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If I know $F$ is a ring/field, and I have $G$ (unkown), but I can find a bijective map: $$\Phi:F\to G$$ such that
$\forall a,b\in F.\;\Phi(a+b) =\Phi(a)+\Phi(b).\;\Phi(ab) = \Phi(a)\Phi(b)$
Does that mean G is automatically a ring/field and is isomorphic to $F$? If so what's the name of the theorem, if not any counter-examples? Thank you.

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So you know that $G$ has two operations, you just don't know that those make $G$ a ring a priori? – Tobias Kildetoft Mar 11 '13 at 15:08
If all you know about $G$ is that there is a bijection $\Phi:F\rightarrow G$, then you can define $+$ and $\cdot$ in $G$ to be just what you have, and then $F$ and $G$ will be isomorphic. – Arthur Mar 11 '13 at 15:11
The is called transport of structure by $\Phi$ from the ring/field $F$ to the set $G$. – Marc van Leeuwen Mar 11 '13 at 15:21
@TobiasKildetoft Yes. – mez Mar 11 '13 at 19:21
@MarcvanLeeuwen I see. thanks. – mez Mar 11 '13 at 19:22
up vote 3 down vote accepted

Since a field is a ring, we will just use the word ring. We can now show that $G$ satisfies the axioms for a ring. For any two elements $\Phi(a), \Phi(b)$ we can form two more $\Phi(a+b), \Phi(ab)$. And we also have the distributivity, so it must be a ring as well. We also have that if $F$ is a field, then $G$ is a field because they are isomorphic rings.

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