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I'm currently stuck at problem 1.1 c) in Hartshorne's algebraic geometry book. I just can't let it go. Setting is as title says (field $k$, variables $x$ and $y$).

Problem 1.1. a) and b) concerns themselves with the special cases $y-x^2$ and $xy - 1$, and classifying the resulting quotient rings (being isomorphic to a polynomial ring in one variable over $k$ in both cases, but allowing negative exponents in the second).

c) asks of me to prove that these are the only two possible outcomes, up to isomorphism. And I just can't. Any help would be appreciated.

A related question I came up with was, if we're in case b), since any element in $k$ can be inverted, and x can be inverted, then surely, any element in the ring can be converted, and we have a field. Is this so?

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It has been two years since I had commutative algebra. I'm sure this is really obvious, I just don't have the right theorems and results in my head at the moment, and I have no access to Atiyah-McDonald either. –  Arthur Sep 3 '11 at 4:19
    
Could anyone explain to me the doubt as shown in my comment below, or should I ask a new question? –  awllower Jun 10 at 5:33
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1 Answer 1

up vote 3 down vote accepted

An irreducible quadratic corresponds to a nonsingular conic in $\mathbb{P}^2_k$.

Choose three points on the conic, and select projective coordinates so that these points are $(1:0:0)$, $(0:1:0)$, and $(0:0:1)$. In these coordinates, the conic has equation $cxy + ayz+bxz=0$ for some nonzero $a,b,c$ (if any of $a$, $b$, or $c$ are zero, then the conic is singular). Rescaling by multiplying $x$ by $a$, $y$ by $b$, and $z$ by $c$ gives that the conic can be written with equation $xy+yz+xz = 0$.

This conic is isomorphic to the projective line $\mathbb{P}^1_k$.

The regular functions on the conic $xy+yz+xz=0$ minus some hyperplane are therefore equivalent to regular functions on $\mathbb{P}^1_k$ minus one or two points. The regular functions on the projective line minus a single point yield $k[x]$, and the regular functions on the projective line minus two points yield $k[x,x^{-1}]$. You get the latter if and only if the conic intersects the line at infinity in exactly one point.

Added. I missed your "related" question. No, it is not true that $k[x,x^{-1}]$ is a field: it's the ring of Laurent polynomials with coefficients in $k$; i.e., expressions of the form $$a_{-m}x^{-m} + \cdots + a_{-1}x^{-1} + a_0 + a_1x + \cdots + a_nx^n$$ for some nonnegative integers $m$ and $n$. But this is not a field. For instance, $x-1$ cannot be inverted in $k[x,x^{-1}]$.

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Not exactly as simple as I had hoped, but I get the basic idea. I do think, however, that projective-just-about-anything are introduced 3 pages after the problem in the book, so I believe there's a more purely ring-theoretic proof to it. About the related question, I knew the inversion couldn't be done in $k[x]$, since that inverse would be the sum of an infinite geometric series. However, the addition of an inverted $x$ made it a bit less intuitive. –  Arthur Sep 3 '11 at 6:02
    
It's a classical problem and there are lots of ways to go about it. If you don't want to talk about geometry at all, the brief summary is "do a change of variable to eliminate terms in the general quadratic equation". –  Hurkyl Sep 3 '11 at 7:34
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@Arthur: Like Hurkyl says: you can always do a change of variable to bring the quadratic into "standard form", of which there are two possibilities. One possibility leads to the polynomial ring, the other to the ring of Laurent polynomials. This is how I solved the problem back when I took the course; it was graded correct (-; (Then again, though Hartshorne was teaching the course, he wasn't doing the grading, so this may not be what he had in mind) –  Arturo Magidin Sep 3 '11 at 17:40
    
Could anyone explain why the conic $xy+yz+xz=0$ is isomorphic to the projective line $\mathbb P_k^1$? I failed to find an obvious isomorphism. Also, I cannot understand why we consider the regular functions of the conic minus some hypersurface: what does it have to do with our question? Thanks in advance! –  awllower Jun 6 at 8:40
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