Vector space over the center of a division ring

Groups, Rings, Domains, Modules, etc, Galois theory
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Alkesk
Posts: 9
Joined: Sat Dec 12, 2015 11:19 am

Vector space over the center of a division ring

#1

Post by Alkesk »

Let \(\displaystyle{R}\) be an associative division ring. Prove that \(\displaystyle{Z\left ( R \right )}\) is a field, where \(\displaystyle{Z\left ( R \right )=\left \{ z\in R:rz=zr , \forall r\in R \right \}}\) and that \(\displaystyle{R}\) can't be a vector space over \(\displaystyle{Z\left ( R \right )}\) with dimension \(\displaystyle{2}\).
Tsakanikas Nickos
Community Team
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Vector space over the center of a division ring

#2

Post by Tsakanikas Nickos »

  • In case \( \displaystyle R \) is a division ring, then every non-zero element of \( \displaystyle R \), and particularly of \( \displaystyle Z(R) \), is invertible. Therefore, \( \displaystyle Z(R) \) is a field:

    Let \( \displaystyle x \in Z(R) \). Then \( \displaystyle \exists x^{-1} \in R \). Let \( \displaystyle r \in R \). Then \[ \displaystyle x^{-1}r = x^{-1}r1_{R} = x^{-1}rxx^{-1} = x^{-1}xrx^{-1} = 1_{R}rx^{-1} = rx^{-1} \] so \( x^{-1} \in Z(R) \).
  • Now, \( \displaystyle R \) can be considered as a vector space over \( \displaystyle Z(R) \) if we define
    \[ * : Z(R) \times R \rightarrow R , (z,r) \mapsto z*r = z \cdot r \] where \( \cdot \) is the multiplication of \( \displaystyle R \). However, \[ \displaystyle dim_{Z(R)}\left(R\right) \neq 2 \; : \]
    (This solution has been proposed to me by a classmate)

    If \( \displaystyle dim_{Z(R)}\left(R\right) = 2 \), then, since \( \displaystyle R \) is finite-dimensional vector space over \( \displaystyle Z(R) \) and \( \{ 1_{R} \} \) is linearly independent, \( \{ 1_{R} \} \) can be extended to a \( \displaystyle Z(R) \)-basis \( \{ 1_{R} , e \} \) of \( \displaystyle R \). Let \( \displaystyle r \in R \). Then \[ \displaystyle r = x \cdot 1_{R} + y \cdot e = x + ye \] for some \( \displaystyle x,y \in Z(R) \). We have that \[ \displaystyle er = e( x +ye ) = ex + eye = xe +yee = (x +ye)e = re \]which means that \( \displaystyle e \in Z(R) \). Since \( \displaystyle 1_{R} \in Z(R) \) as well, it follows that \( \displaystyle R = Z(R) \), which is a contradiction.
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