explanation on a scheme which is not affine scheme - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:52:33Z http://mathoverflow.net/feeds/question/87480 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87480/explanation-on-a-scheme-which-is-not-affine-scheme explanation on a scheme which is not affine scheme Keivan Monfared 2012-02-03T21:39:05Z 2012-02-04T13:04:26Z <p>Hartshorne at the end of page 76 of his Algebraic Geometry book gives an example of a scheme which is not an affine scheme. The scheme is constructed by gluing two affine lines together along their maximal ideals obtained by removing a point P. There's also a figure accompanying the example: </p> <p><em><strong></em>__<em>_</em>__<em>_</em>__<em>_</em>__</strong>:<em><strong></em>__<em>_</em>__<em>_</em>___</strong></p> <p>Can someone please explain how to show that this is not an affine scheme?</p> http://mathoverflow.net/questions/87480/explanation-on-a-scheme-which-is-not-affine-scheme/87482#87482 Answer by Keivan Monfared for explanation on a scheme which is not affine scheme Keivan Monfared 2012-02-03T21:41:29Z 2012-02-03T21:41:29Z <p>Is that only because it's not Hausdorff?</p> http://mathoverflow.net/questions/87480/explanation-on-a-scheme-which-is-not-affine-scheme/87498#87498 Answer by Charles Staats for explanation on a scheme which is not affine scheme Charles Staats 2012-02-04T00:56:23Z 2012-02-04T00:56:23Z <p>Compute the ring $R$ of globally defined regular functions. If the scheme were affine, then there would be a bijective correspondence between closed points of the scheme and maximal ideals of $R$, given by taking a closed point to the ideal of functions that vanish at that point. But the two points represented by the colon in your diagram give the same ideal of $R$, so this "correspondence" is not injective. Therefore, the scheme is not affine.</p> http://mathoverflow.net/questions/87480/explanation-on-a-scheme-which-is-not-affine-scheme/87516#87516 Answer by Georges Elencwajg for explanation on a scheme which is not affine scheme Georges Elencwajg 2012-02-04T09:43:51Z 2012-02-04T13:04:26Z <p>Call $X$ your scheme over the field $k$, $P_1$ and $P_2$ the two special closed points, $A_1$and $A_2$ their respective open complements and $A_{12}=A_1\cap A_2$, so that $A_i\simeq \mathbb A^1_k$ and $A_{12}\simeq\mathbb G_m$, all affine schemes.<br> Here are some (not independent) proofs that $X$ is not affine. </p> <p><strong>Proof 1</strong><br> The point $(P_1,P_2)\in X \times X$ is in the closure of the diagonal $\Delta_X\subset X \times X$, but $(P_1,P_2)\notin \Delta_X$ . So $\Delta_X$ is not closed, hence $X$ is not separated and <em>a fortiori</em> not affine </p> <p><strong>Proof 2</strong><br> The images of the restriction map $\Gamma(A_i,\mathcal O_X)=k[T] \to \Gamma(A_{12},\mathcal O_X)=k[T,T^{-1}]$ are both<br> $k[T]$, and together do not generate $k[T,T^{-1}]$. However, in an affine scheme (or more generally in a separated scheme) the ring of regular sections on the intersection of two open affines <em>is</em> generated by the images of the regular sections on the two opens. </p> <p><strong>Proof 3</strong><br> The two open immersions $\iota_j:\mathbb A^1_k \to X$ with respective image $A_j\subset X$ coincide on the open subscheme $\mathbb G_m\subset \mathbb A^1_k$ but are nevertheless distinct. This couldn't happen if $X$ were affine (or just separated). </p> <p><strong>Proof 4</strong><br> The cohomology vector space $H^1(X,\mathcal O_X)$ is infinite dimensional, whereas the cohomology of a coherent sheaf on an affine scheme vanishes in positive degree.<br> In detail, consider the covering $\mathcal U=\lbrace A_1,A_2\rbrace$ of $X$. It is a Leray covering because $A_1,A_2,A_{12}$ are affine hence acyclic, for the coherent sheaf $\mathcal O_X$ (cf. Proof 2) . Thus Čech cohomology computes genuine cohomology.<br> The Čech complex is the linear map $$C^0=\Gamma(A_1,\mathcal O_X)\times \Gamma(A_2,\mathcal O_X)=k[T]\times k[T]\stackrel {d^0}{\to} C^1=\Gamma(A_{12},\mathcal O^*_X)=k[T,T^{-1}]\to 0$$ given by $$d^0(P(T),Q(T)) =Q(T)-P(T)$$.<br> Hence we get $H^1(X,\mathcal O_X)=k[T,T^{-1}]/k[T]$ </p> <p><strong>Proof 5</strong><br> The Čech complex above proves that $\Gamma(X,\mathcal O_X)=k[T]$ so that the restriction to the strictly smaller open affine subscheme $A_1\subsetneq X$ is bijective: $res: \Gamma(X,\mathcal O_X)\stackrel {\simeq}{\to} \Gamma(A_1,\mathcal O_X)$.<br> This cannot happen for an affine scheme $X$.<br> [In categorical language: $\Gamma$ is an anti-equivalence from the category of affine schemes to that of rings] </p> <p><strong>Proof 6</strong><br> Every global function $P(T)\in \Gamma(X,\mathcal O_X)=k[T]$ (see Proof 5) takes the exact same value at $P_1$ and $P_2$, namely $P(0)\in \kappa(P_1)=\kappa (P_2)=k$.<br> In contrast given two closed points in an affine scheme , there exists a global regular function vanishing at the first one but not at the second. </p>