${\mathbb{F}}_3({\mu }_5)\cap {\mathbb{F}}_3({\mu }_7)$ is larger than ${\mathbb{F}}_3$ although five and seven are coprime

In $(\mathbb{Z}\mathrm{/}5{)}^{\times }$, the class $[3]$ has order $4$, since $3^4=81\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}5)$ and no smaller positive power of $3$ is congruent to $1$ modulo $5$. So [L1] gives $[{\mathbb{F}}_3({\mu }_5):{\mathbb{F}}_3]=4$, hence this splitting field has order $3^4$ and [L5] lets us identify it over ${\mathbb{F}}_3$ with the unique subfield ${\mathbb{F}}_{3^4}$ of $\mathrm{\Omega }$.

In $(\mathbb{Z}\mathrm{/}7{)}^{\times }$, the powers of $[3]$ are $[3],[2],[6],[4],[5],[1]$, so $[3]$ has order $6$. Thus [L1] gives $[{\mathbb{F}}_3({\mu }_7):{\mathbb{F}}_3]=6$, hence this splitting field has order $3^6$ and [L5] lets us identify it over ${\mathbb{F}}_3$ with the unique subfield ${\mathbb{F}}_{3^6}$ of $\mathrm{\Omega }$.

Under the fixed identifications of steps 1.1 and 1.2, both ${\mathbb{F}}_{3^4}$ and ${\mathbb{F}}_{3^6}$ are subfields of $\mathrm{\Omega }={\mathbb{F}}_{3^{12}}$ by [L2], since $4\mid 12$ and $6\mid 12$. Their intersection is then an intermediate field of $\mathrm{\Omega }\mathrm{/}{\mathbb{F}}_3$, so by [L2] it is ${\mathbb{F}}_{3^d}$ for some divisor $d$ of $12$. Because the intersection lies in both fields, [L2] gives $d\mid 4$ and $d\mid 6$, hence $d\mid 2$; and since ${\mathbb{F}}_{3^2}$ lies in both fields, [L2] gives $2\mid d$. Therefore $d=2$ and ${\mathbb{F}}_3({\mu }_5)\cap {\mathbb{F}}_3({\mu }_7)={\mathbb{F}}_{3^2}={\mathbb{F}}_9\mathrm{.}$

Since $\gcd (5,7)=1$, the right-hand side of the refuted identity is $K({\mu }_1)$, which is just $K={\mathbb{F}}_3$ because $t-1$ already splits over $K$. So the claimed equality would read ${\mathbb{F}}_9={\mathbb{F}}_3$, which is false.

The refuted statement therefore fails over the base field ${\mathbb{F}}_3$, even though the rational theorem [L3] is true.