Quotient of Polynomial Ring by Irreducible Polynomial is Simple Extension Let $F$ be a field, and $f$ an irreducible polynomial in $F[x]$.
Is $F[x]/(f)$ necessarily of the form $F(\alpha)$, where $\alpha$ is a root of $f$?
The only examples I know, e.g.
$\mathbb{R}[x]/(x^2+1)\cong\mathbb{R}(i)\cong\mathbb{C}$ seems to support the argument.
What I know is $K=F[x]/(f)$ is a field when $f$ is irreducible, and $[K:F]=\deg f$.
Thanks for any help in proving or disproving.
 A: Yes.
To make your question more precise, you ask if whenever $F\subseteq K$ is a field extension and $f\in F[X]$ is irreducible over $F$, with root $\alpha\in K$, do we have that $F(\alpha)\cong F[X]/f$.
Consider $\phi\colon F[X]\to F[\alpha]$ defined as identity on $F$ and $X\mapsto \alpha$. This is a well-defined homomorphism, and you can check that its kernel is precisely $(f)$. Indeed, if for $g\in F[X]$ we have that $\phi(g)=0$, it means that by definition $g(\alpha)=0$. But then, since $f$ is the minimal polynomial of $\alpha$ (because $\alpha$ is its root and it is irreducible), you must have that $f$ divides $g$, whence $g\in (f)$. Conversely, if $g\in (f)$, clearly $\varphi(g)=f(\alpha)\cdot h(\alpha)=0$. 
By the preceding paragraph, $\phi$ induces an isomorphism $\bar\phi\colon F[X]/(f)\to F[\alpha]$ (it is clearly onto!). Since you know that $F[X]/(f)$ is a field, $F[\alpha]$ is also a field, so $F[\alpha]=F(\alpha)$ and we are done.
A: Consider the homomorphism from $F[X]$ to $F[\alpha]$ that sends $g$ to $g(\alpha)$ has kernel generated by $f$ (as $F$ field,$f$ can be assumed to be monic and it is given to be irreducible).Then the homomorphism gives isomorphism of $F(X)/(f)$ and $F[\alpha]$.This implies $F[\alpha]$ is a field and hence it is $F(\alpha)$.
A: Yes, just let $\alpha = x + (f) \in F[x]/(f)$.
