abstract algebra - Order of an element in a field.

Let $C = \mathbb{F}_2[X] / \langle X^6 + X + 1\rangle$ be a ring.

1. Let $\alpha$ be the class of $X$ in $C$. Prove that $\alpha \in C^\times$ and
   compute the order of $\alpha$ in $C^\times$.




2. Conclude that $C$ is a field.




3. Prove that $P(X)$ is irreducible and primitive.

My main focus here is the first question, because the last two follow from it. My train of thought was to:
1) Prove that $X^6 + X + 1$ is irreducible, from which would follow that $C$ is a field and $C^\times = C \backslash \{0\}$. Proving then that $\alpha \in C^\times$ would be equivalent to proving that $\alpha \neq 0$.

But as you can see, it is only aksed to prove $C$ to be a field in the second question, and that $X^6 + X + 1$ is irreducible in the third. So there must be another, presumedly better, way of answering the first question. Plus even if I go through the trouble of proving $C$ is a field from the first question, I don't see how to find the order of $\alpha$.

Answer

For 1., you want to find a polynomial $g(X)\in\mathbb F_2[X]$ such that $Xg(X)\equiv 1\pmod{X^6+X+1}$. To that end, try $g(X)=X^5+1$.

Now to find the order of $\alpha$, I'm going to cheat slightly. If $C$ is a field then $\alpha$ is a primitive element of $C$ over $\mathbb F_2$, and moreover $C/\mathbb F_2$ is a degree $6$ extension, we expect the order of $\alpha$ to be $2^6-1=63$.

So I "guess" that the order of $\alpha$ will be $63$, therefore you need to calculate $\alpha^{63}$, $\alpha^9$, and $\alpha^{21}$. Calculating $\alpha^{63}$ looks like it will be tedious, but using the identities $\alpha^6=\alpha+1$ and $(a+b)^2=a^2+b^2$ it's not so bad.

Then you will have shown that $C$ contains $63$ units, and $C$ is a commutative ring with $64$ elements, so it must be a field.