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Let $L = \mathbb F_2[X]/\langle X^4 + X + 1 \rangle$ is a field. Show $L^* = L / \{0\} = \langle X \rangle$ is cyclic.

I've proven that $X^4 + X + 1$ is irreducible, so $L$ is a field. I also know that $X^5 + X + 1$ is not irreducible. Also I've proven that $|L| = 16$.

Could someone help me out proving that $L^*$ is cyclic ?

Jyrki Lahtonen
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Shuzheng
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3 Answers3

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since the group is of order 15 any element must have order 1,3,5 or 15.

in any field the equation $x^k = 1 $ can have at most k roots, so the number of elements with order less than $15$ is at most $1 + 3 + 5 = 9$

hence there must be an element of order 15

i will add, after the query from OP, that a slightly more sophisticated form of this argument will show that for any finite field its multiplicative group is cyclic.

David Holden
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  • Can you explain a little more ? Indeed $x^k = 1$ has at most $k$ roots. But in this case it has $15$ since every element to the power of the group is $1$ ? – Shuzheng Nov 28 '13 at 11:58
  • no, i have no desire to start sentences with capital letters except in case the first word is a proper noun, or i am quoting someone verbatim. perhaps you could supply them mentally, since the algorithm required is very simple? – David Holden Nov 28 '13 at 13:25
  • @user111854 set k=1, one root. set k=3 then at most 3 roots (one of which is the k=1 solution) so in fact at most 2 elements of order 3. set k=5, then, likewise, at most 4 elements of order 5. so there must be at least 8 elements of order 15 as this is the only remaining possibility. (note that 8 is the size of the set of +ve numbers <16 which are relatively prime to 15.) – David Holden Nov 28 '13 at 13:31
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    there is a useful discussion of the question on mathoverflow (2011) http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic – David Holden Nov 28 '13 at 17:16
  • Thanks David for your enlightning answer. – Shuzheng Nov 28 '13 at 18:03
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Calculate $X,X^2,\ldots, X^{15}$ (modulo $X^4+X+1$) and show that you get everything in $L^*$.

Casteels
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The multiplicative group of a finite field is always cyclic.

lhf
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    See http://math.stackexchange.com/questions/59903/finite-subgroups-of-the-multiplicative-group-of-a-field-are-cyclic. – lhf Nov 28 '13 at 11:19
  • I have the lemma: Let $F$ be a finite field. Then $|F| = p^n$, where $p$ is prime $n \ge 1$ and there exists an irreducible polynomial of degree $n$ such that $F \cong \mathbb F_p[X]/\langle f \rangle$. Can I use this ? Let $F = \mathbb F_p$ ? – Shuzheng Nov 28 '13 at 11:26
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    That won't help you prove it cyclic. You can either do it directly for this case only, as suggested by Casteels, or you can use the general result that finite subgroups of multiplicative groups of fields are cyclic. – Derek Holt Nov 28 '13 at 11:38