We know that $\mathbb R \times \mathbb R$ forms a field under addition and multiplication defined as $(a,b)+(c,d)=(a+c,b+d)$ ; $(a,b)*(c,d)=(ac-bd,ad+bc)$ ; is there any other way to make $\mathbb R \times \mathbb R$ into a field non-isomorphic to this usual field ?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ What you're asking is to list all the fields with cardinality $2^{\aleph_0}$. There are lots of those. $\endgroup$– Dan ShvedCommented Oct 22, 2014 at 14:18
-
$\begingroup$ Do you have any restrictions? Because if you just want a field structure on the set $\Bbb R^2$ then any field $F$ of the same cardinality has a (set) bijection $\Bbb R^2$, which gives a field structure. $\endgroup$– ArthurCommented Oct 22, 2014 at 14:19
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
There are many ways to do it while preserving addition as it is.
Note that if $F$ is any field of characteristics $0$ and cardinality $2^{\aleph_0}$ then as a vector space over $\Bbb Q$, $F$ and $\Bbb R$ and $\Bbb{R\times R}$ are isomorphic. So we can transport the multiplicative structure from $F$ to $\Bbb R$ or $\Bbb{R\times R}$ while preserving addition.
So really you're asking how many non-isomorphic fields of characteristics $0$ there are on a set of cardinality $2^{\aleph_0}$. Some non-isomorphic examples are $\Bbb R$, the different $p$-adic fields, any transcendental extension of $\Bbb R$ by at most $2^{\aleph_0}$ transcendental elements; any subfield of $\Bbb R$ of the right cardinality, $\Bbb C$, etc.
answered Oct 22, 2014 at 14:20
$\begingroup$
$\endgroup$
10
Yes, there are bijections between $\mathbb{R}$ and $\mathbb{R}\times \mathbb{R}$, and thus we can take any bijection $f:\mathbb{R}\rightarrow \mathbb{R}\times \mathbb{R}$ and declare it a field isomorphism. That is, it forms a field with addition $$ (a,b)+(c,d)=f(f^{-1}(a,b)+f^{-1}(c,d))$$ where the additive identity is $f(0)$, and multiplication $$ (a,b)\cdot (c,d)=f(f^{-1}(a,b)\cdot f^{-1}(c,d))$$ where the multiplicative identity is $f(1)$.
These are all isomorphic (being isomorphic to $\mathbb{R}$), but not to the normal field structure on $\mathbb{R}\times \mathbb{R}$, since the latter is not isomorphic to $\mathbb{R}$.
-
$\begingroup$ But they are all isomorphic. $\endgroup$– Asaf Karagila ♦Commented Oct 22, 2014 at 14:18
-
$\begingroup$ @AsafKaragila Yes, but not to the normal field structure on $\mathbb{R}\times \mathbb{R}$. $\endgroup$– HaydenCommented Oct 22, 2014 at 14:19
-
$\begingroup$ But the question explicitly said "non-isomorphic". $\endgroup$– Asaf Karagila ♦Commented Oct 22, 2014 at 14:21
-
$\begingroup$ @AsafKaragila I read the question as saying "non-isomorphic to the usual field structure on $\mathbb{R}\times \mathbb{R}$". In this sense, I provided such an example, since $\mathbb{R}$ with its usual field structure is not isomorphic to the usual field structure on $\mathbb{R}\times \mathbb{R}$. $\endgroup$– HaydenCommented Oct 22, 2014 at 14:22
-
$\begingroup$ I think that you're being too literal with this. But sure, let's wait to hear from the OP. $\endgroup$– Asaf Karagila ♦Commented Oct 22, 2014 at 14:24
$\begingroup$
$\endgroup$
Assuming you want $\mathbb{R}$ to be a sub-field of the field $\mathbb{R}\times\mathbb{R}$, the answer is no. To see that, note that $\mathbb{C}$ is algebraically closed, and that it is the only possible algebraic extension of $\mathbb{R}$. In other words, once you extend $\mathbb{R}$ by a new number which is algebraic over $\mathbb{R}$, you automatically get something isomorphic to $\mathbb{C}$.
answered Oct 22, 2014 at 14:18