Show that E is a splitting field for these polynomials Show that if $E := \frac{\mathbb{F}_2[x]}{(x^4+x+1)}$, then $E$ is a splitting field for
i) $x^4+x+1$ 
ii) $x^2 +x+1$
So I am trying to study for an exam and I realized that for part a, it has to be the splitting field of the polynomial by construction. Since the polynomial is the same as the one in the modulus. However, I have no idea how to proceed with part ii). Since we are in $\mathbb{F}_2$ and $0$ and $1$ both do not satisfy the polynomial I'm not sure how to find it.
Also, am I correct in assuming that:
$$\{1, x, x^2, x^3\}$$
forms a basis for the field?
 A: First of all, fix an algebraic closure $\overline{\mathbb F}_2$ of $E = \mathbb F_2(\alpha)$, where $\alpha$ is a root of the polynomial $X^4+X+1\in \mathbb F_2[X]$. Using the fact that $\overline{\mathbb F}_2$ contains exactly one subfield of order $2^n$ ($n\in \mathbb N$), namely $\{a\in \overline{\mathbb F}_2\,|\, a^{2^n}-a = 0\}$, we see that $E$ is the only subfield of order $2^{[E:\mathbb F_2]} = 2^4$ (it is a field, since $X^4+X+1$ is irreducible over $\mathbb F_2$). 
Now, if $\beta$ is any other root of $X^4+X+1$, then we have a canonical isomorphism of fields $E=\mathbb F_2(\alpha)\rightarrow \mathbb F_2(\beta)$ sending $\alpha\mapsto \beta$ (inside $\overline{\mathbb F}_2$). Since $\mathbb F_2(\beta)$ is also a subfield of order $2^4$, we conclude $E = \mathbb F_2(\beta)$ by the above discussion. Hence $E$ is a splitting field for $X^4+X+1$.
Now, $E$ contains a subfield $F$ of order $2^2$ (since $2\mid 4$) and by the above discussion, $F$ is unique. Since also $\frac{\mathbb F_2[X]}{(X^2+X+1)}$ is a field with $2^2$ elements, we conclude $\frac{F_2[X]}{(X^2+X+1)} = F\subseteq E$. This shows that $F$ is the splitting field of $X^2+X+1$, i. e. $X^2+X+1$ splits in $E$. (But note, that $E$ is not the splitting field of $X^2+X+1$, since splitting fields are defined to be minimal).
By the way, the whole argumentation carries over to the case of arbitrary finite fields.
And yes, $\{1,x,x^2,x^3\}$ forms a basis of $E$ over $\mathbb F_2$, where $x$ denotes the image of $X$ in $E = \frac{\mathbb F_2[X]}{(X^4+X+1)}$.
A: Hint:
$$x^2+x+1 + \langle x^4+x+1\rangle$$
$$\equiv x^4+x^2 + \langle x^4+x+1\rangle$$
$$x^2(x^2+1)=(x^2)(x+1)(x+1)+\langle x^4+x+1\rangle \quad \text{ over }\Bbb F_2[x]/\langle x^4+x+1\rangle$$
For the basis, consider the fact that $\Bbb F_2[x]/\langle x^4+x+1\rangle =\Bbb F_2(\alpha)$
A: Part i) is answered in the affirmative if you recall a theorem stating that $\Bbb{F}_2[x]/\langle x^4+x+1\rangle$ is a Galois extension. Basically because any extension involving only finite fields is automatically Galois (and hence normal). For full credit you need to be aware that it is NOT a SPLITTING FIELD by construction. For example $\Bbb{Q}[x]/\langle x^3-2\rangle\cong\Bbb{Q}(\root3\of2)$ is well known not to be a splitting field of $p(x)=x^3-2$ over $\Bbb{Q}$ because it only has a single zero of $p(x)$.
Alternatively you can use the Frobenius automorphism (or freshman's dream). So if we denote 
$$
\alpha=x+\langle x^4+x+1\rangle\in K=\Bbb{F}_2[x]/\langle x^4+x+1\rangle,$$
then it is easy to show that in $K[x]$ we have the factorization
$$
x^4+x+1=(x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)
$$
and $x^4+x+1$ thus splits into linear factors over $K$.

For part ii) we observe that $(x^2+x+1)(x-1)=x^3-1$, so zeros of $x^2+x+1$ are roots of unity of order $3$. Because $|K^*|=16-1=15$ Lagrange's theorem tells us that $\alpha^5$ and $\alpha^{10}$ are both of order three (you do need to check that $\alpha^5\neq1_K$). Therefore over $K$ we  have the factorization
$$
x^2+x+1=(x-\alpha^5)(x-\alpha^{10}).
$$
If you get stuck in verifying this, please study this CW answer I prepared for referrals. What I denoted $\alpha$ here is called $\gamma$ there. Sorry about that.

Re your question about basis I would rather say that $\{1,\alpha,\alpha^2,\alpha^3\}$ is a basis. We do have
$$
\alpha^i=x^i+\langle x^4+x+1\rangle
$$
for all natural numbers $i$. But it is IMO better not to write $x$ in place of $\alpha$. It is better to reserve $x$ for the indeterminate. When you consider the elements of the polynomial ring $K[x]$ that $x$ there has nothing to do with $\alpha$ - it is an indeterminate but $\alpha$ is a constant from the field $K$. 
