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Why do the polynomials $x^5$ and $x^6$ in the ring of polynomials without a linear term have no greatest common divisor? Why isn't it $x^3$?

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    $\begingroup$ So you claim: If $p(x)$ is any common divisor of $x^5$ and $x^6$, then $p(x)$ divides $x^3$. See if you can find a counterexample to that. $\endgroup$ – GEdgar Dec 22 '16 at 16:29
  • $\begingroup$ Note that $x^3$ is a maximal common divisor. It is not, however, a greatest common divisor. There's a subtle difference. $\endgroup$ – Arthur Dec 22 '16 at 16:31
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Hint $\,\ x^2\mid x^5,x^6\ $ but $\, x^2\nmid x^3\ $ since $\ x^3/x^2 = x\not\in R$

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Beause in $k[x^2, x^3]$ both $x^2$ and $x^3$ are divisors of $x^5$ and $x^6$. If $x^3$ were a gcd, it must me a multiple of $x^2$, which is not true. The set of common divisors does not have a largest element with respect to the $|$-order.

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