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.

Solution 1:

I would, instead, use the definition of gcd, and observe that

• The factors of $x^5$ are of the form $d_1=ax^n$ with $a\in\Bbb{Q}^*$ and $n\in\{0,2,3,5\}$.
• Similarly the factors of $x^6$ are the monomial of the form $d_2=bx^m$ with $b\in\Bbb{Q}^*$, $m\in\{0,2,3,4,6\}$.
• The common factors are thus non-zero monomials of degree either $0,2$ or $3$.
• But among those there are none that all the common factors would be divisors of: the lower degree monomials are not divisible by $x^3$, and a cubic monomial is not divisible by $x^2$. Thus no greatest common divisor exists.

Solution 2:

Suppose $\,d\,$ is a gcd of $\,x^5,x^6$ in $R.\,$ By the general (universal) definition of the gcd

$\ \ \ c\mid x^5,x^6\! \iff c\mid d.\, $ Taking $c = d\,$ $\rm\color{#0a0}{shows}$ $\,d\mid \color{#c00}{x^5},x^6,\,$ i.e. the gcd is a common divisor.

$x^3\mid x^5,x^6\ \ \Rightarrow\ \ x^3\mid d,\ $ hence $\ d = x^3,\, x^4,\,$ or $\ \color{#c00}{x^5}\, $ (up to unit factors). But none work since

$x^2\mid x^5,x^6\ \ \Rightarrow\ \ \color{#90f}{x^2\mid x^3}\,$ if $\:d=x^3,\ $ $\rm\color{#0a0}{and}$ $\,d=x^4\Rightarrow\, \color{#90f}{x^4\mid x^5},\,$ and $\,d=x^5\Rightarrow\, \color{#90f}{x^5\mid x^6}.\ $

Thus every possibility for $\,d\,$ yields some $\,\color{#90f}{x^i\mid x^{i+1}}$ in $R$ $\,\Rightarrow\, x^{i+1}/x^i = x\in R,\,$ contradiction.

Remark $ $ There are a couple problems in the attempted proof. First, it is easy to show that the elements of $\,\Bbb Q[x^2,x^3]$ are precisely the polynomials $\,f(x)\in \Bbb Q[x]\,$ whose coefficient of $\,x^{1}$ is zero, i.e. $\:x^2\mid f(x)-f(0).\,$ There is no reason given for your assumption that these have the very special form $\,ax^2+bx^3\,$ for $\,a,b\in \Bbb Q\,$ for the both gcd and the Bezout coefficients polynomials.

Second, there is no reason to believe that if the gcd exists then it satisfies a Bezout identity. Indeed, this is proved in $\,\Bbb Q[x]\,$ by using the division algorithm, but that fails dramatically here. We cannot even perform the first step in computing the Bezout identity for $\,\gcd(x^6,x^5)\,$ by the (extended) Euclidean algorithm, i.e. we cannot divide $\,x^6\,$ by $\,x^5\,$ in $R$ to get a smaller degree remainder, since $\,x^6 = x^5 q + r\,\Rightarrow\, r=0\,$ by evaluating at $\,x=0,\,$ so $\,x^6 = x^5 q\Rightarrow\, q = x\in R,\,$ contradiction.

Solution 3:

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?