Equivalent conditions to "$P(x)$ is the minimal polynomial of $u$ over $K$"

Question

Let $K$ be a subfield of $F$, $u \in F $ be an algebraic element over $K$ and $P(x) \in K[x]$ be a monic polynomial. Show that:

$P(x)$ is the minimal polynomial of $u$ over $K$ if and only either $P(x)$ is irreducible over $K$ and $P(u)=0$ or $P(x)$ is the polynomial in $K[x]$ of the least positive degree such that $P(u)=0$
or $ \deg(P(x))=[K(u):K]$ and $P(u)=0$.

I just want you guys show me the connection of the last statement with the 2 previous ones which seem obvious!

Answer 1

Using your notations, here are $3$ statements, which you have to show to be equivalent to each other:

1. $P(x)$ is irreducible over $K$ and $P(u)=0$
2. $P(x)$ is the polynomial in $K[x]$ of the least positive degree such that $P(u)=0$
3. $\text{deg}(P(x))=[K(u):K]$ and $P(u)=0$.

(If $P$ satisfies one of these conditions, then it is called the minimal polynomial of $u$ over $K$).

According to your comment, you only need to handle with $(3)$.

— Let's see $3 \implies 2$. Suppose that $Q \in K[x]$ is a monic polynomial such that $Q(u)=0$, and let's show that its degree is greater or equal to the degree of $P$. To do this, assume that this is not the case: $r:=\text{deg}(Q)<\text{deg}(P)$. Then can you find a basis of $K(u)$ over $K$? Can you proceed from here?

Hint:

Try $\{1,u,u^2,\dots,u^{r-1}\}$. It has $r<\text{deg}(P)$ elements, contradicting $(3)$…

— As for $2 \implies 3$, you only have to find a basis of $K(u)$ over $K$ which has cardinality $n:=\text{deg}(P)$. I let you think about it!