Why is the Endomorphism ring of a finite dimensional central simple algebra isomorphic to the matrix ring over the centre/underlying field having dimension $\displaystyle n^2$, where $\displaystyle n$ is the dimension of the algebra over the field?
Why is the Endomorphism ring of a finite dimensional central simple algebra isomorphic to the matrix ring over the centre/underlying field having dimension $\displaystyle n^2$, where $\displaystyle n$ is the dimension of the algebra over the field?
This is my attempt:
Think of a central simple algebra V is a vector space over K with additional conditions.
More generally,
If $\displaystyle V \cong m_1V^{(1)} \oplus m_2V^{(2)} \cdots \oplus m_kV^{(k)}$, then $\displaystyle \text{End V} \cong \bigoplus_{i=1}^k \text{Mat}_{m_i}$.
Since V is simple, $\displaystyle \text{End V} \cong \text{Mat}_{m_1}$, where $\displaystyle m_1=n$ (since V is a central simple algebra of a dimension n) and $\displaystyle \text{dim (End V)} = n^2$.
Thank you both. My second question is:
As you know, a finite dimensional simple algebra is isomorphic to a matrix ring of a division ring.
If the algebra is central, why is the division ring then central?
Is it because the center of the division ring has to be contained in the centre of the matrix ring? Thus it must be the underlying field. Doesn't hurt to double check!!!
this is just a simple linear algebra fact and it's true for any finite dimensional algebra. in general, for any finite dimensional K-vector space $\displaystyle A$ with $\displaystyle \dim_K A=n,$ we have $\displaystyle End_K (A) \cong M_n(K)$
and so $\displaystyle \dim_K End_K(A)=n^2.$ you might have meant this fact about a K-central simple algebra $\displaystyle A,$ which is far from being trivial: $\displaystyle A \otimes_K A^{op} \cong End_K(A).$