Subfields of irreducible polynomial fields with known dimensions

Question

Let's say we have an irreducible polynomial, $h(x) = x^4 + x + 1 \in \Bbb F_2[x]$, and that L is a field equal to $\Bbb F_2 [x]/(h(x))$. How would I go about finding a subfield K such that $[L : K] = 2$? And how would I prove that the multiplicative group $K^*$ is cyclic?

All I've gotten so far is that $K$ will have four elements, although I think the set ($x^3, x^2, x, 1$) is too simple to be the answer.

Answer 1

Here are some hints/steps/suggestions/observations.

• You know that $L$ has sixteen elements. Therefore $K$ should have four. Therefore $K^*$ has three. What do you know about groups of three elements?
• $L^*$ has fifteen elements, so by Lagrange's theorem all $z\in L^*$ satisfy the equation $z^{15}=1.$ Similarly if $w\in K^*$ it is a solution of $w^3=1$. Can you show that all the solutions $y$ of $y^3=1$ are in $K^*$? Hint: How many solutions (at most) to a cubic equation in a field.
• Assuming that you can do the previous bullet, and given that $z^{15}=1$, for which exponents $m$ can you tell that $(z^m)^3=1$?
• After you have found the correct $m$ in the previous bullet, apply it to $z=x+\langle h(x)\rangle\in L$. Remember that the subfield $K$ has to be closed under addition and multiplication.
• Have you found all the four elements of $K$, and checked that it is a subfield? Time to mark this one as solved! You can also check/verify your solution to my solution here. You are welcome to read through it all, but only the first paragraph afte the last horizontal line is relevant to this problem.