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

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.