{"id":492,"date":"2025-02-13T23:02:57","date_gmt":"2025-02-13T15:02:57","guid":{"rendered":"https:\/\/qwq.cafe\/?p=492"},"modified":"2025-02-27T18:20:10","modified_gmt":"2025-02-27T10:20:10","slug":"%e6%a4%ad%e5%9c%86%e6%9b%b2%e7%ba%bf%e5%88%9d%e6%8e%a2","status":"publish","type":"post","link":"https:\/\/qwq.cafe\/?p=492","title":{"rendered":"\u692d\u5706\u66f2\u7ebf\u521d\u63a2"},"content":{"rendered":"

\u66fe\u7ecf\u5728\u770b\u4e00\u7bc7\u79d1\u666e\u6587\u7ae0\u65f6\u5f97\u77e5\u4e86\u692d\u5706\u66f2\u7ebf\u7684 Weierstrass \u5f62\u5f0f, \u4ece\u90a3\u4e4b\u540e\u4fbf\u5bf9\u692d\u5706\u66f2\u7ebf\u4ea7\u751f\u4e86\u5f88\u5f3a\u7684\u5174\u8da3. \u5982\u4eca\u7ec8\u4e8e\u5b66\u4e60\u4e86\u53ef\u4ee5"\u4e0e\u4e4b\u4e00\u6218"\u7684\u6570\u5b66\u5de5\u5177—Riemann-Roch \u5b9a\u7406. \u672c\u6587\u5c06\u8bc1\u660e\u692d\u5706\u66f2\u7ebf\u90fd\u540c\u6784\u4e8e\u5f62\u5982 $Y^2T=X(X-T)(X-\\lambda T)$ \u7684\u65b9\u7a0b\u5728 $P^2$ \u4e2d\u51b3\u5b9a\u7684\u66f2\u7ebf.<\/p>\n

<\/p>\n

\u5b9e\u9645\u4e0a, \u672c\u6587\u662f\u5728\u5c1d\u8bd5\u7ed9\u51fa Daniel Perrin \u6240\u8457\u7684 Algebraic Geometry – An Introduction<\/em> \u5728\u7b2c VIII \u7ae0\u7684\u7b2c 4 \u9053\u4e60\u9898\u7684\u4e00\u4e2a\u89e3\u7b54. \u672c\u6587\u4e2d\u4e0d\u52a0\u89e3\u91ca\u6240\u4f7f\u7528\u7684\u7b26\u53f7\u5747\u4e0e\u4e66\u4e2d\u76f8\u540c. \u672c\u89e3\u7b54\u53ef\u80fd\u6709\u8bef, \u5982\u679c\u53d1\u73b0\u6b22\u8fce\u6307\u51fa.<\/p>\n

\u672c\u6587\u4e2d\u7ea6\u5b9a $k$ \u662f\u4e00\u4e2a\u7279\u5f81\u4e0d\u4e3a 2 \u7684\u4ee3\u6570\u95ed\u57df, $P^2$ \u4e3a $k$ \u4e0a\u7684\u5c04\u5f71\u5e73\u9762, $\\mathcal O_{P^2}$ \u4e3a\u5176\u4e0a\u7684\u5c42.<\/p>\n

\u9996\u5148, \u6211\u4eec\u7ed9\u51fa\u692d\u5706\u66f2\u7ebf\u7684\u5b9a\u4e49.<\/p>\n

Def.<\/strong> \u692d\u5706\u66f2\u7ebf\u4e3a\u4e8f\u683c\u4e3a 1 \u7684\u4e0d\u53ef\u7ea6\u5149\u6ed1\u5c04\u5f71\u66f2\u7ebf.<\/p>\n

\u6211\u4eec\u9700\u8981\u4f7f\u7528\u7684 Riemann-Roch \u5b9a\u7406\u7684\u5f62\u5f0f\u5982\u4e0b:<\/p>\n

