Is there a field structure on $mathbb{R}^3$ that keeps its structure as a vector space over $mathbb{Q}$？
$begingroup$
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:
$(a,b,c)+(d,e,f)=(a+d,b+e,c+f)$.
abstract-algebra
asked Jan 19 at 4:32
NEneNEne
233
$endgroup$
add a comment |
$begingroup$
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:
$(a,b,c)+(d,e,f)=(a+d,b+e,c+f)$.
abstract-algebra
asked Jan 19 at 4:32
NEneNEne
233
$endgroup$
$begingroup$
It is not clear what kind of appropriateness the question requires, but as $mathbb{R}$ as a $mathbb{Q}$-vector space has uncountable rank, $mathbb{R}^3$ is also uncountable-dimensional over $mathbb{Q}$, and I'm pretty sure (has no example though) there exists a division algebra which has uncountable rank over $mathbb{Q}$.
$endgroup$
– cjackal
Jan 19 at 4:43
$begingroup$
A possible property that you might neglect (if there was no problem in the final): If you demand the continuity of multiplication operation(with respect to the natural topology of $mathbb{R}$ and $mathbb{Q}$) then $mathbb{Q}$-linearity promotes to $mathbb{R}$-linearity for free, so you can use Frobenius' result.
$endgroup$
– cjackal
Jan 19 at 4:50
$begingroup$
@cjackal: Obviously $Bbb R, Bbb C$ and $Bbb H$ have the same dimension, as $Bbb Q$-vector spaces, as $Bbb R^3$, so ...
$endgroup$
– Torsten Schoeneberg
Jan 19 at 5:00
$begingroup$
@TorstenSchoeneberg Oh, I desperately need a cover to hide in right now :(
$endgroup$
– cjackal
Jan 19 at 5:02
$begingroup$
Thank you! But I'm pretty sure that there is no other demand. I find this problem strange, and some of my classmates also got stuck on the $mathbb{Q}$-linearity. Your view of the uncountable rank helps a lot. Perhaps I could construct an example via linear isomorphism to $mathbb{R}$...?
$endgroup$
– NEne
Jan 19 at 5:10
add a comment |
$begingroup$
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:
$(a,b,c)+(d,e,f)=(a+d,b+e,c+f)$.
abstract-algebra
asked Jan 19 at 4:32
NEneNEne
233
$endgroup$
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:
$(a,b,c)+(d,e,f)=(a+d,b+e,c+f)$.
abstract-algebra
abstract-algebra
asked Jan 19 at 4:32
NEneNEne
233
asked Jan 19 at 4:32
NEneNEne
233
edited Jan 19 at 4:44
NEne
asked Jan 19 at 4:32
NEneNEne
233
asked Jan 19 at 4:32
NEneNEne
233
asked Jan 19 at 4:32
NEneNEne
233
233
$begingroup$
It is not clear what kind of appropriateness the question requires, but as $mathbb{R}$ as a $mathbb{Q}$-vector space has uncountable rank, $mathbb{R}^3$ is also uncountable-dimensional over $mathbb{Q}$, and I'm pretty sure (has no example though) there exists a division algebra which has uncountable rank over $mathbb{Q}$.
$endgroup$
– cjackal
Jan 19 at 4:43
$begingroup$
A possible property that you might neglect (if there was no problem in the final): If you demand the continuity of multiplication operation(with respect to the natural topology of $mathbb{R}$ and $mathbb{Q}$) then $mathbb{Q}$-linearity promotes to $mathbb{R}$-linearity for free, so you can use Frobenius' result.
$endgroup$
– cjackal
Jan 19 at 4:50
$begingroup$
@cjackal: Obviously $Bbb R, Bbb C$ and $Bbb H$ have the same dimension, as $Bbb Q$-vector spaces, as $Bbb R^3$, so ...
$endgroup$
– Torsten Schoeneberg
Jan 19 at 5:00
$begingroup$
@TorstenSchoeneberg Oh, I desperately need a cover to hide in right now :(
$endgroup$
– cjackal
Jan 19 at 5:02
$begingroup$
Thank you! But I'm pretty sure that there is no other demand. I find this problem strange, and some of my classmates also got stuck on the $mathbb{Q}$-linearity. Your view of the uncountable rank helps a lot. Perhaps I could construct an example via linear isomorphism to $mathbb{R}$...?
