# Field of roots of polynomial [closed]

Let $f(x)=(x^2+x+1)(x^5+x+1)\in \mathbb Z_2[x]$. Find the field of roots of this polynomial.

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## closed as off-topic by Gerry Myerson, Belgi, Fly by Night, Mathmo123, Najib IdrissiAug 3 at 14:13

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You can't do any part of this question? You can't find the roots of $x^2+x+1$? You don't know what "field of roots" means? Come on, give us a hand here, tell us what you know about the problem. Get us started, we'll take you the rest of the way. –  Gerry Myerson Aug 3 at 13:05
Chiming in with Gerry. Can you for example find the irreducible factors of $x^2+x+1$ and $x^5+x+1$? –  Jyrki Lahtonen Aug 3 at 13:09
For the roots to be in $\Bbb{C}$ you would need first to realize the field $\Bbb{Z}_2$ as a subfield of $\Bbb{C}$. That is not possible. The field containing the roots of a polynomial must also contain a copy of the field of coefficients. Have you covered methods for extending a field by adjoining roots of irreducible polynomials? Comes in handy here. Also you need basics of finite fields. Review those first. –  Jyrki Lahtonen Aug 3 at 13:17
There are no fields of the form $\Bbb{Z}_k$ that contain $\Bbb{Z}_2$ as a subfield (no rings either, but that's besides the point). –  Jyrki Lahtonen Aug 3 at 13:23
What do you know about finite fields? It is unthinkable that this would have been given as a problem without covering the basics. For example, do you know that $p(x)\in\Bbb{Z}_2[x]$ is irreducible, then $\Bbb{Z}_2[x]/\langle p(x)\rangle$ is a field. Do you know that up to isomorphism a finite field is determined by its cardinality? Do you know which finite fields can appear as subfields of another? Without those bits this exercise is IMO nearly impossible. Gotta rush, sorry. –  Jyrki Lahtonen Aug 3 at 13:28

also one might note that in any field a root of $x^2+x+1=0$ is a cube root of unity, hence satisfies $x^5=x^2$. this is reflected in the factorization: $$x^5+x+1 = (x^2+x+1)(x^3-x^2+1)$$ which gives Nicky's form in $\mathbb{Z}_2$