3

I have that $\exists p(x) \in F[x]$ with $p(x)=0$ then $p(a^2)=0$

if I take $p(x):=b_0+b_1x+b_2x^2+...+b_nx^n$ then $p(a^2):=b_0+b_1a^2+b_2a^4+...+b_na^{2n}=0$.

I want to prove that $\exists q(x) \in F[x]$ with $q(x)=0$ so $q(a)=0$ if I take $q(x):=c_0+c_1x+c_2x^2+...+c_nx^n$ then $q(a):=c_0+c_1a+c_2a^2+...+c_na^n$.

I asked to my Prof and he told me that it´s the way to prove it, but I don´t know how it could be $c_0+c_1x+c_2x^2+...+c_nx^n=0$.

Also, I know that $a^2$ is root of $p(x)$ but I don´t know how prove that $a$ is root of $p(x^2)$.

Estif Emmanuel Rodriguez
  • 33
  • 5

2 Answers2

4

Hint : Take $q(x) = b_0 + b_1 x^2 + \cdots + b_n x^{2n}$ and check $q(\alpha)$

Infinity_hunter
  • 5,369
  • 1
  • 10
  • 30
2

You have a polynomial that $\alpha ^2$ is a root of, using your notation $p(x):=b_0+b_1x+b_2x^2+...+b_nx^n$ so $p(a^2):=b_0+b_1a^2+b_2a^4+...+b_na^{2n}=0$

So just define a new polynomial such that all the odd coefficients are 0 and all the even coefficients match up to the coefficients from $p$

$q(x):=b_0+b_1x^2+b_2x^4+...+b_nx^{2n}$

Then $q(\alpha)=p(\alpha ^2)=0$

Alan
  • 16,582