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If $F_1$ is field with cardinality $2^9$ and $F_2$ is a field with cardinality $2^6$ then what is the cardinality of $F_1\cap F_2$. My answer is $2^n$ such that $n|9$ and $n|6$ That gives me $n=1$ and $n=3$ .Which one is correct can somebody help

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    the multiplicative group will contain all those elements that are prime to $512$ and $64$ because the groups are $U(512)$ and $U(64)$ .am i rightDon Anotonio?. – 2017-02-07
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    To answer this, you need the following statement, which is the converse of the statement you are already using in your thoughts: If $d|n$, we have $\mathbb F_{p^d} \subset \mathbb F_{p^n}$. You can proof it like this: You can find a copy of $\mathbb F_{p^d}$ in $\mathbb F_{p^n}$ as the fixed points of $x \mapsto x^{p^d}$. – 2017-02-07
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    since the prime subfield $Gf(p)$ is contained in any other subfield then the answer should be $GF(2^3)$ – 2017-02-07
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    This is correct. – 2017-02-07

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