A finite field $F$ such that $F[\sqrt{2}]$ and $F[\sqrt{3}]$ are not isomorphic (as fields) Assuming $F$ is a finite field such that $F[\sqrt{2}]$ and $F[\sqrt{3}]$ are both fields, I am trying to prove that they must both be isomorphic. Or is there a counterexample? Is there a counterexample where $F$ has prime order? I have looked hard for one and cannot find any. 
For the avoidance of doubt $F[\sqrt{2}] := F[X]/(X^2-2)$, so assume that $X^2 -2$ and $X^2-3$ are irreducible over $F$. 

If neither $2$ nor $3$ are squares in the finite field $F$, then both $F[\sqrt{2}]$ and $F[\sqrt{3}]$ are extensions of $F$ having degree $2$, so both have $|F|^2$ elements. Hence they're isomorphic: if $p$ is a prime, then any two fields of cardinality $p^n$ ($n>0$) are isomorphic, because both are the splitting field of $x^{p^n}-x$.
The only case in which they aren't isomorphic, is when one among $2$ and $3$ is a square and the other isn't.