Thm (Riemann-Roch).<\/strong> \u8bbe $C$ \u662f\u4e00\u4e2a\u4e8f\u683c\u4e3a $g$ \u7684\u4e0d\u53ef\u7ea6\u5149\u6ed1\u5c04\u5f71\u66f2\u7ebf, $D$ \u4e3a $C$ \u4e0a\u7684\u9664\u5b50, \u5219\u6211\u4eec\u6709
\n$$h^0\\mathcal O_C(D)-h^1\\mathcal O_C(D)=\\deg(D)+1-g.
\n$$
\n\u63a5\u4e0b\u6765\u6211\u4eec\u5f00\u59cb\u8fdb\u884c\u8bc1\u660e. \u4efb\u53d6\u4e00\u70b9 $P_0\\in C$, \u8bb0 $C'=C-\\{P_0\\}$. \u540e\u6587\u6211\u4eec\u8bb0 $v$ \u4e3a\u79bb\u6563\u8d4b\u503c\u73af $\\mathcal O_{C,P_0}$ \u4e0a\u7684\u8d4b\u503c\u51fd\u6570.<\/p>\n

Step 1.<\/strong> \u8bc1\u660e $\\forall n\\in \\mathbb N^*,h^0\\mathcal O_C(nP_0)=n$.<\/p>\n

Proof.<\/p>\n

\u7531 Riemann-Roch \u516c\u5f0f\u5f97
\n$$\\begin{align}
\nh^0\\mathcal O_C(nP_0)&=h^1\\mathcal O_C(nP_0)+\\deg(nP_0)+1-g\\\\
\n&=h^1\\mathcal O_C(nP_0)+n+1-1\\\\
\n&=h^1\\mathcal O_C(nP_0)+n.
\n\\end{align}
\n$$
\n\u7531\u4e8e $n>0=2g-2$, \u4ece\u800c $h^1\\mathcal O_C(nP_0)=0$, \u56e0\u6b64 $h^0\\mathcal O_C(nP_0)=n$. $\\square$<\/p>\n

Step 2.<\/strong> \u8bc1\u660e\u5b58\u5728 $x,y\\in K(C)$, \u4f7f\u5f97 $1,x$ \u6784\u6210 $H^0\\mathcal O_C(2P_0)$ \u7684\u4e00\u7ec4\u57fa, $1,x,y$ \u6784\u6210 $H^0\\mathcal O_C(3P_0)$ \u7684\u4e00\u7ec4\u57fa (\u4f5c\u4e3a $k$ – \u7ebf\u6027\u7a7a\u95f4).<\/p>\n

Proof.<\/p>\n

\u7531\u4e8e $h^0\\mathcal O_C(2P_0)=2$, \u4e14 $1\\in H^0\\mathcal O_C(2P_0)$, \u56e0\u6b64\u5b58\u5728 $x\\in H^0\\mathcal O_C(2P_0)$ \u4f7f\u5f97 $1,x$ \u6784\u6210 $H^0\\mathcal O_C(2P_0)$ \u7684\u4e00\u7ec4\u57fa. \u53c8\u7531\u4e8e $h^0\\mathcal O_C(P_0)=1$, \u6211\u4eec\u5f97 $v(x)$ \u53ea\u80fd\u4e3a $-2$. <\/p>\n

\u7531\u4e8e $1,x$ \u5728 $H^0\\mathcal O_C(3P_0)$ \u4e2d\u7ebf\u6027\u65e0\u5173, \u4e14 $h^0\\mathcal O_C(3P_0)=3$, \u4ece\u800c\u5b58\u5728 $y\\in H^0\\mathcal O_C(3P_0)$ \u4f7f\u5f97 $1,x,y$ \u6784\u6210 $H^0\\mathcal O_C(3P_0)$ \u7684\u4e00\u7ec4\u57fa. \u800c\u4e14\u5bb9\u6613\u5f97\u5230 $v(y)$ \u53ea\u80fd\u4e3a $-3$. $\\square$<\/p>\n

Step 3.<\/strong> \u8bc1\u660e $1,x,y,x^2,xy,y^2,x^3$ \u5173\u4e8e $k$ \u662f\u7ebf\u6027\u76f8\u5173\u7684. \u4ee4 $P(x,y)$ \u4e3a\u8fd9\u4e2a\u76f8\u5173\u5173\u7cfb. \u8bc1\u660e\u5728 $P(x,y)$ \u4e2d $y^2$ \u548c $x^3$ \u7684\u7cfb\u6570\u90fd\u4e0d\u4e3a $0$.<\/p>\n

Proof.<\/p>\n

\u7531\u4e8e $v(x^2)=-4,v(xy)=-5,v(y^2)=-6,v(x^3)=-6$, \u4ece\u800c $1,x,y,x^2,xy,y^2,x^3\\in H^0\\mathcal O_C(6P_0)$.<\/p>\n

