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

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

#1

Post by Grigorios Kostakos » Mon Jan 18, 2016 3:52 am

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 \(\mathbb{R}\)?
Justify your answer.

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

#2

Post by Tsakanikas Nickos » Mon Jan 18, 2016 3:53 am

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!
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Re: $\mathbb{R}^5$ over $\mathbb{R}$

#3

Post by S.F.Papadopoulos » Thu Sep 08, 2016 2:05 pm

We have
1)Every finite extension is algebraic
2)The algebraic closure of real numbers are
the complex numbers
3)Every finite extension of real numbers are
the complex numbers.
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