Characterizing elementary embeddings of $L$ and $L_\alpha$ under 0# - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:13:28Z http://mathoverflow.net/feeds/question/73251 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73251/characterizing-elementary-embeddings-of-l-and-l-alpha-under-0 Characterizing elementary embeddings of $L$ and $L_\alpha$ under 0# Norman Lewis Perlmutter 2011-08-19T23:38:05Z 2011-08-21T19:20:46Z <p>Suppose 0# exists.</p> <p>It is clear that every order preserving map from the indiscernibles to the indiscernibles gives an elementary embedding from $L$ to $L$. Furthermore, following lemmas 18.7 and 18.8 of Jech, if $\alpha$ is an <i> <strike>infinite </strike> infinite limit</i> ordinal, an increasing map from alpha to beta gives an elementary embedding from $L_{i_\alpha}$ to $L_{i_\beta}$, where $i_\alpha$ is the $\alpha$-th indiscernible. This is because $L_{i_\alpha}$ equals the Skolem hull in itself of the first $\alpha$ indiscernibles. However, I am not clear on the following points.</p> <p>1) Is it the case that for a <i><strike>finite</strike> successor</i> ordinal, n, $L_{i_n}$ is necessarily equal to the Skolem hull in $L_{i_n}$ of the first n indiscernibles? Jech only proves this result for infinite ordinals. </p> <p>2) Is it possible that there could be an elementary embedding from $L$ to $L$, or from $L_{i_\alpha}$ to $L_{i_\beta}$ ($\alpha, \beta$ may be finite or infinite), that does not always map indiscernibles to indiscernibles? This sounds weird, but I'm not convinced it's impossible. As far as I know, there's no formula in $L$ that defines "$\alpha$ is a Silver indiscernible." (In fact there is no such formula -- see Andreas Blass's comment below.)</p> http://mathoverflow.net/questions/73251/characterizing-elementary-embeddings-of-l-and-l-alpha-under-0/73292#73292 Answer by Andreas Blass for Characterizing elementary embeddings of $L$ and $L_\alpha$ under 0# Andreas Blass 2011-08-20T19:39:38Z 2011-08-20T19:39:38Z <p>The answer to question 1 is no. Let $n$ be a finite ordinal, and consider the structure $M$ with universe <code>$L_{i_n}$</code>, with constant symbols for the smaller Silver indiscernibles <code>$i_0,\dots,i_{n-1}$</code> as well as symbols for the membership relation $\in$ and the usual, $L$-definable Skolem functions. This structure $M$ is constructible. (This is where it's essential that $n$ is finite.) So the Skolem hull, the smallest elementary substructure $N$ of $M$, is <em>constructibly</em> countable. But $M$ itself is very large in the sense of $L$, since Silver indiscernibles like $i_n$ are constructibly inaccessible (and much more). Therefore $N$ is not all of $M$.</p>