Is it true that $F(\alpha)\cong F[x]/(p(x))$? Let $K$ be a field and let $F$ be a subfield of $K$. Let $p(x)$ be an irreducible polynomial in $F[x]$ and let $p(\alpha)=0$, where $\alpha\in K$. Is it true that $F(\alpha)\cong F[x]/(p(x))$?
I see that $F(\alpha)=F[\alpha]=\operatorname{sp}\{1,\alpha,...,\alpha^{n-1}\}$, where $n=\operatorname{deg}p(x)$. ($\operatorname{sp} A$ means the span of $A$ in a vector space $K$ over $F$.) So if it's guaranteed that $\{1,\alpha,...\alpha^{n-1}\}$ is linearly independent, then we can define a function $\psi:F(\alpha)\to F[x]/(p(x)),a_0+a_1\alpha+...+a_{n-1}\alpha^{n-1}\mapsto \overline {a_0+a_1x+...+a_{n-1}x^{n-1}}$ and it is easy to prove that $\psi$ is an isomorphism. However, the problem is, since I don't know if $\{1,\alpha,...,\alpha^{n-1}\}$ is linearly independent, I don't know if $\psi$ is well-defined.
 A: Here's an important "philosophical" point to consider here: when you are trying to prove an isomorphism between an algebraic object $A$ and an algebraic object that is a quotient $B/\Phi$, it is almost always preferable to do so by defining a morphism from $B$ to $A$ that has kernel $\Phi$. This will allow you to use the Fundamental Theorem of Homomorphisms, and completely bypass the issue of establishing that maps are well defined.  So when you encounter a problem like this, your first inclination should be to find a map from $B$ to $A$ (ideally one that is "obvious" and will have the correct kernel) rather than trying to find a map from $A$ to $B/\Phi$. 
This is a case in point, as indicated in my comment. You are trying to prove an isomorphism between $F(\alpha)$ and the quotient $F[x]/(p(x))$. Rather than trying to define a morphism from $F(\alpha)$ to $F[x]/(p(x))$ and then trying to prove that it is an isomorphism, it is much easier to define a map from $F[x]$ onto $F(\alpha)$ that has $(p(x))$ as its kernel, and then invoke the Fundamental Theorem of Homomorphisms to immediately get an isomorphism between $F[x]/(p(x))$ and $F(\alpha)$. 
Here in fact you are further aided by several universal properties, which take care of a lot of the technical details. To define a morphism from $F[x]$ to $F(\alpha)$, you just need to say what happens to $F$ and what happens to $x$: map $F$ to itself via the identity, and map $x$ to $\alpha$. The universal property of the polynomial ring gives you a homomorphism $e\colon F[x]\to F(\alpha)$ given by $e(q(x)) = q(\alpha)$ (the "evaluation at $\alpha$" map). The kernel consists precisely of the polynomial $q(x)$ that have $\alpha$ as a root; if $q(x)\in \mathrm{ker}(e)$, then dividing by $p(x)$ with remainder gives $q(x) = p(x)m(x) + r(x)$ with $r=0$ or $\mathrm{deg}(r)<\mathrm{deg}(p)$. Evaluating at $\alpha$ we obtain that $r(\alpha)=0$, and so the irreducibility of $p(x)$ gives that $r=0$. Thus, $\mathrm{ker}(e) = (p(x))$. Hence we have an isomorphism of $F[x]/(p(x))$ with its image under $e$, contained in $F(\alpha)$.
To finish off the proof, note that since $p(x)$ is irreducible and $F[x]$ is a PID, it follows that $(p(x))$ is a maximal ideal, and hence $F[x]/(p(x))$ is a field. Therefore, $e(F[x]/(p(x)))$ is a field. It contains $F$ and contains $\alpha$ (as $\alpha=e(x)$), and hence contains the smallest subfield of $K$ that contains $F$ and $\alpha$, namely $F(\alpha)$. This gives $F(\alpha)\subseteq e(F[x]/(p(x))) \subseteq F(\alpha)$, giving the desired equality and establishing the isomorphism.
Of course, there is a lot of detail above that one can gloss over if you've already been playing with minimal polynomials and the like. But the important point is the one raised in the first paragraph: it is usually a lot easier to prove an isomorphism between an object like $F(\alpha)$ and a quotient like $F[x]/(p(x))$ by mapping from the second object $F[x]$ to the first $F(\alpha)$ with appropriate kernel, than to try to map in the other direction.
