$\textbf {Definition} :$ Let $\Bbb K$ be a base field. Let $A$ be a $\Bbb K$-algebra. Then $A$ is said to be a simple extension of $\Bbb K$ if $\exists$ $x \in A$ such that $A=\Bbb K[x]$ as $\Bbb K$-algebras. In this case we say that $x$ is a primitive element of $A$ over $\Bbb K.$
$\textbf {Question} :$ Is $\Bbb K^2$ a simple extension of $\Bbb K$?
To get an answer of the above question we need only to ensure whether or not there exists an element $x = (a,b) \in \Bbb K^2$ such that $\Bbb K^2 = \Bbb K[(a,b)]$ as $\Bbb K$-algebras. Am I correct? I know that since $\Bbb K$ is a field I can always find two elements $a,b \in \Bbb K$ with $a \neq b.$ Will this two distinct elements in $\Bbb K$ work?
Any suggestion will be highly appreciated. Thank you very much.
EDIT $:$ I observe that for any $F \in \Bbb K[X]$ we have $F((a,b)) = (F(a),F(b)).$ So we need only to find out two polynomials $F,G \in \Bbb K[X]$ such that $F(a)=1,F(b)=0$ and $G(a)=0,G(b)=1.$ Since $\Bbb K^2 = \text {span}\ \{(1,0),(0,1) \}.$ Let $a=1,b=0$ and let us take $F(X) = X$ and $G(X) = (X-1)^2.$ Then we are through.