A minimal polynomial problem that's bothering me

Question

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?

Answer 1

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 $\alpha \in E$.

Answer 2

Yes, if it is normed and has minimal degree and the root is algebraic.