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Find the greatest common divisor of the polynomials $f(x)=x^{4}+ix^{2}+1$ and $g(x)=ix^{2}+1$ over the field $\mathbb{C}$.

Here I am getting some difficulty because I think both are prime and hence $\gcd =1$ , but I am not sure. Any help

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  • $\begingroup$ None is prime: the only irreducible polynomials in $\mathbf C[x]$ are linear polynomials. $\endgroup$ – Bernard Apr 13 '17 at 21:29
  • $\begingroup$ I mean both are prime to each other in the sense of factors $\endgroup$ – M. A. SARKAR Apr 13 '17 at 21:39
  • $\begingroup$ Coprimes, yes Performing the Euclidean algorthm, you find a constant as gcd. $\endgroup$ – Bernard Apr 13 '17 at 22:29
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The gcd= 1.

$x^4+ix^2+1=(ix^2+1)(\dfrac{x^2}{i}+2) - 1$

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First note that $$\gcd(x^4+ix^2+1,ix^2+1)=\gcd(x^4,ix^2+1)$$ and that $x^4$ and $ix^2 + 1$ don't have a common linear factor since they don't have common zeroes. All irreducible polynomials over $\Bbb C$ are linear, so those polynomials don't have common irreducible factors. We conclude that GCD is $1$.

Note. This a bit lengthy explanation, but easier than long division. Alternatively, note that $$1+x^4 = (1+ix^2)(1-ix^2).$$

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