goodar2006 wrote: prove that for each natural number $n$ there exist a polynomial with degree $2n+1$ with coefficients in $\mathbb{Q}[x]$ such that it has exactly $2$ complex zeros and it's irreducible in $\mathbb{Q}[x]$.(20 points) I have a question "complex zero" mean if $i$ is the complex number than $P(i)=0$???

a complex zero means $z$ such that $p(z)=0$ and $z\in \mathbb C$.

goodar2006 wrote: a complex zero means $z$ such that $p(z)=0$ and $z\in \mathbb C$. ok I understand u!

please help me!!!!!!!!! i think have a easy solution but dont know if its true assume p(x)=(x^3+2)(x^2-a1)(x^2-a2)...(x^2-an-1) such that "ai"s is not perfect square of a Q[x]. so the first term have two complex zero and one real zero (not Q) and all other term have real zero(not Q) so p(x) is irreducible in Q[x] !!!

exiri wrote: please help me!!!!!!!!! i think have a easy solution but dont know if its true assume p(x)=(x^3+2)(x^2-a1)(x^2-a2)...(x^2-an-1) such that "ai"s is not perfect square of a Q[x]. so the first term have two complex zero and one real zero (not Q) and all other term have real zero(not Q) so p(x) is irreducible in Q[x] !!! how you conclude that $p(x)$ is irreducible in $\mathbb Q[x]$?? for example $x^4-5x^2+6$ has all it's zeros in reals (and not in $\mathbb Q$), but it's reducible over $\mathbb Z[x]$.

exiri wrote: please help me!!!!!!!!! i think have a easy solution but dont know if its true assume p(x)=(x^3+2)(x^2-a1)(x^2-a2)...(x^2-an-1) such that "ai"s is not perfect square of a Q[x]. so the first term have two complex zero and one real zero (not Q) and all other term have real zero(not Q) so p(x) is irreducible in Q[x] !!! something intresting!!! you've factorized $p(x)=(x^3+2)(x^2-a_1)....(x^2-a_{n-1})$!!! how can you come with this idea that it's irreducible???

goodar2006 wrote: prove that for each natural number $n$ there exist a polynomial with degree $2n+1$ with coefficients in $\mathbb{Q}[x]$ such that it has exactly $2$ complex zeros and it's irreducible in $\mathbb{Q}[x]$.(20 points) I think You mean "...with coefficients in $\mathbb{Q}$", not "... in $\mathbb{Q}[x]$ ". Search for a polynomial $P(x)$ as $P(x) = \frac{1}{p}x^{2n+1} + T_{2n-1}(x) + \frac{1}{q}$ , where: $T_m(x)$ is the Chebyshev plynomial of degree $m$ and $p,\, q$ are big enough prime numbers. $p \geq 2$ and big enough $q$ ensure that $P(x)$ has at least $2n-1$ real roots, big enough $p$ will ensure that $P(x)$ will have exactly $2n-1$ real roots. Irreducibility of $P(x)$ in $\mathbb{Q}[x]$ follows by Eisenstein's criterion.

dgrozev wrote: goodar2006 wrote: prove that for each natural number $n$ there exist a polynomial with degree $2n+1$ with coefficients in $\mathbb{Q}[x]$ such that it has exactly $2$ complex zeros and it's irreducible in $\mathbb{Q}[x]$.(20 points) I think You mean "...with coefficients in $\mathbb{Q}$", not "... in $\mathbb{Q}[x]$ ". Search for a polynomial $P(x)$ as $P(x) = \frac{1}{p}x^{2n+1} + T_{2n-1}(x) + \frac{1}{q}$ , where: $T_m(x)$ is the Chebyshev plynomial of degree $m$ and $p,\, q$ are big enough prime numbers. $p \geq 2$ and big enough $q$ ensure that $P(x)$ has at least $2n-1$ real roots, big enough $p$ will ensure that $P(x)$ will have exactly $2n-1$ real roots. Irreducibility of $P(x)$ in $\mathbb{Q}[x]$ follows by Eisenstein's criterion. Why do P(x) have exactly 2n-1 real root?l don't really understand that.

$T_{2n-1}$ has exactly $2n-1$ different real roots in $(-1,1), T_{2n-1}(-1)=-1, T_{2n-1}(1)=1$ and it alternatively takes values $-1,1$ at $2n+1$ different points - its extremal values in $(-1,1)$ and at -1 and 1. Hence $T_{2n-1}$ commands the sign of $P$ in those $2n+1$ points, providing $p,q$ are large enough.

Cool,thank you so much!

That was also my idea, starting from a polynomial $Q$ with exactly $2n-1$ real roots and then affect just a bit it by adding some small (uniformly in some interval) polynomial $\Delta(x)$, so that $Q(x)+\Delta(x)$ still has exactly $2n-1$ real roots. But $Q:=(x-1)^{2n-1}$ is not a good starting point. Because we generally cannot guarantee $Q+\Delta$ still has exactly $2n-1$ real roots. However, in case $Q$ has articulated roots where it goes well up and well down (something like Chebyshev polynomial does) disturbing it a bit doesn't affect the number of roots.