\u53c8\u7531\u4e8e $h^0\\mathcal O_C(6P_0)=6$, \u4ece\u800c $1,x,y,x^2,xy,y^2,x^3$ \u7ebf\u6027\u76f8\u5173.<\/p>\n

\u5047\u8bbe $1,x,y,x^2$ \u7ebf\u6027\u76f8\u5173. \u7531\u4e8e $1,x,y$ \u7ebf\u6027\u65e0\u5173, \u5219\u5b58\u5728 $a_1,a_2,a_3\\in k$ \u4f7f\u5f97 $x^2=a_1+a_2x+a_3y$, \u4f46\u662f $v(a_1+a_2x+a_3y)\\ge \\inf\\{v(1),v(x),v(y)\\}=-3$, \u4e0e $v(x^2)=-4$ \u4ea7\u751f\u77db\u76fe. \u56e0\u6b64 $1,x,y,x^2$ \u7ebf\u6027\u65e0\u5173, \u540c\u7406\u53ef\u5f97, $1,x,y,x^2,xy$ \u7ebf\u6027\u65e0\u5173. <\/p>\n

\u7531\u4e8e $1,x,y,x^2,xy$ \u7ebf\u6027\u65e0\u5173, \u5728 $P(x,y)$ \u4e2d $y^2$ \u548c $x^3$ \u7684\u7cfb\u6570\u4e0d\u80fd\u540c\u65f6\u4e3a $0$. \u518d\u4e0e\u4e0a\u9762\u540c\u7406\u53ef\u5f97 $y^2$ \u548c $x^3$ \u7684\u7cfb\u6570\u90fd\u4e0d\u80fd\u4e3a $0$. $\\square$<\/p>\n

Step 4.<\/strong> \u8bc1\u660e\u5728\u57fa\u53d8\u6362\u7684\u610f\u4e49\u4e0b, \u6211\u4eec\u53ef\u4ee5\u5047\u5b9a $P(x,y)$ \u5177\u6709 $y^2-x(x-1)(x-\\lambda)$ \u7684\u5f62\u5f0f, \u5176\u4e2d $\\lambda\\ne 0,1$.<\/p>\n

Proof.<\/p>\n

\u4e0d\u59a8\u8bbe $P(x,y)=y^2+2a_0y+a_0^2+a_1xy-f(x)$, \u5176\u4e2d $a_0,a_1\\in k$, $f(x)$ \u662f\u5173\u4e8e $x$ \u7684 $3$ \u6b21\u591a\u9879\u5f0f.<\/p>\n

\u5219\u901a\u8fc7\u4ee3\u6362 $y'=y+a_0$, \u6211\u4eec\u53ef\u4ee5\u5047\u5b9a $P(x,y)$ \u5177\u6709 $y^2+a_1xy-f(x)$ \u7684\u5f62\u5f0f, \u5176\u4e2d $f(x)$ \u662f\u5173\u4e8e $x$ \u7684 $3$ \u6b21\u591a\u9879\u5f0f..<\/p>\n

\u518d\u901a\u8fc7\u4ee3\u6362 $y'=y+(a_1\/2)x$, \u6211\u4eec\u53ef\u4ee5\u5047\u5b9a $P(x,y)$ \u5177\u6709 $y^2-f(x)$ \u7684\u5f62\u5f0f, \u5176\u4e2d $f(x)$ \u662f\u5173\u4e8e $x$ \u7684 $3$ \u6b21\u591a\u9879\u5f0f.<\/p>\n

\u8981\u8bc1\u53ef\u4ee5\u901a\u8fc7 $x'=ax+b$ \u5f62\u5f0f\u7684\u53d8\u6362\u5c06 $P(x,y)$ \u53d8\u4e3a\u9898\u76ee\u6240\u6c42\u7684\u5f62\u5f0f, \u53ea\u9700\u8981\u8bc1\u660e $f(x)$ \u6ca1\u6709\u91cd\u6839.<\/p>\n

\u4e0d\u59a8\u8bbe $0$ \u662f $f$ \u7684\u91cd\u6839. \u7531\u4e8e $P_0$ \u662f $x$ \u7684\u6781\u70b9, \u56e0\u6b64\u5fc5\u7136\u5b58\u5728 $Q\\in C',x(Q)=0$. \u518d\u7531 $P(x,y)\\equiv 0$ \u53ef\u5f97 $y(Q)=0$.<\/p>\n

