The Elusive Field Structure of $\mathbb{R}^n$ Beyond $n=2$

[Math Amateur]

\(\displaystyle \mathbb{R}^n\) 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

 

[Opalg]

\(\displaystyle \mathbb{R}^n\) 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.)

 

[Math Amateur]

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]

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 $n>2$.