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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.

user26857
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Dominic Wynter
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  • For my own interest: what is $k[x]_x$? – MT_ Jul 18 '16 at 22:32
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    I'd suggest showing that $k[x,y]/(1-xy)$ satisfies the universal property of the localization $k[x]_x$. – carmichael561 Jul 18 '16 at 22:33
  • @MichaelTong $k[x]_x = k[x, x^{-1}]$. See here for the general context – Alex G. Jul 18 '16 at 22:35
  • $k[x]_x:={\frac{f(x)}{x^n} \mid f\in k[x], n\in\mathbb{N}}\subset k(x)$ is the localization at $x$ of $k[x]$. – Dominic Wynter Jul 18 '16 at 22:35
  • @AlexG. Ah okay, that's what I thought but wasn't sure. Been a while since I read Harris. – MT_ Jul 18 '16 at 22:36
  • Compare to the more general case here when $,k,$ is a commutative ring. – Bill Dubuque Jul 18 '16 at 22:41
  • Could you explain why this implies $k[x]_x$ is a field? I feel like you're hiding something. Anyway, it's very easy to show that in general when you have $f \in A$ then $A_f \simeq A[X]/(fX - 1)$. If you unwind the universal properties of localization, polynomial rings, and quotients that are involved you get the same thing. Or define maps both ways and show they are inverse. – Hoot Jul 18 '16 at 22:41
  • @Hoot As OP pointed out, $k [x]_x $ is not a field – Alex G. Jul 18 '16 at 22:50
  • @AlexG. I mean as a matter of contradiction. I know it's not a field. It's hard to refer to things in the middle of such proofs, I guess! – Hoot Jul 18 '16 at 23:08
  • I'm adding this question by me here, another elementary approach to see this. $k[x,y] \rightarrow k[x,y], x \mapsto x+y, y\mapsto y$ gives $k[x,y] \rightarrow k[t,\frac{1}{t}], x \mapsto t+\frac{1}{t}, y\mapsto \frac{1}{t}$.

    https://math.stackexchange.com/q/4099793/677131

     – Varadharajan R Apr 12 '21 at 21:21

3 Answers3

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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.

egreg
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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 $\phi$.

user26857
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    I don't understand the first comment. What I was suggesting was an elementary proof (of course "elementary" is undefined...) In the U.S. they only teach universal properties of anything in graduate school (except for a few places like Berkeley and Harvard); I wouldn't call that "elementary." Checking that something is well-defined by referring directly to the equivalence relations is "elementary." –  Jul 18 '16 at 23:02
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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 $k[x,y] / (xy-1) \to k[x]_x$.