\u7531\u4e8e $\\Gamma(C',\\mathcal O_C)$ \u4e3a\u6700\u591a\u53ea\u6709 $P_0$ \u8fd9\u4e00\u4e2a\u6781\u70b9\u7684\u6709\u7406\u51fd\u6570\u6784\u6210\u7684\u73af, \u53c8\u56e0\u4e3a $h^0\\mathcal O_C(nP_0)=n$, \u4ece\u800c $x,y$ \u6784\u6210\u5176\u4f5c\u4e3a $k$ – \u4ee3\u6570\u7684\u751f\u6210\u5143. \u518d\u7ed3\u5408\u4e4b\u524d\u7684\u7ebf\u6027\u76f8\u5173\u6027\u53ef\u77e5 $\\Gamma(C',\\mathcal O_C)\\cong k[X,Y]\/(Y^2-f(X))$.<\/p>\n

\u5219\u5728 $Q$ \u70b9\u5904, \u6211\u4eec\u6709
\n$$\\frac{\\partial P}{\\partial x}=f'(x)=0, \\frac{\\partial P}{\\partial y}=2y=0.
\n$$
\n\u56e0\u6b64\u5728\u5207\u7a7a\u95f4 $T_QC$ \u4e2d $\\frac{\\partial}{\\partial x}$ \u548c $\\frac\\partial {\\partial y}$ \u7ebf\u6027\u65e0\u5173, \u4ece\u800c $\\dim T_QC\\ge 2$ \u4e0e\u66f2\u7ebf\u5149\u6ed1\u77db\u76fe. \u56e0\u6b64 $f(x)$ \u6ca1\u6709\u91cd\u6839. $\\square$<\/p>\n

Step 5.<\/strong> \u8003\u8651\u6620\u5c04 $\\varphi:C'\\to k^2,Q\\mapsto (x(Q),y(Q))$. \u8bc1\u660e $\\varphi$ \u662f $C'$ \u4e0e\u65b9\u7a0b $Y^2=X(X-1)(X-\\lambda)$ \u5728 $k^2$ \u4e2d\u51b3\u5b9a\u7684\u66f2\u7ebf\u7684\u540c\u6784. \u800c\u56e0\u6b64 $\\varphi$ \u53ef\u4ee5\u5ef6\u62d3\u6210 $C$ \u4e0e\u65b9\u7a0b $Y^2T=X(X-T)(X-\\lambda T)$ \u5728 $P^2$ \u4e2d\u51b3\u5b9a\u7684\u66f2\u7ebf\u7684\u540c\u6784.<\/p>\n

Proof.<\/p>\n

$\\varphi$ \u8bf1\u5bfc\u4e86\u540c\u6001 $\\varphi^*:k[X,Y]\/(Y^2-X(X-1)(X-\\lambda))\\to \\Gamma(C',\\mathcal O_C)$, \u6613\u5f97 $\\varphi^*(X)=x,\\varphi^*(Y)=y$, \u7ed3\u5408\u7b2c 3 \u6b65\u7684\u7ebf\u6027\u76f8\u5173\u6027\u53ef\u77e5 $\\varphi^*$ \u4e3a\u73af\u540c\u6784. \u56e0\u6b64 $\\varphi$ \u662f $C'$ \u4e0e\u65b9\u7a0b $Y^2=X(X-1)(X-\\lambda)$ \u5728 $k^2$ \u4e2d\u51b3\u5b9a\u7684\u66f2\u7ebf\u7684\u540c\u6784. $\\square$<\/p>\n","protected":false},"excerpt":{"rendered":"

\u66fe\u7ecf\u5728\u770b\u4e00\u7bc7\u79d1\u666e\u6587\u7ae0\u65f6\u5f97\u77e5\u4e86\u692d\u5706\u66f2\u7ebf\u7684 Weierstrass \u5f62\u5f0f, \u4ece\u90a3\u4e4b\u540e\u4fbf\u5bf9\u692d\u5706\u66f2\u7ebf\u4ea7\u751f\u4e86\u5f88\u5f3a\u7684\u5174\u8da3. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[14,11,8],"class_list":["post-492","post","type-post","status-publish","format-standard","hentry","category-6","tag-14","tag-11","tag-8"],"_links":{"self":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/492","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=492"}],"version-history":[{"count":3,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/492\/revisions"}],"predecessor-version":[{"id":495,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/492\/revisions\/495"}],"wp:attachment":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=492"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=492"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=492"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}