At the moment I'm trying to understand the concept of localizations of rings / modules. I have done some exercises (using the book of Atiyah / MacDonald) and I will do some more, but a more practical question took my attention.

I started with an easy example and considered the ring $k[x]_x$. Localizing at $x$ means that $x$ and its powers become units in the localization. Hence $$k[x]_x\cong k[x,x^{-1}]\cong k[x,y]/\langle xy-1\rangle,$$ am I right there for a start?

Then I wanted to compute the localization of $k[x,y]/\langle xy-1\rangle$ at $\langle x-1,y-1\rangle$. This should be a maximal ideal in the quotient, since $\langle xy-1\rangle\subset\langle x-1,y-1\rangle$. (Geometrically, what does this mean here? The maximal ideal corresponds to the point $p=(1,1)$ on the hyperbola $xy-1=0$, so we "gather only local information around $p$" in some way?)

Localization commutes with the quotient, thus $$(k[x,y]/\langle xy-1\rangle)_{\langle x-1,y-1\rangle}\cong k[x,y]_{\langle x-1,y-1\rangle}/\langle xy-1\rangle_{\langle x-1,y-1\rangle},$$ and here already I am stuck. Is there a general way to compute localized rings of this form, or at least some plan that often works in a case like this? Edit: Can we guess what this should be isomorphic to when looking at it geometrically?

Thank you for your help / hints in advance!


Everything you wrote is correct and you have the right attitude : aggressively attacking simple examples and trying to see what is going on geometrically.

As for localization, you can use your last isomorphism if you want: localization does indeed commute with quotients.
But it is simpler to exploit your preceding isomorphism $ k[x,y]/\langle xy-1\rangle \cong k[x,x^{-1}]$ instead and to write $$(k[x,y]/\langle xy-1\rangle)_{\langle x-1,y-1\rangle}\cong k[x,x^{-1}]_{{\langle x-1,x^{-1}-1\rangle}}=k[x,x^{-1}]_{{\langle x-1\rangle}}=k[x]_{{\langle x-1\rangle}} =k[x-1]_{{\langle x-1\rangle}}$$

Geometrical interpretation (very important!)
You are studying a hyperbola in the plane near the point $(1,1)$ by projecting it on the $x$-axis and you obtain an isomorphism with the affine line punctured at zero.
The projection sends $(1,1)$ to the point $x=1$ on the punctured line and the local ring $(k[x,y]/\langle xy-1\rangle)_{\langle x-1,y-1\rangle}$ of the hyperbola at $(1,1)$ isomorphically to the local ring of the line at $1$, namely $k[x-1]_{{\langle x-1\rangle}}$.
Note carefully that the puncture of the affine line plays no role in these local questions: the point $x=1$ only sees its immediate vicinity and doesn't care what happens at $x=0$.

  • 1
    $\begingroup$ I wish to check if I understood every equality here: (1) is due to $x^{-1}-1=-x^{-1}(x-1)$; (2) holds since $x$ is invertible in the localization, with inverse being $\frac{x-1}{x(x-1)}\in k[x]_{\langle x-1\rangle}$!? I don't really get the last one, it's not clear to me why you can just exchange $x$ with $x-1$ there, but not in the ideal. But that may be because the object $k[x-1]$ confuses me a bit. Thanks for your answer! $\endgroup$ – InvisiblePanda Feb 9 '12 at 10:04
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    $\begingroup$ Dear @Rand:yes, (1) is what you say. For (2) just remember that $k[x]_{{\langle x-1\rangle}}$ contains every fraction that can be written with a denominator not divisible by $x-1$. Obviously $1/x$ qualifies. Finally you have $k[x]=k[x-a]$ for every $a\in k$: this has nothing to do with localization but results from the calculation $x^n=((x-a)+a)^n=(x-a)^n+...+a^n$, etc. $\endgroup$ – Georges Elencwajg Feb 9 '12 at 10:32
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    $\begingroup$ Hello @Georges! Oh, seems like I confused localizing at $x-1$ with localizing at $\langle x-1\rangle$. Of course you are right, $x-1$ and multiples are explicitly not allowed as denominators, so it is clear that $\frac{1}{x}$ lies in there. Thank you very much for your help! I guess I'll post some more questions on math.stackexchange as they occur, I like it here already ;) $\endgroup$ – InvisiblePanda Feb 9 '12 at 10:39
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    $\begingroup$ Dear @Rand: I'm happy you already like it here. A very warm "welcome" in the name of all of us! $\endgroup$ – Georges Elencwajg Feb 9 '12 at 16:14

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