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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.

1 Answers1

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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.
Jyrki Lahtonen
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