$\mathbb{R}^5$ over $\mathbb{R}$
- 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 \(\mathbb{R}\)?
Justify your answer.
NOTE: I don't have a solution to this.
Justify your answer.
NOTE: I don't have a solution to this.
Grigorios Kostakos
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- Community Team
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- Joined: Tue Nov 10, 2015 8:25 pm
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!
\[ \displaystyle 5 = \left[ \mathbb{R}^5 : \mathbb{R} \right ] = \left[ \mathbb{R}^5 : \mathbb{R}(\alpha) \right ] \left[ \mathbb{R}(\alpha) : \mathbb{R} \right ] \]
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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- Posts: 16
- Joined: Fri Aug 12, 2016 4:33 pm
Re: $\mathbb{R}^5$ over $\mathbb{R}$
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.
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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