Artin Schreier Theorem for Rings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:00:03Z http://mathoverflow.net/feeds/question/10449 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10449/artin-schreier-theorem-for-rings Artin Schreier Theorem for Rings Jose Capco 2010-01-02T02:30:00Z 2010-01-26T06:44:17Z <p>This has been in my mind for quite some time. Looking at Artin Schreier Theorem for fields:</p> <p>If L is a field and K its algebraic closure and if 1&lt; [K:L] &lt; infinity then L=K[i] and L is a real closed field (Thus L has characteristic 0. Here i is just the square root of -1).</p> <p>I was wondering if a "generalized" Artin Schreier exist or if someone could refer to me to some paper that attempts this. There is a concept of real closedness and "algebraic closedness" of reduced commutative rings, but I doubt that the statement would hold.</p> <p>So one has the following conjecture:</p> <p>If L is a reduced commutative ring and K is its total integral closure (this is an equivalent notion of algebraic closure if K and L were fields) and if 1&lt;[K:L]&lt; infinity (here I mean that K is a finite L-module that is not the same as L) then L is real (thus its characteristic is 0.. and one can add that L is real closed in the sense of reduced commutative rings).</p> <p>Can one easily show this, even at least prove that L has characteristic 0?</p>