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Consider the irreducible polynomial $g = X^4 + X + 1$ over $F_2$ and let $E$ be the extension of $F_2 =$ {0, 1} with a root $α$ of $g.$

Could $E$ have a subfield of order 8

When working through my work Ive come across this question which has stumped me and I cant get my head aound if this is true of not so any help will be appreciated.

  • 0
    Do you know the relationship between the degrees of field extensions given $F\subset E\subset K $? – 2017-01-17
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    What is the source of these problems? There have been a lot of more or less identical problems along these lines posted lately. Are they homework? An exam? – 2017-01-17
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    I hope the professor who is assigning all these problems about $\mathbb{F}_{16}$ is okay with their students posting their homework online. – 2017-01-17

2 Answers 2

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If $F$ is a subfield of $E$ and both fields are finite, we always have $|E|=|F|^n$ where $n$ is the degree of the field extension. The reason is that $E$ has structure as an $F$-vector space.

In particular a proper subfield of order $8$ must live in a field with at least $64$ elements...

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$E$ has $2^4=16$ elements and so the multiplicative group $E^\times$ has order $15$.

A subfield of order $8$ would yield a multiplicative group of order $7$, which does not divide $15$, as required by Lagrange's theorem.

Therefore, there is no subfield of size $8$.

This argument implies that the only possible sizes for proper subfields of $E$ are $2$ and $4$, since the proper divisors of $15$ are $1,3,5$ and $6=5+1$ is not a power of $2$.