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## $\mathbb{R}^5$ over $\mathbb{R}$

Linear Algebra, Algebraic structures (Groups, Rings, Modules, etc), Galois theory, Homological Algebra
Grigorios Kostakos
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### $\mathbb{R}^5$ over $\mathbb{R}$

It is possible to define a multiplication in $\mathbb{R}^5$ such that, with the usual addition and multiplication between elements of $\mathbb{R}$, the $\mathbb{R}^5$ becomes a field which contains the $\mathbb{R}$?

NOTE: I don't have a solution to this.
Grigorios Kostakos
Tsakanikas Nickos
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### Re: $\mathbb{R}^5$ over $\mathbb{R}$

Suppose that a multiplication can be defined in $\displaystyle \mathbb{R}^5$ so that it becomes a field extension of $\displaystyle \mathbb{R}.$ Then $\displaystyle \mathbb{R}^5$ can be viewed as a vector space over $\displaystyle \mathbb{R}$ and obviously $\displaystyle \dim_{\mathbb{R}}\left( \mathbb{R}^5 \right) = 5.$ Therefore the degree $\displaystyle \left[ \mathbb{R}^5 : \mathbb{R} \right ]$ of the field extension $\displaystyle \mathbb{R}^5 / \mathbb{R}$ equals 5. Let $\displaystyle \alpha \in \mathbb{R}^5 \smallsetminus \mathbb{R}.$ Then we have that
$\displaystyle 5 = \left[ \mathbb{R}^5 : \mathbb{R} \right ] = \left[ \mathbb{R}^5 : \mathbb{R}(\alpha) \right ] \left[ \mathbb{R}(\alpha) : \mathbb{R} \right ]$
which means that either
$\displaystyle \left[ \mathbb{R}^5 : \mathbb{R}(\alpha) \right ] = 5 \text{ and } \big[ \mathbb{R}(\alpha) : \mathbb{R} \big ] = 1 \; (*)$
or
$\displaystyle \left[ \mathbb{R}^5 : \mathbb{R}(\alpha) \right ] = 1 \text{ and } \big[ \mathbb{R}(\alpha) : \mathbb{R} \big ] = 5 \; (**)$

- If $\displaystyle (*)$ is true, then $\displaystyle \left[ \mathbb{R}(\alpha) : \mathbb{R} \right ] = 1$ means that $\displaystyle \mathbb{R}(\alpha) = \mathbb{R}$, which is a contradiction, since $\displaystyle \alpha \notin \mathbb{R}.$

- If $\displaystyle (**)$ is true, then $\displaystyle \left[ \mathbb{R}(\alpha) : \mathbb{R} \right ] = 5$ means that $\displaystyle irr\left( \alpha , \mathbb{R} \right)$ is of odd degree. However, this cannot be true, since it is known that every polynomial of odd degree over $\displaystyle \mathbb{R}$ has a real root and, thus, $\displaystyle \partial \left( irr\left( \alpha , \mathbb{R} \right) \right) \leq 4 .$

Hence the desired multiplication cannot be defined.

In fact, with a similar argument we can prove that there is no field extension of $\displaystyle \mathbb{R}$ of odd degree stricktly greater that 1, which means that $\displaystyle \mathbb{R}$ is the only field extension of $\displaystyle \mathbb{R}$ of odd degree!