So I have just got into Algebra and I intuitively understand the idea of an extension. However, I am struggling in answering this certain question. In $GF(2)$, it is known that $x^2+x+1$ is irreducible. However, we can extend this field such that $x^2+x+1$ has a root. A trivial extension would be the complex field and we would get two complex roots using the qudratic formula. However, if we were to extend our base field to $GF(2^4) = GF(16)$ and then reduce it by $x^4 + x +1$. What would the roots of $x^2+x+1$ be in this extension field?
EDIT:
Thank to Wuestenfux, I do understand extension field better now. However, how do you figure out that $\beta^5$ and $\beta^{10}$ are roots of the polynomial in the extension field?
The complex field would not be an extension of $GF(2)$, since the complex field does not contain $GF(2)$ as a subfield.
The polynomial $x^2+x+1$ is irreducible over $GF(2)$ and it splits in $GF(4)$ into linear factors. For this, let $\alpha$ be a root. Then $\alpha^2=\alpha+1$ and so $GF(4)=\{0,1,\alpha,\alpha^2\}$.
The field $GF(16)$ contains $GF(4)$ as a subfield. If $\beta$ is a root of the (primitive) irreducible polynomial $x^4+x+1$ over $GF(2)$, the elements of $GF(16)$ are $0,1,\beta,\ldots,\beta^{14}$. In particular, $0,1,\beta^5,\beta^{10}$ are the elements of the subfield $GF(4)$.