$endgroup$
– NEne
Jan 19 at 5:10
add a comment |
$begingroup$
It is not clear what kind of appropriateness the question requires, but as $mathbb{R}$ as a $mathbb{Q}$-vector space has uncountable rank, $mathbb{R}^3$ is also uncountable-dimensional over $mathbb{Q}$, and I'm pretty sure (has no example though) there exists a division algebra which has uncountable rank over $mathbb{Q}$.
$endgroup$
– cjackal
Jan 19 at 4:43
$begingroup$
A possible property that you might neglect (if there was no problem in the final): If you demand the continuity of multiplication operation(with respect to the natural topology of $mathbb{R}$ and $mathbb{Q}$) then $mathbb{Q}$-linearity promotes to $mathbb{R}$-linearity for free, so you can use Frobenius' result.
$endgroup$
– cjackal
Jan 19 at 4:50
$begingroup$
@cjackal: Obviously $Bbb R, Bbb C$ and $Bbb H$ have the same dimension, as $Bbb Q$-vector spaces, as $Bbb R^3$, so ...
$endgroup$
– Torsten Schoeneberg
Jan 19 at 5:00
$begingroup$
@TorstenSchoeneberg Oh, I desperately need a cover to hide in right now :(
$endgroup$
– cjackal
Jan 19 at 5:02
$begingroup$
Thank you! But I'm pretty sure that there is no other demand. I find this problem strange, and some of my classmates also got stuck on the $mathbb{Q}$-linearity. Your view of the uncountable rank helps a lot. Perhaps I could construct an example via linear isomorphism to $mathbb{R}$...?
$endgroup$
– NEne
Jan 19 at 5:10
$begingroup$
It is not clear what kind of appropriateness the question requires, but as $mathbb{R}$ as a $mathbb{Q}$-vector space has uncountable rank, $mathbb{R}^3$ is also uncountable-dimensional over $mathbb{Q}$, and I'm pretty sure (has no example though) there exists a division algebra which has uncountable rank over $mathbb{Q}$.
$endgroup$
– cjackal
Jan 19 at 4:43
$begingroup$
It is not clear what kind of appropriateness the question requires, but as $mathbb{R}$ as a $mathbb{Q}$-vector space has uncountable rank, $mathbb{R}^3$ is also uncountable-dimensional over $mathbb{Q}$, and I'm pretty sure (has no example though) there exists a division algebra which has uncountable rank over $mathbb{Q}$.
$endgroup$
– cjackal
Jan 19 at 4:43
$begingroup$
A possible property that you might neglect (if there was no problem in the final): If you demand the continuity of multiplication operation(with respect to the natural topology of $mathbb{R}$ and $mathbb{Q}$) then $mathbb{Q}$-linearity promotes to $mathbb{R}$-linearity for free, so you can use Frobenius' result.
$endgroup$
– cjackal
Jan 19 at 4:50
$begingroup$
A possible property that you might neglect (if there was no problem in the final): If you demand the continuity of multiplication operation(with respect to the natural topology of $mathbb{R}$ and $mathbb{Q}$) then $mathbb{Q}$-linearity promotes to $mathbb{R}$-linearity for free, so you can use Frobenius' result.
$endgroup$
– cjackal
Jan 19 at 4:50
$begingroup$
@cjackal: Obviously $Bbb R, Bbb C$ and $Bbb H$ have the same dimension, as $Bbb Q$-vector spaces, as $Bbb R^3$, so ...
$endgroup$
– Torsten Schoeneberg
Jan 19 at 5:00
$begingroup$
@cjackal: Obviously $Bbb R, Bbb C$ and $Bbb H$ have the same dimension, as $Bbb Q$-vector spaces, as $Bbb R^3$, so ...
$endgroup$
– Torsten Schoeneberg
Jan 19 at 5:00
$begingroup$
@TorstenSchoeneberg Oh, I desperately need a cover to hide in right now :(
$endgroup$
– cjackal
Jan 19 at 5:02
$begingroup$
@TorstenSchoeneberg Oh, I desperately need a cover to hide in right now :(
$endgroup$
– cjackal
Jan 19 at 5:02
$begingroup$
Thank you! But I'm pretty sure that there is no other demand. I find this problem strange, and some of my classmates also got stuck on the $mathbb{Q}$-linearity. Your view of the uncountable rank helps a lot. Perhaps I could construct an example via linear isomorphism to $mathbb{R}$...?
