# Question about algebraic field extensions

If I have a subfield $F$ of a field $E$, and an algebraic (over $F$) $\alpha\in E$, I can form $F(\alpha)$ which is isomorphic to $F[x]/\langle f(x)\rangle$ for $f(x) = irr(\alpha, F)$. That is, $f(x)$ is the minimal degree and monic element of $F[x]$ such that $f(\alpha) = 0$.

The book I'm using defines $F(\alpha)$ officially as the image of $F[x]$ under $\phi_{\alpha}$, the map $f(x)\mapsto f(\alpha)$, and the isomorphism mentioned above comes from the fact that its kernel is $\langle f(x)\rangle$.

My question is: if there are other roots of the polynomial $f(x)$ in $E$, then are they necessarily contained in $F(\alpha)$? My intuitive guess is yes, since the field $F[x]/\langle f(x)\rangle$ doesn't know the difference between distinct roots of $f(x)$. But if this is correct then why?

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No. Take $F = \mathbb{Q}, E = \mathbb{C}$ and $\alpha = \sqrt[3]{2}$. Then $F(\alpha)$ lies in $\mathbb{R}$ but $f(x) = x^3 - 2$ has two complex roots which do not lie in $\mathbb{R}$, hence cannot lie in $F(\alpha)$.

The fact that this can be false motivates the definition of a normal extension.

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The other roots are not necessarily contained in $F(\alpha)$. For example let $\alpha=2^{\frac{1}{3}}$. $F=Q$, the rationals, then $F(\alpha)$ is contained in the real numbers. However the other roots of the minimal polynomial are not real.
Since others have provided nice examples, let me address your last point about $F[x]/ \langle f(x)\rangle$ not knowing the difference between the roots of $f$.
Suppose $a_1,a_2, \cdots ,a_n$ are roots of $f$, an irreducible polynomial in $F$. Since $F[x]/ \langle f(x)\rangle$ doesnt know the difference between the roots of $f$ (i.e. the $a_i$s) this will mean that all the fields $F(a_1),F(a_2), \cdots F(a_n)$s are isomorphic. This is as good as it gets though.