Let $P$ be a non-zero polynomial with real algebraic coefficients; prove or disprove the following
"All real roots of $P$ are algebraic numbers"
Asked
Active
Viewed 68 times
-2
-
What do you mean with "with real algebraic factors"? – Hagen von Eitzen Mar 31 '18 at 13:09
-
Can you give us an example of what you are talking about? – lulu Mar 31 '18 at 13:11
-
2You probably mean, "with real algebraic coefficients"? If so this is a near-duplicate of On the meaning of being algebraically closed – hmakholm left over Monica Mar 31 '18 at 13:12
-
Real Sorry! just editted: factors----->coefficients – K. Sadri Mar 31 '18 at 15:19
1 Answers
2
If the polynomial is $p(x)=a_nx^n+...+a_1x+a_0$ has all $a_i$ algebraic over the rationals. Let $K:\mathbb{Q}$ be an algebraic extension in which $K$ is a splitting field of all the minimal polynomials of the $a_i$. Then $$Q(x)=\prod_{\sigma\in Aut(K:\mathbb{Q})}(\sigma(a_n)x^n+...+\sigma(a_1)x+\sigma(a_0))$$
is a polynomial with rational coefficients, since it is invariant under $Aut(K:\mathbb{Q})$. All roots of $p(x)$ are roots of $Q(x)$, since $p(x)$ is one of its factors.
geometryfan
- 141