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Since the polynomial $p=x^4−2$ is irreducible over $\mathbb{Q}$, the factor ring $K=\mathbb{Q}[x]/(p)$ is a field.

I'd like to factor the polynomial $q=y^4−2$ in $K[y]$ into a product of irreducible polynomials, so that I can prove all factors in this decomposition are indeed irreducible. How should I factor the polynomial out?

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  • $\begingroup$ $y^4-2$ is a constant in $K$ because it does not mention $x$. So I do not understand the question. Please can you explain? $\endgroup$ – apt1002 Nov 19 '13 at 3:11
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In $K$, $x$ is a root of $p$, so $y-x$ is a factor of $q$. $y+x$ is also a factor, So $y^2-x^2$ is a factor. Now divide $q$ by $y^2-x^2$ to get the remaining factor (not forgetting to use $x^4=2$ somewhere along the way). You should get $$q=(y-x)(y+x)(y^2+x^2)$$

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    $\begingroup$ For short: in $K$ we have $2=x^4$ and then $y^4-2=y^4-x^4$. $\endgroup$ – user89712 Nov 19 '13 at 8:58
  • $\begingroup$ How can I show that $y^2 + x^2$ is irreducible over $K$? $\endgroup$ – PandaMan Dec 7 '13 at 7:14
  • $\begingroup$ Good question. If it were not irreducible, then $K$ would be normal over the rationals; adjoining one root of $x^4-2$ to the rationals would adjoin them all. But clearly adjoining a real root will not get you either of the nonreal roots. $\endgroup$ – Gerry Myerson Dec 8 '13 at 0:58
  • $\begingroup$ Thanks Gerry! That makes sense. $\endgroup$ – PandaMan Dec 8 '13 at 22:55

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