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Let $R = \mathbb{Z}[x^2, x^3]$. Then $R$ contains all integer polynomials that lack the $x$ term. That is, $R$ contains all polynomials of form $a_0 + a_2 x^2 + a_3x^3 + \ldots + a_n x^n$ for $a_i \in \mathbb{Z}$.

Question: What is $\gcd(x^2, x^3)$?

  1. $\gcd(x^2, x^3) = d$ iff $d \mid x^2, x^3$ and if $s \mid x^2, x^3$ then $s \mid d$ with $s,d \in R$.

  2. Then isn't $\gcd(x^2, x^3) = x^2$, or am I missing some tricky reason why this is not the case?

Similarly, isn't $\gcd(x^5, x^6) = x^5$?

Techn1cal
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  • $x^2$ does not divide $x^3$ in $\mathbb Z[x^2, x^3]$, so cannot be a gcd for $x^2$ and $x^3$. – Dustan Levenstein Dec 03 '13 at 09:33
  • Yes -- and I just realized neither does $x^5 \mid x^6$ since we can't say $x^6 = (x^5)(x)$. I wasn't seeing how $x \notin R$ was really impacting things here. – Techn1cal Dec 03 '13 at 09:36

1 Answers1

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Note that $R\subset \mathbb{Z}[x]$, so if $d\mid x^2$ in $R$, then $d\mid x^2$ in $\mathbb{Z}[x]$. Hence (by unique factorization in $\mathbb{Z}[x]$ if you will) $$d \in \{\pm 1, \pm x, \pm x^2\}\cap R = \{\pm 1, \pm x^2\}$$

Now you can check that $x^2 \nmid x^3$ in $R$, and hence $d = \pm 1$, whence $$ \gcd(x^2,x^3) = 1 $$

Prahlad Vaidyanathan
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