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Thread: Show that the affine plane without the origin is not isomorphic to the affine plane

  1. #1
    Nov 2010

    Show that the affine plane without the origin is not isomorphic to the affine plane

    For anyone unfamiliar with algebraic geometry, let me just define a couple of things. (This is fairly basic geometry, as far as that goes; no scheme theory or anything here.) All of our points will come from some algebraically closed field $\displaystyle k$. Let $\displaystyle X,Y$ be two varieties (affine, quasi-affine, projective, quasi-projective; doesn't matter). A regular function on $\displaystyle Y$ is a mapping $\displaystyle \frac{g}{h}:Y\longrightarrow k$ such that $\displaystyle \frac{g}{h}$ is defined at every point in $\displaystyle Y$. A morphism is a continuous (with respect to the Zariski topology) function $\displaystyle \phi:X\longrightarrow Y$ with the property that, for any regular function $\displaystyle f:Y\longrightarrow k$, the composition $\displaystyle f\circ \phi:X\longrightarrow k$ is a regular function on $\displaystyle X$. An isomorphism of varieties is just a morphism for which there exists an inverse morphism.

    The rings of regular functions on $\displaystyle X,Y$ are denoted $\displaystyle \mathcal{O}(X),\mathcal{O}(Y)$ respectively.

    Now, let $\displaystyle X=\mathbb{A}^2\setminus\{(0,0)\}$ (which is a quasi-affine variety).

    The problem: show that $\displaystyle X\ncong \mathbb{A}^2$. (in which case we would say $\displaystyle X$ is not affine).

    First, just a couple of facts:

    1. I have shown that $\displaystyle \mathcal{O}(X)=\mathcal{O}(\mathbb{A}^2)=k[x,y]$.

    2. We have this theorem: if $\displaystyle X$ is any variety and $\displaystyle Y$ an affine variety, then there is a bijection between the set of morphisms from $\displaystyle X$ to $\displaystyle Y$ and $\displaystyle k$-algebra homomorphisms from $\displaystyle \mathcal{O}(Y)$ to $\displaystyle \mathcal{O}(X)$. This correspondence is given by $\displaystyle \phi\leftrightarrow\tilde{\phi}$ where, if $\displaystyle f\in \mathcal{O}(Y)$, we have $\displaystyle \tilde{\varphi}(f)=f\circ\varphi$. Under this, isomorphisms of varieties correspond to $\displaystyle k$-algebra isomorphisms.


    So, number (2) says that if there were an isomorphism of varieties $\displaystyle \varphi:X\rightarrow\mathbb{A}^2$, there would be a corresponding $\displaystyle k$-algebra isomorphism $\displaystyle \tilde\varphi$, or equivalently, a corresponding $\displaystyle k-$algebra automorphism of $\displaystyle k[x,y]$. This is where I'm stuck; showing that there is no such $\displaystyle k$-algebra automorphism which would give rise to an isomorphism on the varieties. I found something online which says

    " isomorphisms would have to correspond to automorphisms of $\displaystyle k[x,y]$ , but this is just the set of invertible linear transformations of $\displaystyle x$ and $\displaystyle y$; none of these yield an isomorphism."

    I guess the first thing I don't quite get is what is meant by an invertible linear transformation... a function of the form $\displaystyle x\mapsto p(x,y),y\mapsto q(x,y)$? And, either way, how to prove that each invertible linear transformation (whatever they are) cannot possibly give a corresponding isomorphism on the varieties.
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  2. #2
    Nov 2010
    Algebraic geometry can be kinda tricky, so it's not surprising if there aren't many people around who could help with this sort of stuf...

    Even if you can't answer this question, does anyone know a good site I could go to with lots of algebraic geometry resources, maybe worked problems and stuff. I don't usually have trouble picking stuff up pretty easily, especially in algebra... but I really struggle with this and I don't know why.

    Doesn't help that I have to take a preliminary exam on this stuff on Tuesday lol..
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