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Under which conditions on the ring $A$ do we have the isomorphism $A[x]_x\cong A[x,y]/(xy-1)$, and why does this even hold?

I am asking because of the following isomorphism: $(k[x_1,\ldots,x_n]/\mathfrak a)_f\cong k[x_1,\ldots,x_n,y]/(\mathfrak a,fy-1)$ with $f\in k[x_1,\ldots,x_n]$ and $\mathfrak a$ an ideal in $k[x_1,\ldots,x_n]$ (described in the excellent answer of M. Brandenburg). He says that this is "by general nonsense", what is meant by this concretely? What are the general isomorphisms used here? Is this supposed to be obvious? because it is not obvious to me

Community
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2 Answers2

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I would describe this intuitively as follows. $A[x]_{x}$ is (very vaguely) the ring made from $A[x]$ forcing $x$ to be invertible while only adding the necessary relations to make a ring. $A[x,y]/(xy-1)$ fits this description as well, so one would expect them to be isomorphic. More concretely we have:

Define the map $\varphi : A[x,y]/(xy-1) \rightarrow A[x]_{x}$ by $x \rightarrow x$ and $y \rightarrow \frac{1}{x}$. It is well-defined because it sends $xy - 1$ to $0$. Now one can check it is a ring homomorphism. It is also clearly surjective. Now suppose that $f(x,y)$ is sent to $0$. Write

$$f(x,y) = \sum_{i,j} a_{ij} x^iy^j$$

Then $$\varphi(f) = \sum_{i,j} a_{ij}x^{i-j} = \sum_k\left(\sum_{k = i-j} a_{ij}\right)x^k \equiv 0$$

Thus we get for each fixed $k$ we get $\sum_{k = i-j} a_{ij} = 0$. Now use the relation $xy = 1$ to write $f$ as

$$f = \sum_{0 = i - j} a_{ij} + \sum_{i > j} a_{ij}x^{i-j} + \sum_{j > i} a_{ij}y^{j-i}$$

Finally, we may regroup and note that each sum is zero. For example, for $i > j$ we have

$$\sum_{i > j} a_{ij}x^{i-j} = \sum_{k > 0}\left(\sum_{k = i-j} a_{ij}\right)x^k$$

This proves $\varphi$ is injective and hence an isomorphism.

RghtHndSd
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Since $R = A[x,y]/(xy-1)\,$ is a ring $\supset A[x]$ where $x$ is a unit, and every element has the form $f(x)/x^n,\, f\in A[x],\,$ it follows from the universal property of localization that $\,R \cong A[x]_x.\ $ Said universal property is the "general nonsense" being referred to, and it makes the result trivial.

Here is the relevant result Theorem from Atiyah & MacDonald, Commutative Algebra, p. 39.

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Bill Dubuque
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  • so elegant, thanks a lot! –  Jan 29 '14 at 00:12
  • to check the conditions, is the following ok? (i) $x^n$ is unit since inverse is $y^n$. (ii) if $f=0$ in the quotient, $f=(xy-1)g$ for some $g$; but $f$ depends only on $x$ hence $f=0$. (iii) given any element of the quotient $\sum a_{nm} x^{n-m}$ say, this is equal to some polynomial in $x$ times $x^{-d}$, with $d$ the $y$ degree of the initial element. thanks! –  Jan 29 '14 at 20:47