Algebraic extensions and sub rings

For the converse we have :

Suppose $\mathbb{L}$ is NOT an algebraic extension of $\mathbb{K}$. Then we have $ x\in \mathbb{L} $ that is not algebraic over $\mathbb{K}$ (i.e. transcendental).

We consider $A=\mathbb{K}[x]$.

If $A=\mathbb{K}[x]$ is a field we have,

$\frac{1}{x}\in \mathbb{K}[x]$ this means, $\exists P\in \mathbb{K}[X], \frac{1}{x}=P(x)$ because $\mathbb{K}[x]$ contains every polynomial expression of $x$ and since $\mathbb{K}[x]$ is a field $\frac{1}{x}$ exists in $\mathbb{K}[x]$.

But then, this means,$ x$ is a solution of $XP(X)-1=0$. This is impossible since $x$ is transcendental over $\mathbb{K}$ !

For the direct implication we have:

Suppose $\mathbb{L}$ is an algebraic extension of $\mathbb{K}$, we consider $A$ a sub ring of $\mathbb{L}$ that contains $\mathbb{K}$. \begin{equation} \forall x \in A\backslash \lbrace{0}\rbrace, \exists P \in \mathbb{K}[X], P(x)=0 \tag{1} \end{equation}

we consider $P$ of minimal degree $n$, it is monic irreducible over $\mathbb{K}$.

Let $G\in \mathbb{K}[X]$ such that, $P(X)=XG(X)+a_0$ such a $G$ exists since $\mathbb{K}[X]$ every polynomial expressions in $X$.

$G$ is of degree $n$ and this forces $a_0 \neq 0$.

Otherwise we would get, $P(x)=xg(x)+0=0 \implies g(x)=0$ because $A$ is a sub ring of a field, it is an integral domain. And $P$ is the polynomial of minimal degree that verifies $(1)$. This is absurd so $a_0 \neq 0$.

Finally we get, \begin{equation} P(x)=g(x)x+a_0=0 \\ g(x)x=-a_0 \\ x(\frac{g(x)}{-a_0})=1 \end{equation}

We have that, $\frac{g(x)}{-a_0}\in A$ and $a_0 \neq 0$. Thus $x$ is invertible in $A$ which proves that $A$ is a field that contains $\mathbb{K}$.