Field with both $\mathbb{Z}/{5}$ and $\mathbb{Z}/{7}$ as subfields? Is it possible for a field to exist containing both $\mathbb{Z}/{5}$ and $\mathbb{Z}/{7}$ as subfields?
I know that there is a result that says that every field must have a subfield isomorphic to either $\mathbb{Z}/{p}$ for some prime $p$ or $\mathbb{Q}$, but it doesn't tell us for which primes $p$, or whether it can have two subfields of prime order.
$\mathbb{Z}/{35}$ contains both $\mathbb{Z}/{5}$ and $\mathbb{Z}/{7}$, but it's not a field. 
Could anyone point me in the right direction here?
Thank you for your time and patience.
 A: The characteristic of a field is the smallest positive integer $n$ such that $n\times 1=0$, or $\infty$ if there is no such integer. If it is not infinite, then it has to be a prime number.
A couple of relevant results:


*

*$\Bbb F_p$ has characteristic $p$ for any prime $p$.

*If $K$ is a subfield of $L$, then they both have the same characteristic.


As a consequence, there can be no field containing $\Bbb F_5$ and $\Bbb F_7$ as subfields.
Edit: $\Bbb F_p$ here is just the field notation for $\Bbb Z_p$, when you want to emphasize that you are talking about the field rather than the group.

Edit to include the answers to your questions in comments:
Proof of 2. above. Call $\phi$ the injective field homomorphism $K\to L$. Then $\phi(n\times 1_K)=n\times 1_L$ for any $n\in \Bbb N$. This implies, on the one hand, that if $n\times 1_K=0$ then $n\times 1_L=0$. On the other hand, since $\phi$ is injective, it also implies  that if $n\times 1_L=0$ then $n\times 1_K=0$.
An infinite field may have finite characteristic. The field of fractions of $\Bbb Z_2[X]$ has characteristic $2$, but it is infinite.
A: The ring $\def\Z{\mathbb{Z}}\Z/35\Z$ contains $\Z/5\Z$ and $\Z/7\Z$ as subrings only if you don't require subrings share the identity with the ring.
However, a subfield of a field must share the identity, because fields have a single nonzero idempotent, namely $1$: indeed, from $x^2=x$ we can deduce, in a field, $x=0$ or $x=1$.
On the other hand, there is at most one unital ring homomorphism $\chi\colon\Z\to R$, for every ring $R$ (with $1$, of course), because the map $\chi$ is determined by $\chi(1)=1$, being a group homomorphism.
Actually, $\chi$ exists, but this is not needed for the argument here.
If both $\Z/5\Z$ and $\Z/7\Z$ are (unital) subrings of $R$, there would be two distinct homomorphisms $\Z\to R$. Contradiction.
A: Hint: If a field has an element $x\neq 0$ with $px=0$ then for all elements $y$ of the set $py=0$.
