Reverse mathematics strength of identically zero polynomials are the zero polynomial - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T05:16:40Z http://mathoverflow.net/feeds/question/44410 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44410/reverse-mathematics-strength-of-identically-zero-polynomials-are-the-zero-polyno Reverse mathematics strength of identically zero polynomials are the zero polynomial Ricky Demer 2010-11-01T03:25:52Z 2010-12-17T16:44:31Z <p>According to wikipedia, the statement "every polynomial over a countable field that is not the zero polynomial has only finitely many roots" is equivalent to RCA0 over RCA0* (which is called ERCA-0 in <a href="http://mathoverflow.net/questions/30904/weakest-subsystems-of-second-order-arithmetic-for-mathematical-logic" rel="nofollow">this answer</a>). I've been interested in reverse math for 1-2 years, so when I got to working with polynomial rings, I got curious as to the strength of the equivalence of their representations as functions and formal polynomials. Let izizp (identically zero implies zero polynomial) be the following statement:</p> <p>For all (countably) infinite fields $F$, for all members $c_0,...,c_n$ of $F$, if $\; \displaystyle\sum_{m=0}^n \; \; c_m\cdot x^m \;$ is identically zero, then $c_0,...,c_n$ are all zero.</p> <p><br></p> <p>Is izizp a theorem of RCA0*?</p> <p>Is izizp equivalent to RCA0 over RCA0*?</p>