Show that $a^3+a+1$ is in the subfield $L$ with $4$ elements of the field $K=F_2[X]/(X^4+X^3+1)$ The field $K$ is constructed in the following way: $K=F_2[X]/(X^4+X^3+1)$, where $F_2$ is short for $\mathbb{Z}$/$2$$\mathbb{Z}$.
Let $a$ be the class of $X$ in $K$ (so $a=X+(X^4+X^3+1))$. 
The field $K$ contains a unique subfield $L$ with $4$ elements.
I need to show that $a^3+a+1$ is in that subfield $L$.
I know that a subfield of field $K$ is a subset containing $0,1$ and that it is closed with respect to addition, multiplication and inverses. But I don't know how to show that the element $a^3+a+1$ is in $L$. How can I do this?

We just need to perform a little of linear algebra to check that $a^3+a+1$ is an element of order $3$. In the given field, $a^4=a^3+1$, so:
$$ (a^3+a+1)^2 = a^6+a^2+1 = a^2(a^3+1)+a^2+1 = a+a^3,$$
$$ (a^3+a+1)^3 = (a^3+a)^2+(a^3+a) = a^6+a^2+a^3+a = 1 $$ 
and the set $\{0,a^3+a+1,a^3+a,1\}$ is a subfield with four elements.
