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?

The field $K$ that you have has $2^4=16$ elements. 
Let $b=a^3+a+1$, then 
$$b^2=(a^3+a+1)^2=a^6+a^2+1=a^2(a^3+1)+a^2+1=a^5+1=a(a^3+1)+1=a^3+a.$$
This gives,
$$b^2=b+1$$
Furthermore, we get
$$b^4=b \implies b^3=1.$$
Thus $b$ has order $3$. Now we know that $K$ has a unique subfield $L$ of size $4$, so $|L^{\times}|=3$ (cyclic group of order $3$) and by uniqueness it should be generated by $b$ (element of order $3$). 
