Section 2.2 Examples
\begin{equation*}
i_{\Z}\colon\Z\to R\quad\text{given by}\quad n\mapsto n\cdot 1_R=\underbrace{1_R+\cdots+1_R}_{n\text{-times}}
\end{equation*}
Example 2.2.2. (Ring homomorphism onto the trivial ring).
\begin{equation*}
R\to\{0\}\quad\text{given by}\quad r\mapsto 0\,(=1).
\end{equation*}
Example 2.2.3. (Epimorphism from \(\Z\) onto \(\Z/n\Z\)).
\begin{equation*}
i\mapsto [i].
\end{equation*}
Verify that \(\varphi\) is a ring epimorphism.Example 2.2.4. (Embedding of \(F\) into an \(F\)-algebra).
\begin{equation*}
\varphi(\alpha)=\varphi(\alpha 1_F)=\alpha\cdot\varphi(1_F)=\alpha\cdot 1_A\text{.}
\end{equation*}
The kernel of \(\varphi\text{,}\) \(\ker(\varphi)\subsetneq F\) is a two-sided ideal of \(F\text{.}\) Since every nonzero element of \(F\) is invertible and \(\varphi\) is a nonzero map, \(\ker(\varphi)=\{0\}\text{.}\) Thus \(\varphi\colon F\to A\) given by \(\alpha\mapsto\alpha\cdot 1_A\) is a monomorphism.Example 2.2.5. (Inclusion as a monomorphism of rings).
\begin{equation*}
i\colon S\to R\quad\text{given by}\quad s\mapsto s
\end{equation*}
is a monomorphism of rings.Example 2.2.6. (Homomorphisms from simple rings).
In particular a ring homomorphism from a division ring or a field to a nonzero ring with unity is a ring monomorphism.
Example 2.2.7. (Conjugation).
\begin{equation*}
\sigma\colon\C\to\C\quad\text{given by}\quad a+ib\mapsto a-ib
\end{equation*}
is an \(\R\)-algebra automorphism.Example 2.2.8. (Isomorphism between matrices and its opposite algebra).
\begin{equation*}
A\mapsto \left(A^t\right)^{\rm op}.
\end{equation*}
Note that
\begin{equation*}
\varphi(AB)=\left((AB)^t\right)^{\rm op}=\left(B^tA^t\right)^{\rm op}=\left(A^t\right)^{\rm op}\left(B^t\right)^{\rm op}=\varphi(A)\varphi(B).
\end{equation*}
This is an \(F\)-algebra homomorphism. The kernel of \(\varphi\text{,}\) \(\ker(\varphi)\) is a two-sided ideal of \(M_n(F)\text{.}\) By Example 1.6.9, \(\ker(\varphi)\) is either \(M_n(F)\) or \(\{0\}\text{.}\) Since \(\varphi\) is a nonzero map we must have \(\ker(\varphi)=\{0\}\text{,}\) i.e., \(\varphi\) is an \(F\)-algebra monomorphism. In particular \(\varphi\) is also an injective \(F\)-linear map. As \(\dim_F\left(M_n(F)^{\rm op}\right)=\dim_F\left(M_n(F)\right)\) we get that \(\varphi\) is surjective. Therefore, \(\varphi\) is an isomorphism of \(F\)-algebras.Example 2.2.9. (Matrix ring over a matrix ring).
