Regarding the structure of certain minimal polynomials

Question

My question is: is the following statement true? If so, how might I go about proving it?

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Let $m$ be an integer not less than 1, let $F = GF(2^{6m})$, and let $L = GF(2^{2m})$. Let $\gamma$ be a primitive element of $L$, and let $\beta \in F$ be such that $\beta^3 = \gamma$. If $\phi_a(x) = c_0x^{6m}+c_1x^{6m-1}+...+c_{6m-1}x+c_{6m}$ is the minimal polynomial of an element $a \in \beta L$, then $$c_i = 0 \hspace{5mm} \text{if $i \not\equiv 0 \pmod{3}$}$$

Answer 1

The following sketch is the best I can muster right away. It probably isn't the cleanest route to the destination, but it leads to a proof.

Assume that $z=\gamma^j$ is an arbitrary non-zero element of $L$. Then

• $\beta z=\beta^{1+3j}$, and
• $(\beta z)^3=\gamma^{1+3j}\in L$.

Let $\phi(x)$ be the minimal polynomial of $\gamma^{1+3j}\in L$. Then

• $a=\beta z$ is a zero of the polynomial $\phi(x^3)$, and
• the coefficients of terms of $\phi(x^3)$ of degree $\not\equiv0\pmod3$ are all zero.

This proves your observation, if we can show that $\phi(x^3)$ has the correct degree ($=\deg\phi_a(x)$). Because $\deg\phi(x)$ is always a factor of $2m$, we are done, if we have that extra bit of information that $\deg\phi_a(x)=6m$.

Even if that is not the case we can make the following deductions (leaving them sketchy as per your request):

• We have $\deg\phi(x)$ is equal to the degree of the extension $GF(2)[a^3]$ over $GF(2)$.
• Similarly $\deg\phi_a(x)$ is equal to the degree of the extension $GF(2)[a]$ over $GF(2)$.
• Here $GF(2)[a^3]$ is a subfield of $L$.
• But $GF(2)[a]$ is not a subfield of $L$. It is a subfield of $F$ though.
• Clearly $GF(2)[a^3]$ is a subfield of $GF(2)[a]$.
• Putting these bits together shows that $GF(2)[a]$ is a cubic extension of $GF(2)[a^3]$.
• Therefore $\deg\phi_a(x)=3\deg\phi(x)$, and the claim follows as above.

The next to last bullet is a bit subtler than the rest. Remember that the cardinality of a subfield of $F$ determines it uniquely.