A minimal polynomial problem that's bothering me Let $F$ be a field and $P$ an irreducible polynomial in $F[x]$. If I find a root of $P$ in an extention of $F$, does that make $P$ a minimal polynomial?
 A: Yes, if it is normed and has minimal degree and the root is algebraic.
A: If $E$ is an extension of $F$ in which an irreducible polynomial $P(x) \in F[x]$ has a root $\alpha$, then $P(x)$ is indeed a polynomial of least degree such that
$P(\alpha) = 0; \tag 1$
furthermore, we may without loss of generality assume $P(x)$ is monic, that is, the leading coefficient of
$P(x) = \displaystyle \sum_0^{\deg P} p_k x^k, \; p_k \in F, \; 0 \le k \le n, \tag 2$
$p_{\deg P} = 1$; certainly this assumption will not effect the validity of (1), since it may be effected by simply dividing $P(x)$ through by $p_{\deg P} \ne 0$; under such circumstance we indeed term $P(x)$ the minimal polynomial of $\alpha \in E$.
We may see that $\deg P(x)$ is least amongst all polynomials $Q(x) \in F[x]$ such that $Q(\alpha) = 0$ by assuming to the contrary that there exists some such $Q(x) \in F[x]$ with $\deg Q(x) < \deg P(x)$; then by the Euclidean division algorithm for polynomials, we may write
$P(x) = q(x) Q(x) + r(x); \; q(x), r(x) \in F[x], \tag 3$
where either $r(x) = 0$ or
$0 \le \deg r(x) < \deg Q(x); \tag 4$
if $r(x) \ne 0$ we may thus write
$r(x) = P(x) - q(x) Q(x), \tag 5$
which yields
$r(\alpha) = P(\alpha) - q(\alpha) Q(\alpha) = 0 - q(\alpha) \cdot 0 = 0; \tag 6$
but by (4) this contradicts our selection of $Q(x)$ as having least degree in $F[x]$ with $Q(\alpha) = 0$; thus $r(x) = 0$ and 
$P(x) = q(x) Q(x); \tag 7$
but now since $\deg Q(x) < \deg P(x)$, this equation asserts that $P(x)$ is reducible in $F[x]$, contrary to hypothesis.  Therefore there is no such $Q(x) \in F[x]$ and thus $P(x)$ is indeed a minimal polynomial for $\alpha \in E$.