\begin{equation*}
\begin{pmatrix}S_{11}\amp S_{12}\amp \cdots\amp S_{1n}\\S_{21}\amp S_{22}\amp \cdots\amp S_{2n}\\\vdots\amp \vdots\amp \ddots\amp \vdots\\S_{n1}\amp S_{n2}\amp \cdots\amp S_{nn}\end{pmatrix}\quad\text{where}\quad S_{ij}\in S=M_m(R)
\end{equation*}
Hence the above matrix has \(nm\) entries from \(R\) in each row as well as in each column. Assume that \(S_{ij}=\left(s_{ij}(pq)\right)_{1\leq p,q\leq m}\in M_m(R)\) and \(T_{ij}=\left(t_{ij}(pq)\right)_{1\leq p,q\leq m}\in M_m(R)\text{.}\) So the above matrix becomes:
\begin{equation}
\begin{pmatrix}
s_{11}(11)\amp \cdots\amp s_{11}(1m)\amp s_{12}(11)\amp \cdots\amp s_{12}(1m)\amp \cdots\amp s_{1n}(11)\amp \cdots\amp s_{1n}(1m)\\s_{11}(21)\amp \cdots\amp s_{11}(2m)\amp s_{12}(21)\amp \cdots\amp s_{12}(2m)\amp \cdots\amp s_{2n}(21)\amp \cdots\amp s_{2n}(1m)\\
\vdots\amp \ddots\amp \vdots\amp \vdots\amp \ddots\amp \vdots\amp \cdots\amp \vdots\amp \ddots\amp \vdots\\
s_{11}(m1)\amp \cdots\amp s_{11}(mm)\amp s_{12}(m1)\amp \cdots\amp s_{12}(mm)\amp \cdots\amp s_{2n}(m1)\amp \cdots\amp s_{2n}(mm)\\
\vdots\amp \ddots\amp \vdots\amp \vdots\amp \ddots\amp \vdots\amp \cdots\amp \vdots\amp \ddots\amp \vdots\\
\vdots\amp \ddots\amp \vdots\amp \vdots\amp \ddots\amp \vdots\amp \cdots\amp \vdots\amp \ddots\amp \vdots\\
s_{n1}(11)\amp \cdots\amp s_{n1}(1m)\amp s_{n2}(11)\amp \cdots\amp s_{n2}(1m)\amp \cdots\amp s_{nn}(11)\amp \cdots\amp s_{nn}(1m)\\s_{n1}(21)\amp \cdots\amp s_{n1}(2m)\amp s_{n2}(21)\amp \cdots\amp s_{n2}(2m)\amp \cdots\amp s_{nn}(21)\amp \cdots\amp s_{nn}(1m)\\
\vdots\amp \ddots\amp \vdots\amp \vdots\amp \ddots\amp \vdots\amp \cdots\amp \vdots\amp \ddots\amp \vdots\\
s_{n1}(m1)\amp \cdots\amp s_{n1}(mm)\amp s_{n2}(m1)\amp \cdots\amp s_{n2}(mm)\amp \cdots\amp s_{nn}(m1)\amp \cdots\amp s_{nn}(mm)
\end{pmatrix}\tag{2.2.1}
\end{equation}
The matrix multiplication of \((S_{ij}),(T_{ij})\in M_n(S)\) is the following matrix in \(M_n(S)\text{.}\)
\begin{equation}
(S_{ij})(T_{ij})=\left(\sum_{k}S_{ik}T_{kj}\right)=\left(\sum_{k=1}^n\left(\sum_{\ell=1}^ms_{ik}(p\ell)\cdot t_{kj}(\ell q)\right)_{1\leq p,q\leq m}\right)\tag{2.2.2}
\end{equation}
Define a map \(\varphi\colon M_n\left(M_m(R)\right)\to M_{nm}(R)\) given by \((S_{ij})\mapsto\text{ matrix given in eq.}\) (2.2.1). This is additive group homomorphism and eq. (2.2.2) shows that \(\varphi\left((S_{ij})(T_{ij})\right)=\varphi\left(S_{ij}\right)\varphi\left(T_{ij}\right)\text{,}\) and \(\varphi\) maps unity to unity. Therefore, \(\varphi\) is a nonzero ring homomorphism. Verify that \(\varphi\) is a ring isomorphism.Example 2.2.10. (Inner automorphism).
\begin{equation*}
\Int(a)\colon A\to A\quad\text{given by}\quad x\mapsto axa^{-1}.
