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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.

Xam
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user425181
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    I think it's fine nothing is wrong – Deepesh Meena Aug 13 '17 at 19:27
  • @sharding4 Well if it is 1 would also arrive at an absurd because $9 = deg (x ^ 5p (x) + x ^ 6q (x)) \neq deg (1) = 0$ or not? – user425181 Aug 13 '17 at 19:39
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    I think using Bezout identity as a test is a bit questionable, because there is no reason the gcd would be given by such a recipe in a non-PID. I would use the definition of a gcd. The crux is that both $x^2$ and $x^3$ are common factors, but there is no common factor divisible by both of them. – Jyrki Lahtonen Aug 13 '17 at 19:47
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    @James There are a couple unrecoverable errors in the attempted proof - see the Remark in my answer. – Bill Dubuque Aug 13 '17 at 22:59

3 Answers3

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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.
Jyrki Lahtonen
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    Identifying the factors is simplified by the fact that a factor in $R$ is also a factor in $\Bbb{Q}[x]$, which is a UFD with a well understood theory of divisibility. – Jyrki Lahtonen Aug 13 '17 at 20:17
  • I do not understand the last part of your answer, could you explain that more clearly? Thank you. – user425181 Aug 13 '17 at 20:40
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    I am using the definition from Wikipedia: If R is a commutative ring, and a and b are in R, then an element d of R is called a common divisor of a and b if it divides both a and b (that is, if there are elements x and y in R such that d·x = a and d·y = b). If d is a common divisor of a and b, and every common divisor of a and b divides d, then d is called a greatest common divisor of a and b. And I show that none of the common divisors have the required property that all the common divisors divide it. – Jyrki Lahtonen Aug 13 '17 at 20:42
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    (cont'd) Basically this: Up to a unit factor the common divisors are $1,x^2,x^3$. None of them is divisible by all the others, so there is no greatest common divisor. I guess it is possible that someone somewhere uses a different definition of gcd, but then you need to share your definition with us @user425181. – Jyrki Lahtonen Aug 13 '17 at 20:44
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    @user425181 I explain it slightly differently in my answer, emphasizing how the universal definition of the gcd is employed. – Bill Dubuque Aug 13 '17 at 21:33
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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.

Bill Dubuque
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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?

Ross Millikan
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  • It is a polynomial that divides both $x ^ 5$ and $x ^ 6$, am I right?This polynomial is in $R$ , then has the form $ax ^ 2 + bx ^ 3$. R is an integral domain? How do I test it? – user425181 Aug 13 '17 at 19:45
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    R includes $x^4, x^5$ and all higher powers of $x$ because it has to be closed under products. To be an integral domain you have to have a ring such that no product of nonzero elements is zero, which is true here. The Wikipedia page says Bézout's identity is not valid even in all integral domains, so I don't know if you can use that for your argument. – Ross Millikan Aug 13 '17 at 19:51
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    Yes $x^3$ is a common divisor. I just don't know how you are defining the greatest. Your argument seems to claim that $x^5$ and $x^6$ do not have a common divisor, but they do have $1,x^2,x^3$. The other way to fail to have a gcd is for there always to be a greater one, but I don't see that here – Ross Millikan Aug 13 '17 at 19:52
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    @BillDubuque: I thought it was an answer to what was wrong with the argument, showing that a common divisor exists and suggesting that the Bézout identity is not true in all rings, so needed to be established in this one. – Ross Millikan Aug 13 '17 at 20:06
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    @Ross I see, It might help to explicitly clarify that you are only addressing the attempted proof, not giving any hints as to how to proceed. – Bill Dubuque Aug 13 '17 at 20:14
  • @user425181 Polynomials in $R$ need not have that special form, and the Bezout gcd identity need not hold (in fact the Division algorithm may fail). See the Remark in my answer. – Bill Dubuque Aug 13 '17 at 23:01
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    @RossMillikan: Generally speaking, the ordering is by divisibility: $x^4$ is strictly less than $x^8$ since $x^4 \mid x^8$ but $x^8 \not\mid x^4$, and $x^4$ is incomparable with $1+x^6$ since neither divides the other. (although, it turns out in this case that between comparable nonzero elements, the ordering by divisibility coincides with the ordering by degree) –  Aug 13 '17 at 23:10