1
$\begingroup$

Consider a Desarguesian affine space $\mathcal{A}$. Choose a fixed point $o$ and a fixed line $\mathbb{K}$ through $o$. Select another, arbitrary point $e$ on $\mathbb{K}$. For each point $x \in \mathbb{K}$, we write $\vec{x}$ instead of $\vec{ox}$.

Define the sum in $\mathbb{K}$ as follows: $\forall a,b\in \mathbb{K}:$ $a+b=c$ if $\vec{a}+\vec{b}=\vec{c}$, which is equivalent with $\tau_{\vec{a}} \tau_{\vec{b}} = \tau_{\vec{c}} $ ($\tau$ is a translation along the specified vector, also note the exponential notation!).

Define the product in $\mathbb{K}$ as follows: $o \cdot a = o = a \cdot o$, for all $a\in \mathbb{K}$ and $a \cdot b = c,$ $o \notin \{a,b\},$ if $h_{e,b}^{-1} \tau_{\vec{a}} h_{e,b} = \tau_{\vec{c}}$ ($h_{e,b}$ is the homothety with center $o$ that maps $e$ to $b$). We can also say that $a \cdot b = c \iff h_{e,a}h_{e,b} = h_{e,c}$.

Now the statement is

With the notation used above, $\mathbb{K},+,\cdot$ is a skew field.

Proof:

$\mathbb{K},+$ is a commutative group with identity $o$ (because the group of translation in $\mathcal{A}$ is commutative and the definition of the sum of elements in $\mathbb{K}$ is totally based on translations along vectors).

$\mathbb{K^*},\cdot$ is a group (follows from the last sentence of the given description, which states that multiplication of points on $\mathbb{K}$ is equivalent with a composition of homotheties - which also form a group, when considering homotheties with the same center).

We want to proof that $(a+a')\cdot b = a\cdot b + a'\cdot b$. The left term is determined by the translation $$ h_{e,b}^{-1}\tau_{\vec{a}+\vec{a}'}h_{e,b}=h_{e,b}^{-1}\tau_{\vec{a}} \tau_{\vec{a}'}h_{e,b} = h_{e,b}^{-1}\tau_{\vec{a}} h_{e,b}h_{e,b}^{-1}\tau_{\vec{a}'}h_{e,b} = \tau_{\vec{a\cdot b}}\tau_{\vec{a'\cdot b}}.$$ Which proofs what we needed.

How do I proof that $a\cdot (b+b') = a\cdot b + a \cdot b'$? I'll have to introduce a new definition for the summation of homotheties with center $o$. Then I must prove that this definition corresponds to summation in $\mathbb{K}$. How do I find a new definition?

Or maybe somebody can give another way to prove the last part.

Thanks.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.