ring of fractions.

Answer 2

An obvious thing to try is to define $\psi:k[x]_x \rightarrow k[x,y]/(xy-1)$, $\psi(\frac{f(x)}{x^n}) = \overline{f(x)y^n}$ (the RHS is an equivalence class in the quotient ring). You will have to prove that $\psi$ is well-defined, a homomorphism, and it is the inverse of $\phi$.

Answer 3

Suppose $f(x,y) \in k[x,y]$ is a polynomial such that $f(x, \frac{1}{x}) = 0$.

Then $f(x,y)$ is a polynomial in $y$ over the unique factorization domain $k[x]$ that has a root at $y = \frac{1}{x}$, and is thus divisible by $(xy-1)$.

Consequently, the set of all polynomials $f(x,y) \in k[x,y]$ such that $f(x, \frac{1}{x}) = 0$ is precisely $(xy-1)$, and so $f(x,y) \mapsto f(x,\frac{1}{x})$ gives an isomorphism $k[x,y] / (xy-1) \to k[x]_x$.