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The polynomial is: $P(x)=x^6+x^4-x^3+x^2+1$. I need to factor it over $C, Q, R$ if one complex root is $\sqrt[3]{1}$. Also find all fields in which $P$ is reducible.

Now, I know how to find one factor of P using given complex root, and I end up with $(x^2+x+1)(x^4-x^3+x^2-x+1)$, but how should I go about $(x^4-x^3+x^2-x+1)$?

How can I find all fields in which $P$ is reducible?

J.Doe
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1 Answers1

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A start: To factor the quartic, it may be useful to find its roots.

Rewrite the quartic as $x^2(x^2+\frac{1}{x^2}-\left(x+\frac{1}{x}\right)+1)$.

Let $t=x+\frac{1}{x}$. Then to solve our quartic we first solve $$t^2-2-t+1=0.$$

Another way: Our quartic is $\frac{x^5+1}{x+1}$. So its roots are certain $10$-th roots of unity.

André Nicolas
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  • thanks...what about second part, about finding all fields? – J.Doe Jul 11 '16 at 18:31
  • Technically it is reducible in all these fields, by the factorization you found. But maybe you mean expressible as a product of linear factors. Then we need a primitive tenth root of unity, and a primitive sixth root of unity (for the $x^2+x+1$ part). I don't know what all fields means. Any such field contains a primitive $30$-th root of unity, and any field that contains such a primitive $30$-th root will do the job. – André Nicolas Jul 11 '16 at 18:37
  • it just says that: find all fields in which $P$ is raducible – J.Doe Jul 11 '16 at 20:18
  • It depends on the definition of reducible. If you look here, you will be told that sometimes a polynomial which is not irreducible is called reducible. If we use that definition, then our polynomial is reducible over all fields. But they may mean completely reducible, that is, expressible as a product of linear factors. In the above comment, I described all subfields of the complex numbers over which our polynomial is completely reducible. To describe all fields, including the ones of non-zero characteristic, may be complicated. – André Nicolas Jul 11 '16 at 20:28