$endgroup$
– NEne
Jan 19 at 5:10
$begingroup$
Thank you! But I'm pretty sure that there is no other demand. I find this problem strange, and some of my classmates also got stuck on the $mathbb{Q}$-linearity. Your view of the uncountable rank helps a lot. Perhaps I could construct an example via linear isomorphism to $mathbb{R}$...?
$endgroup$
– NEne
Jan 19 at 5:10
add a comment |
1 Answer
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$begingroup$
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).
answered Jan 19 at 4:58
Noah SchweberNoah Schweber
125k10150287
$endgroup$
1
$begingroup$
Thanks! I'm not quite familiar with infinite dimensional vector spaces, and your answer helps a lot. Now I have enough courage to argue with my professor XD.
$endgroup$
– NEne
Jan 19 at 5:28
add a comment |
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$begingroup$
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).
answered Jan 19 at 4:58
Noah SchweberNoah Schweber
125k10150287
$endgroup$
1
$begingroup$
Thanks! I'm not quite familiar with infinite dimensional vector spaces, and your answer helps a lot. Now I have enough courage to argue with my professor XD.
$endgroup$
– NEne
Jan 19 at 5:28
add a comment |
$begingroup$
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).
answered Jan 19 at 4:58
Noah SchweberNoah Schweber
125k10150287
$endgroup$
1
$begingroup$
Thanks! I'm not quite familiar with infinite dimensional vector spaces, and your answer helps a lot. Now I have enough courage to argue with my professor XD.
$endgroup$
– NEne
Jan 19 at 5:28
add a comment |
$begingroup$
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).
answered Jan 19 at 4:58
Noah SchweberNoah Schweber
125k10150287
$endgroup$
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).
answered Jan 19 at 4:58
Noah SchweberNoah Schweber
125k10150287
answered Jan 19 at 4:58
Noah SchweberNoah Schweber
125k10150287
answered Jan 19 at 4:58
Noah SchweberNoah Schweber
125k10150287
answered Jan 19 at 4:58
Noah SchweberNoah Schweber
125k10150287
125k10150287
1
$begingroup$
Thanks! I'm not quite familiar with infinite dimensional vector spaces, and your answer helps a lot. Now I have enough courage to argue with my professor XD.
$endgroup$
– NEne
Jan 19 at 5:28
add a comment |
1
$begingroup$
Thanks! I'm not quite familiar with infinite dimensional vector spaces, and your answer helps a lot. Now I have enough courage to argue with my professor XD.
$endgroup$
– NEne
Jan 19 at 5:28
1
1
$begingroup$
Thanks! I'm not quite familiar with infinite dimensional vector spaces, and your answer helps a lot. Now I have enough courage to argue with my professor XD.
$endgroup$
– NEne
Jan 19 at 5:28
$begingroup$
Thanks! I'm not quite familiar with infinite dimensional vector spaces, and your answer helps a lot. Now I have enough courage to argue with my professor XD.
$endgroup$
– NEne
Jan 19 at 5:28
add a comment |
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$begingroup$
It is not clear what kind of appropriateness the question requires, but as $mathbb{R}$ as a $mathbb{Q}$-vector space has uncountable rank, $mathbb{R}^3$ is also uncountable-dimensional over $mathbb{Q}$, and I'm pretty sure (has no example though) there exists a division algebra which has uncountable rank over $mathbb{Q}$.
$endgroup$
– cjackal
Jan 19 at 4:43
$begingroup$
A possible property that you might neglect (if there was no problem in the final): If you demand the continuity of multiplication operation(with respect to the natural topology of $mathbb{R}$ and $mathbb{Q}$) then $mathbb{Q}$-linearity promotes to $mathbb{R}$-linearity for free, so you can use Frobenius' result.
$endgroup$
– cjackal
Jan 19 at 4:50
$begingroup$
@cjackal: Obviously $Bbb R, Bbb C$ and $Bbb H$ have the same dimension, as $Bbb Q$-vector spaces, as $Bbb R^3$, so ...
$endgroup$
– Torsten Schoeneberg
Jan 19 at 5:00
$begingroup$
@TorstenSchoeneberg Oh, I desperately need a cover to hide in right now :(
$endgroup$
– cjackal
Jan 19 at 5:02
$begingroup$
Thank you! But I'm pretty sure that there is no other demand. I find this problem strange, and some of my classmates also got stuck on the $mathbb{Q}$-linearity. Your view of the uncountable rank helps a lot. Perhaps I could construct an example via linear isomorphism to $mathbb{R}$...?
$endgroup$
– NEne
Jan 19 at 5:10