Let $E$ be an algebraic extension of $k$.

Question

Let $\alpha$ be an element of $E$, let $p(X)$ be its irreducible polynomial over $k$, and let $E'$ be the subfield generated by all the roots of $p(X)$ which lie in E. Then $E'$ is a finite extension of $k$.

I'm not understanding the relation between $E'$ be generated by all roots of $p(X)$ and he being a finite extension of $k$, to be especific i'm not understanding how $k\subseteq E'$.

I know this must be really simple, but, since i'm really new at algebra as a whole, i'm have a problem with this kind of detail.

Answer 1

The subfield $E'$ of $E$ generated $\color{red}{\text{by }k}$ and all roots of $p(X)$ that belong to $E$.

(otherwise $E'$ might not be an extension of $k$).

If the roots are $\alpha=\alpha_1,\alpha_2,\dots,\alpha_k$, then each field $K(\alpha_i)$ is a finite extension of $K$, where $K$ is any extension field of $k$ contained in $E$ (why?).

Now consider $E'=k(\alpha_1,\dots,\alpha_k)=k(\alpha_1,\dots,\alpha_{k-1})(\alpha_k)$ and remember the dimension formula.