Past exam of the algebra course of the University of Oxford / 2011 / 1

The center of an associative algebra is the subalgebraFor a pair $(z_1,z_2)$, commuting with every $(a_1,a_2)$ means $z_1{a}_1=a_1{z}_1$ and $z_2{a}_2=a_2{z}_2$. Varying the two coordinates independently proves both inclusions, hence

Use the usual unital convention for an algebra over a field. For $a,b,x\in A$ and $c\in K$,Passing to matrices in one fixed basis preserves these identities: $M_{a+cb}=M_a+c{M}_b$, $M_{ab}=M_a{M}_b$ and $M_1=I$. Thus $\psi$ is an algebra homomorphism. If $M_a=0$, evaluating ${\varphi }_a$ at $1_A$ gives $a=0$. Its kernel is zero, so the left regular representation is faithful.

No. Take $A=M_2(K)$ and its diagonal subalgebra $B$. The algebra over a field $B$ is commutative, so $Z(B)=B$. But $e_{11}\in B$ does not commute with $e_{12}$:Hence $e_{11}\notin Z(A)$, although $e_{11}\in Z(B)$. Commutation inside a subalgebra places no condition on elements outside it.

The Artin–Wedderburn theorem says that a finite-dimensional semisimple algebra over $\mathbb{C}$ is a finite product ${\prod }_j{M}_{d_j}(\mathbb{C})$: the division algebras in the general theorem reduce to $\mathbb{C}$ because the base field is algebraically closed. Each factor has center of an associative algebra $\mathbb{C}{I}_{d_j}$. Consequently a central element corresponds to $({\lambda }_j{I}_{d_j}{)}_j$. If it is nilpotent, then ${\lambda }_j^n=0$ for every $j$, whence every ${\lambda }_j=0$. Therefore $\boxed{z=0}$. The centrality assumption matters: a full matrix algebra has nonzero nilpotent elements.