# Subring generated by $x$ is an integral domain iff it is a field iff the minimal polynomial of $x$ is irreducible

Let $R$ be a ring, $K$ a subfield of $R$, and $x \in R$. Let $F(X)$ be the minimal polynomial of $x$ over $K$.

I want to prove that:

$K[x]$ is a field $\iff K[x]$ is an integral domain $\iff F(X)$ is irreducible,

using the following lemmas:

1. If $B$ is an integral domain and $A$ a subring of $B$ such that $B$ is integral over $A$, then one has equivalence $A$ is a field $\iff B$ is a field.

2. $K[X]/(F(X))$ and $K[x]$ are isomorphic.

I need help especially with the last equivalence. How could the minimal polynomial not be irreducible?

• Under item(2), do you perhaps mean to ask if $K[X]/(F(X)$ and $K(X)$ are isomorphic? – Robert Lewis Feb 22 '15 at 19:52

## 1 Answer

The minimal polynomial of say a matrix needn't be irreducible. Consider the minimal polynomial of $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ in $R=M_2(\Bbb R)$.

Note that since $k[x]= k[X]/(f(X))$, $k[x]$ is always integral over $k$, for $k$ is integral over $k$; trivially, and $x=\bar X$ is a root of $f$. Hence $k[x]$ is a field if and only if it is a domain, since $k$ is a field. But $f$ irreducible implies $(f)$ is prime, and conversely, since $k[X]$ is a PID, in particular a UFD.

You could also use that $k[X]$ is a PID directly, which entails that nonzero primes are maximal, i.e. PIDs (and more generally PIRs) have dimension $1$.