\end{equation*}
Example 2.2.11. (Evaluation map).
\begin{equation*}
\ev_a\colon F[X]\to F\quad\text{given by}\quad \sum_i\alpha_iX^i\mapsto \sum_i\alpha_i\,a^i
\end{equation*}
More generally, consider the polynomial ring \(F[X_1,\ldots,X_n]\) over \(F\) in \(n\) variables. Fix \(a_1,a_2,\ldots,a_n\in F\text{.}\) The evaluation map at \(a_1,\ldots,a_n\) given below is an \(F\)-algebra homomorphism.
\begin{equation*}
\ev_{(a_1,\ldots,a_n)}\colon F[X_1,\ldots,X_n]\to F\quad\text{given by}\quad \sum \alpha_{i_1\cdots i_n}X^{i_1}X^{i_2}\cdots X^{i_n}\mapsto \sum \alpha_{i_1\cdots i_n}a_1^{i_1}a_2^{i_2}\cdots a_n^{i_n}
\end{equation*}
Example 2.2.12. (Homomorphism from algebra to the endomorphism ring).
\begin{equation*}
L\colon A\to\End_F(A)\quad\text{given by}\quad a\mapsto L_a
\end{equation*}
\begin{equation*}
R\colon A^{\rm op}\to\End_F(A)\quad\text{given by}\quad a^{\rm op}\mapsto R_a
\end{equation*}
Indeed, - Maps \(L\) and \(R\) are \(F\)-linear. For \(\alpha,\beta\in F\) and \(a,b\in A\) we have\begin{align*} L_{\alpha\cdot a+\beta\cdot b}(x)\amp=(\alpha\cdot a+\beta\cdot b)x\\ \amp=\alpha\cdot ax+\beta\cdot bx\\ \amp=\alpha L_a(x)+\beta L_b(x) \end{align*}Similarly, using (1.4.1) and\begin{equation*} \alpha\cdot a^{\rm op}+\beta\cdot b^{\rm op}=(\alpha\cdot a)^{\rm op}+(\beta\cdot b)^{\rm op}=(\alpha\cdot a+\beta\cdot b)^{\rm op} \end{equation*}we get the following.\begin{align*} R\left({\alpha\cdot a^{\rm op}+\beta\cdot b^{\rm op}}\right)(x)\amp=x(\alpha\cdot a+\beta\cdot b)\\ \amp=x(\alpha\cdot a)+x(\beta\cdot b)\\ \amp=\alpha\cdot (xa)+\beta\cdot (xb)\\ \amp=\alpha R_{a}(x)+\beta R_{b}(x) \end{align*}
- Maps \(L\) and \(R\) are ring homomorphisms. Indeed,\begin{align*} L(ab)(x)\amp=L_{ab}(x)\\ \amp=abx\\ \amp=L_a\circ L_b(x) \end{align*}Hence, \(L(ab)=L(a)L(b)\text{.}\) We also have the following.\begin{align*} R(a^{\rm op}b^{\rm op})(x)\amp=R\left({(ba)^{\rm op}}\right)(x)\\ \amp=xba\\ \amp=R_a\circ R_b(x) \end{align*}Hence, \(R(a^{\rm op}b^{\rm op})=R(a^{\rm op})R(b^{\rm op})\text{.}\) We also have \(L(1)=\unit_A\) and \(R(1)=\unit_A\text{.}\)
Example 2.2.13. (Projection maps).
\begin{equation*}
\pi_k\colon\prod_iR_i\to\prod_iR_i\quad\text{given by}\quad (x_i)_{i\in I}\mapsto (\delta_{ik}\,x_i)_{i\in I},
\end{equation*}
where for each \(i\in I\text{,}\)
\begin{equation*}
\delta_{ik}=\begin{cases}0_i\amp\text{if }i\neq k\\1_i\amp\text{if }i=k\end{cases}
\end{equation*}
Verify that \(\pi_k\) is a ring homomorphism, and that
\begin{equation*}
\pi_k^2=\pi_k\quad\text{and}\quad\pi_k\circ\pi_\ell=\pi_\ell\circ\pi_k=0\quad\text{for}\;k\neq\ell.
\end{equation*}
Thus, \(\pi_k\) are orthogonal idempotents of the ring \(\End_\Rings(\prod_iR_i)\text{.}\)
