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New article: Inductive lattices
author | Dylan Holmes <ocsenave@gmail.com> |
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date | Tue, 01 Nov 2011 01:55:26 -0500 |
parents | b4de894a1e2e |
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Algebraically,184 this amounts to saying that each line is not only a commutative ring185 but a <b>field</b>, as well. This marks our first departure from euclidean186 geometry, as our first axiom denies that each line is a field.187 </p>190 </div>192 <div id="outline-container-1-1" class="outline-3">193 <h3 id="sec-1-1"><span class="section-number-3">1.1</span> The first anti-euclidean axiom </h3>194 <div class="outline-text-3" id="text-1-1">196 <p>A point in a ring is called <b>nilpotent</b> if its square is197 zero. Normally (that is, in \(\mathbb{R}^n\)), only \(0\) is198 nilpotent. Here, as a consequence of the following axiom, there will199 exist other elements that are nilpotent. These elements will200 encapsulate our intuitive idea of “infinitesimally small” numbers.201 </p>202 <blockquote>204 <p><b>Axiom 1:</b> Let \(R\) be the line, considered as a commutative ring, and205 let \(D\subset R\) be the set of nilpotent elements on the line. Then for any206 morphism \(g:D\rightarrow R\), there exists a unique \(b\in R\) such that207 </p>210 \(\forall d\in D, g(d) = g(0)+ b\cdot d\)212 <p>213 Intuitively, this unique \(b\) is the slope of the function \(g\) near214 zero. Because every morphism \(g\) has exactly one such \(b\), we have the215 following results:216 </p>217 <ol>218 <li>The set \(D\) of nilpotent elements contains more than219 just 0. Indeed, suppose the contrary: if \(D=\{0\}\), then for any \(g\), <i>every</i> \(b\in R\) has the220 property described above;—\(b\) isn't uniquely defined.221 </li>222 <li>Pick \(b_1\) and \(b_2\) in \(R\). If every nilpotent \(d\) satisfies \(d\cdot223 b_1 = d\cdot b_2\), then \(b_1\) and \(b_2\) are equal.224 </li>225 </ol>228 </div>230 </div>232 <div id="outline-container-1-2" class="outline-3">233 <h3 id="sec-1-2"><span class="section-number-3">1.2</span> The first axiom \(\ldots\) in terms of arrows </h3>234 <div class="outline-text-3" id="text-1-2">237 <p>238 Define \(\xi:R\times R\rightarrow R^D\) by \(\xi:(a,b)\mapsto (d\mapsto239 a+b\cdot d)\). The first axiom is equivalent to the statement240 “ξ is invertible (i.e., a bijection)”241 </p>242 <p>243 We give \(R\times R\) the structure of an \(R\)-algebra by defining244 multiplication: \( (a_1,b_1)\star(a_2,b_2) = (a_1\cdot a_2,\quad245 a_1\cdot b_2 + a_2\cdot b_1)\). This is called <b>dual-numbers multiplication</b>, and is similar to muliplication of complex numbers.246 </p>248 </div>250 </div>252 <div id="outline-container-1-3" class="outline-3">253 <h3 id="sec-1-3"><span class="section-number-3">1.3</span> Ex </h3>254 <div class="outline-text-3" id="text-1-3">256 <ol>257 <li>If \(a\) and \(b\) are nilpotent, then \(ab\) is nilpotent.258 </li>259 <li>Even if \(a\) and \(b\) are nilpotent, the sum \(a+b\) may not be.260 </li>261 <li>Even if \(a+b\) is nilpotent, either summand \(a\), \(b\) may not be.262 </li>263 <li>264 </li>265 </ol>269 </blockquote>271 </div>272 </div>273 </div>274 <div id="postamble">275 <p class="date">Date: 2011-08-15 22:42:41 EDT</p>276 <p class="author">Author: Dylan Holmes</p>277 <p class="creator">Org version 7.6 with Emacs version 23</p>278 <a href="http://validator.w3.org/check?uri=referer">Validate XHTML 1.0</a>279 </div>280 </div>281 </body>282 </html>