Welcome to our community

Be a part of something great, join today!

Could R^n be a field ...?

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,910
\(\displaystyle \mathbb{R}^n\) is a vector space but not a field because it lacks a suitable multiplication operation between pairs of its elements ...

Why don't mathematicians define a multiplication operation between a pair of elements and investigate the resulting field ...

For example ... why not define multiplication as X where


\(\displaystyle (x_1, x_2, \ ... \ ... \ , x_n) \ X \ (y_1, y_2, \ ... \ ... \ , y_n) = (x_1 y_1, x_2 y_2 , \ ... \ ... \ , x_n y_n)\) ...


Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,693
\(\displaystyle \mathbb{R}^n\) is a vector space but not a field because it lacks a suitable multiplication operation between pairs of its elements ...

Why don't mathematicians define a multiplication operation between a pair of elements and investigate the resulting field ...

For example ... why not define multiplication as X where


\(\displaystyle (x_1, x_2, \ ... \ ... \ , x_n) \ X \ (y_1, y_2, \ ... \ ... \ , y_n) = (x_1 y_1, x_2 y_2 , \ ... \ ... \ , x_n y_n)\) ...


Peter
You can define a multiplication operation, but if it is to give rise to a field then it has to obey the field axioms.

For example, you need to check that there is a zero element and an identity element, and each nonzero element of the space needs to have a multiplicative inverse. In the case of the operation that you suggest, the zero element is $(0,0,\ldots,0)$ and the identity element is $(1,1,\ldots,1)$. The element $(1,0,\ldots,0)$ is nonzero but does not have an inverse. So your proposed multiplication does not make \(\displaystyle \mathbb{R}^n\) into a field.

In fact, when $n>2$ there is no way to make \(\displaystyle \mathbb{R}^n\) into a field. (The nearest you can get is the quaternion multiplication in \(\displaystyle \mathbb{R}^4\), which has many nice properties but is not commutative. So it does not make \(\displaystyle \mathbb{R}^4\) into a field.)
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,910
You can define a multiplication operation, but if it is to give rise to a field then it has to obey the field axioms.

For example, you need to check that there is a zero element and an identity element, and each nonzero element of the space needs to have a multiplicative inverse. In the case of the operation that you suggest, the zero element is $(0,0,\ldots,0)$ and the identity element is $(1,1,\ldots,1)$. The element $(1,0,\ldots,0)$ is nonzero but does not have an inverse. So your proposed multiplication does not make \(\displaystyle \mathbb{R}^n\) into a field.

In fact, when $n>2$ there is no way to make \(\displaystyle \mathbb{R}^n\) into a field. (The nearest you can get is the quaternion multiplication in \(\displaystyle \mathbb{R}^4\), which has many nice properties but is not commutative. So it does not make \(\displaystyle \mathbb{R}^4\) into a field.)


Thanks for that reply, Opalg ... really informative ...

Now ... you write the following:

" ... ... In fact, when $n>2$ there is no way to make \(\displaystyle \mathbb{R}^n\) into a field. ... ... "

Intriguing ...

Now how would you go about formally and rigorously proving that ... can you indicate the broad nature of the proof ...

Thanks again ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,693
Now ... you write the following:

" ... ... In fact, when $n>2$ there is no way to make \(\displaystyle \mathbb{R}^n\) into a field. ... ... "

Intriguing ...

Now how would you go about formally and rigorously proving that ... can you indicate the broad nature of the proof ...
This is not an area that I am really familiar with. I believe that the result goes back to Frobenius in the 19th century.

When $n=2$ it is of course possible to make the space $\Bbb{R}^2$ into a field, with the complex number multiplication that is given by identifying $\Bbb{R}^2$ with $\Bbb{C}$. But $\Bbb{C}$ is an algebraically closed field, which implies that it has no proper algebraic extensions. That somehow implies that it is the end of the road: it cannot be embedded in any larger field. In particular, that stops $\Bbb{R}^n$ from having a field structure when $n>2$.