Splitting a polynomial with one root - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:39:48Z http://mathoverflow.net/feeds/question/69997 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69997/splitting-a-polynomial-with-one-root Splitting a polynomial with one root Dtseng 2011-07-11T07:30:45Z 2011-07-11T11:17:16Z <p>Suppose we have an irreducible polynomial $f\in K[x]$. Is there some way to sometimes tell whether $f$ splits completely after adjoining just one root of $f$ to $K$?</p> <p>I am mostly interested in the case where $K$ is a function field <code>$\mathbb{F}_{q}(t_{1},\ldots,t_{m})$</code> over some finite field, so it might not be feasible to explicitly compute roots.</p> http://mathoverflow.net/questions/69997/splitting-a-polynomial-with-one-root/70005#70005 Answer by Georges Elencwajg for Splitting a polynomial with one root Georges Elencwajg 2011-07-11T11:17:16Z 2011-07-11T11:17:16Z <p>Here is the best I can come up with. Consider an algebraic closure $\bar K$ of $K$ and a root $\alpha \in \bar K$. </p> <p>The number of roots of $f$ in $K(\alpha)$ doesn't depend on $\alpha$: call it $r(f)$<br> Moreover call $s(f)$ the number of the <em>different</em> subfields $K(\alpha)\subset \bar K$ obtained by adjoining roots of $f$ to $K$. Then you have the pleasant equality $$deg(f)=r(f).s(f)$$ This shows in particular that the number of roots that you get by just adjoining one root divides the degree $deg(f)$ of your polynomial.<br> For example if $K=\mathbb Q$ and $f(x)=X^8-2$ you have $r(f)=2$ and $s(f)=4$, since the fields you get by adjoining roots of $f$ to $\mathbb Q$ are [with $\omega =\frac{1}{\sqrt 2}(1+i)$]:<br> $\mathbb Q(\sqrt[4]2)$<br> $\mathbb Q(\pm \omega \sqrt[4]2)$<br> $\mathbb Q(\pm \bar{\omega} \sqrt[4]2)$<br> $\mathbb Q(\pm i \sqrt[4]2)$</p> <p>These results are due to <a href="http://math.arizona.edu/~aprl/publications/rootsinquanta/perlis_rootsinquanta_08Oct2003.pdf" rel="nofollow">Perlis</a> , and although not difficult have found their way in exactly zero books, as far as I am aware.</p>