Let $\alpha\in L$ be algebraic over $K$ in a field extension $L\supseteq K$, i.e. there exists a polynomial $0\neq f\in K[X]$ which suffices $f(\alpha)=0$.
Is there an elementary way to prove that in this case, $\alpha^2+\alpha$ is also algebraic?
Asked
Active
Viewed 241 times
3
-
3You can prove more generally that if $\alpha, \beta$ are algebraic then so are $\alpha + \beta$ and $\alpha \beta$; these are both good exercises. Then by induction it follows that if $\alpha$ is algebraic then so is any polynomial $g(\alpha)$ in it. – Qiaochu Yuan Jan 14 '21 at 23:52
-
https://math.stackexchange.com/q/155122/ – Jan 14 '21 at 23:55
-
Yes. Since $K(\alpha^2 + \alpha) \subset K(\alpha)$, the dimension of $K(\alpha^2 + \alpha)$ as a $K$ vector space must be finite. – hardmath Jan 14 '21 at 23:56
1 Answers
3
Hint Let $\beta=\alpha^2+\alpha$.
$\alpha$ is algebraic implies that $$[K(\alpha):K]=n< \infty$$ for some $n$. Now show that $1,\beta, \beta^2,..., \beta^n \in K[\alpha]$ cannot be linearly independent over $K$.
N. S.
- 132,525
-
Seems to me that the same proof applies if $,\beta,$ is any polynomial in $,\alpha.$ – Somos Jan 15 '21 at 00:30
-