Does there exist a field containing $\mathbb{Z}_5$ and $\mathbb{Z}_7$ as subfields?

Question

I think no because if we assume $\mathbb{F}$ is of size $p^t$ where $p$ is prime and $t\in \mathbb{Z}^+$ then both 5 and 7 would need to divide $p^t$ and I don't think that could happen. Is this correct?

Answer 1

A helpful fact here is that the intersection of all subfields of a field is again a field. It is easy to prove and shows that no field as described exists.

Answer 2

If $E$ is a subfield of $F$, then the multiplicative identity (aka. "$1$") of $E$ coincides with the multiplicative identiy of $F$. Hence for a field having $\Bbb F_5$ as subfield, we have $1+1+1+1+1=0$, and for a field having $\Bbb F_7$ as subfield, we have $1+1+1+1+1+1+1=0$. From these two, we easily obtain $1=0$, contradiction.

Answer 3

If there existed such a field, it would both have characteristic $5$ and $7$. As $5$ and $7$ are coprime, this would imply $1=0$.