1
$\begingroup$

this is a question from an exercise I am looking these days.

suppose $x=\mathrm{e}^{\mathrm{i}}$, define $$P=\{a_nx^n+\cdots+a_1x+a_0\mid a_k\in\mathbb{Z} \textrm{ and } a_k\ge0\}$$. Separate $P$ into two non-empty subsets $A$ and $B$ so that $A\cap B=\emptyset$ and $P=A\cup B$, with $A$, $B$ and $P$ are congruent to each other.

suppose the least non-zero power of polynomial $f(x)=a_nx^n+\cdots+a_1x+a_0$ is $\mathrm{ged}(f(x))$(donot know if there is a better notation, so i just reverse "deg"). then $$A=\{f(x)\in P\mid \mathrm{ged}(f) \textrm{ is even}\}, B=P\setminus A$$ then $xA=B$, but $P=(1+x^{-1})B$ scaled $B$ is not a congruent transformation. many modification tried but still no solution, so what have i do to continue or there is another method?

$\endgroup$
1
$\begingroup$

As $e^i$ is transzendental, it is save to assume that the relevant congruences $P\to A$, $P\to B$ will be of the form $z\mapsto e^{ni}z+m$ (corresponding to $f\mapsto x^nf+m$ on the level of polynomials). Wlog. $0\in A$, $0\notin B$. Then necessarily, $m_A=0$ and $n_A\ne0$. Then $A\cap\Bbb Z=\{0\}$, hence $1\in B$ and $m_B=1$. As also $2\in B$, $n_N=0$, i.e., $$ B=P+1=\{\,f\in P\mid a_0>0\,\}.$$ Then $e^i\in A$, hence $n_A=1$, i.e., $$ A=e^iP=\{\,f\in P\mid a_0=0\,\}. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ +1, so easy solution I didn't notice, I'm also thinking about the linear transform formula, but still no advance for such a long time, there must be something happened to me. so many thx for your ans. $\endgroup$ – Larry Eppes Jun 8 '18 at 14:56
  • $\begingroup$ @LarryEppes The trick to find (or in fact to be forced to find) the conruences was to look for $P\to A$ and $P\to B$ instead of $A\to B$ $\endgroup$ – Hagen von Eitzen Jun 12 '18 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.