Is there a field structure on $\mathbb{R}^3$ that keeps its structure as a vector space over $\mathbb{Q}$？ This was a problem from my abstract algebra final. My professor wanted us to show that there is NO multiplication on $\mathbb{R}^3$ appropriately defined that makes it a field together with the natural addition in the form of vectors. I've seen a lot about the fact that only when n=1,2,4,8 exists division algebras, but they all require $\mathbb{R}^n$ to inherit its structure as a vector space over $\mathbb{R}$. However, I could only deduce $\mathbb{Q}$-linear structure from the conditions given. I think that's not enough. Is there any crucial property I neglected, or is there something wrong with this problem?
By natural addition in the form of vectors I mean the operation:
$(a,b,c)+(d,e,f)=(a+d,b+e,c+f)$.
 A: Unless I'm missing something, this problem is incorrect.
Specifically, I claim that as groups $\mathbb{R}^3$ and $\mathbb{R}$ are isomorphic. From this, we can lift the usual multiplication on $\mathbb{R}$ to an appropriate multiplication on $\mathbb{R}^3$ which provides a counterexample to the problem.
This is a situation where an idea from linear algebra - specifically, that of bases - is extremely useful. Both $\mathbb{R}^3$ and $\mathbb{R}$ are in fact $\mathbb{Q}$-vector spaces, and this subsumes their group structures. But two vector spaces of the same dimension over the same field are isomorphic. Now $\mathbb{R}^3$ and $\mathbb{R}$ have the same cardinality and are each uncountable, so they have the same dimension as $\mathbb{Q}$-vector spaces and hence are isomorphic.
Now, any isomorphism of $\mathbb{Q}$-vector spaces between $\mathbb{R}$ and $\mathbb{R}^3$ will be really weird (in particular, we actually need the axiom of choice to prove they exist in the first place! this goes back to the role of choice in establishing basic (hehe) facts about bases and isomorphism), and the multiplication on $\mathbb{R}^3$ we get by pushing the usual multiplication on $\mathbb{R}$ along such an isomorphism will correspondingly be weird. If we add further conditions to the problem which prevent this, then the argument above breaks down and indeed the modified statement may well be true (most naturally, the statement becomes true if we demand that our multiplication be continuous, and I think even Borel suffices).
