Double centralizer theorem

Remarks

The proof views $A$ as a module over the simple algebra $B \otimes_K A^{\mathrm{op}}$. Then $C_A(B)$ is the endomorphism ring of this module, a matrix algebra over a division algebra. Counting dimensions gives clause 2. Since $B \subseteq C_A(C_A(B))$ and clause 2 applied twice gives equal dimensions, clause 3 follows. Example: let $A = M_n(K)$ and $B = K[x]/(p)$, embedded through the companion matrix $C$ of an irreducible polynomial $p$ of degree $n$. Then $C_A(B)$ consists of the polynomials in $C$, so it equals $B$, and indeed $n \cdot n = n^2$. Example: for $A = M_2(K) \otimes_K M_3(K)$ and $B = M_2(K) \otimes 1$, the centralizer is $1 \otimes M_3(K)$, and $4 \cdot 9 = 36$. Simplicity of $B$ is needed for clause 2: in $A = M_3(K)$ let $B$ be spanned by $I$ and the matrix unit $E_{13}$, a subalgebra with the nilpotent ideal $K E_{13}$. A matrix commutes with $E_{13}$ exactly when its entries in positions $(2,1)$, $(3,1)$, $(3,2)$ vanish and its entries in positions $(1,1)$ and $(3,3)$ agree, so $\dim_K C_A(B) = 5$ and $2 \cdot 5 \ne 9$.

Going deeper

For a student

The main use is to subfields: if $L$ is a subfield of $A$ containing $K$, then $[L : K] \cdot \dim_K C_A(L) = \dim_K A$, so $[L : K]$ divides the degree of $A$. When $C_A(L) = L$, this gives $[L : K]^2 = \dim_K A$, so $L$ is a maximal subfield and splits $A$. Clause 4 shows that a central simple subalgebra always splits off as a tensor factor. With the Skolem–Noether theorem it underlies the crossed product description of the Brauer group.