In this section we assume that \(F\) is a field. We give several examples of \(F\)-algebras. In the following examples checking that the underlying abelian group structures for a ring and a vector space are the same will be easy, and it is left to the reader. In some instances we only verify (1.4.1).
Suppose that \(E\) is a field containing \(F\text{.}\) Then \(F\) is a subspace as well as a subring of \(E\text{.}\) We also get that \(E\) is an \(F\) algebra because \(E\) is commutative.
Consider \(n\times n\) matrices over \(F\text{,}\)\(M_n(F)\text{.}\) Recall that \(M_n(F)\) is a vector space over \(F\) via the scalar multiplication defined by
Verify that \(\alpha\cdot\left((x_{ij})(y_{ij})\right)=\left(\alpha\cdot(x_{ij})\right)(y_{ij})=(x_{ij})\left(\alpha\cdot (y_{ij})\right)\text{.}\) Thus \(M_n(F)\) is an \(F\)-algebra.
Assume that \(A\) is an \(F\)-algebra. Consider the opposite ring \(A^{\rm op}\) (see Example 1.2.9). We claim that \(A^{\rm op}\) is an \(F\)-algebra. We make \(A^{\rm op}\) a vector space over \(F\) via the following scalar multiplication. For any \(\alpha\in F\) and \(x^{\rm op}\in A^{\rm op}\) we define
\begin{equation*}
\alpha\cdot x^{\rm op}=({\alpha\cdot x})^{\rm op},\quad\text{where }\alpha\cdot x\;\text{is scalar multiplication defined on }A
\end{equation*}
Suppose that \(\alpha\in F\) and \(x^{\rm op},y^{\rm op}\in A^{\rm op}\text{.}\) Then we have
\begin{align*}
\alpha\cdot(x^{\rm op}y^{\rm op})\amp=\alpha\cdot(yx)\amp\\
\amp=(\alpha\cdot y)x\amp\text{Since }A\text{ is an }F\text{-algebra}\\
\amp=x^{\rm op}(\alpha\cdot y)^{\rm op}\amp\\
\amp=x^{\rm op}(\alpha\cdot y^{\rm op})\amp
\end{align*}
Similarly we get \(\alpha\cdot(x^{\rm op}y^{\rm op})=(\alpha\cdot x^{\rm op})y^{\rm op}\text{.}\) Hence \(A^{\rm op}\) is an \(F\)-algebra.
Let \(V\) be a vector space over \(F\text{.}\) By Example 1.2.11, \(\End_F(V)\) is a ring and we have seen that it is also a vector space. It is easy to verify that \(\End_F(V)\) is an \(F\)-algebra.
is an \(F\)-algebra. This is a subring of \(A\) (refer to Example 1.6.1). Verify that \(C_A(S)\) is a vector subspace of \(A\text{.}\) Since \(A\) is an \(F\)-algebra the identity (1.4.1) is satisfied by \(C_A(S)\text{.}\) Hence, centralizer is an \(F\)-subalgebra.
