Is it true that given a field $E$, any finite algebraic extension $F$ of $E$ is obtained by adding to $E$ a root of some irreducible polynomial over $E$? I know this is true in finite fields but I wonder if it still holds also in general.
EDIT: I added "finite" to algebraic extension.
On
It is not. If $\alpha$ is a root of the irreducible polynomial $\mu_\alpha$, then $[E(\alpha):E]=\deg\mu_\alpha$. So such an extension always has finite degree. But not all algebraic extensions have finite degree. For instance, $\overline{\mathbb Q}/\mathbb Q$, where $\overline{\mathbb Q}$ is the algebraic closure of $\mathbb Q$, does not have finite degree, even though it is algebraic.
As you state it, it is false; it is even false for finite fields, as an extension given by a single element can only be finite, but algebraic extensions may be infinite.
The exact statement, in full generality, is due to Artin, generalizing the classical Primitive Element Theorem.
The classical Primitive Element Theorem, which seems to be what you are alluding to, is that if $F/E$ is finite and separable, then $F=E(\alpha)$ for some $\alpha\in F$. This applies to the cases where $E$ is a perfect field (so all extensions are separable), which would include all finite fields and all fields of characteristic $0$.
For an example where this does not happen, let $E=\mathbb{F}_p(u,v)$ be the function field on two indeterminates over the finite field of order $p$, and let $F=E(u^{1/p},v^{1/p})$. This extension is not primitive (i.e., given by a single element). To verify this, show that for every $\alpha\in F$, $\alpha^p\in E$. That means every extension of the form $E(\alpha)$ has degree at most $p$ (in fact, $1$ or $p$), but $F$ has degree $p^2$ over $E$.