Let $R= \mathbb{Q}[x^2 ,x^3 ]$, the set of all polynomials over $\mathbb{Q}$ with no $x$ term. Show that the $\gcd$ of $x^5$ and $x^6$ does not exist 
Let $R= \mathbb{Q}[x^2 ,x^3 ]$, the set of all polynomials over $\mathbb{Q}$ with no $x$ term. Show that a gcd of $x^5$ and $x^6$ does not exist in $R$.

I am reasoning for the absurd and I am assuming that $gcd (x ^ 5, x ^ 6) = ax ^ 2 + bx ^ 3\in\mathbb{Q}[x^2 ,x^3 ] $, then, for the Bezout identity there exists $p (x) = cx ^ 2 + dx^3$ and  $q(x)= ex ^ 2 + fx ^ 3$ in $\mathbb{Q}[x^2 ,x^3 ]$ such that $x ^ 5p (x) + x ^ 6q (x) = ax ^ 2 + bx ^ 3$, which is absurd since $$9 = deg (x ^ 5p (x) + x ^ 6q (x)) \neq  deg (ax ^ 2 + bx ^ 3) = 3.$$
Is it okay what I did? Thank you very much.

What ordering are you using to define greatest?  If it is by degree, what is wrong with $x^3?$  It certainly divides $x^5$ and $x^6$, while $x^4$ and $x^5$ do not divide both of them.  Using the Bézout identity presumes that you have the proper kind of integral domain.  Have you shown that?
