Rings and Ring Homomorphisms

Purpose
The purpose of this page is to give examples of rings.

Examples of Rings

Example 1: Let \mathbb{R}[x] be the set of polynomials with coefficients in \mathbb{R}. Let f(x)=a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1x + a_0 and g(x)=b_mx^m+ b_{m-1}x^{m-1} + \cdots + b_1x + b_0, assume that m<n, then we define addition and multiplication by:

    \[f(x)+g(x)=a_nx^n+ \cdots +a_{m+1}x^{m+1} +(a_m+b_m)x^m + \cdots + (a_1+b_1)x + (a_0+b_0)\]


    \[f(x)g(x)=a_nb_m x^{m+n} + (a_nb_{m-1}+a_{n-1}b_m)x^{m+n-1} + \cdots + a_0b_0\]


In other words, they are exactly what you think they should be. Clearly, by adding and multiplying polynomials, we obtain a polynomial. Thus, the subset is closed under this operation. The additive identity is the zero polynomial: 0(x)=0. For any polynomial f(x), it’s additive inverse is given by:

    \[-f(x)=-a_nx^n -a_{n-1}x^{n-1} - \cdots - a_1x -a_0\]


The multiplicative identity is the constant polynomial 1(x)=1.

Example 2: Let \mathbb{R}(x) be the set of rational functions with coefficients in \mathbb{R}, i.e. elements are of the form r(x)=\frac{p(x)}{q(x)}, where p(x) and q(x) are polynomials and q(x)\not=0. Let r_1(x)=\frac{p_1(x)}{q_1(x)} and r_2(x)=\frac{p_2(x)}{q_2(x)}, then we define addition and multiplication by:

    \[r_1(x)+r_2(x)=\frac{p_1(x)}{q_1(x)}+\frac{p_2(x)}{q_2(x)}=\frac{p_1(x)q_2(x)+p_2(x)q_1(x)}{q_1(x)q_2(x)}\]


    \[r_1(x)\cdot r_2(x)=\frac{p_1(x)}{q_1(x)}\cdot\frac{p_2(x)}{q_2(x)}=\frac{p_1(x)p_2(x)}{q_1(x)q_2(x)}\]


The additive identity is 0(x)=0 and multiplicative identity 1(x)=1 and for any r(x)=\frac{p(x)}{q(x)}, the additive inverse is -r(x)=-\frac{p(x)}{q(x)} and the multiplicative inverse (if r(x)\not=0) is r(x)^{-1}=\frac{q(x)}{p(x)}.

Example 3: Let R be the set of real-valued functions with domain D, i.e. f:D \rightarrow \mathbb{R}. For any two functions f,g \in R, we define addition and multiplication by:

    \[(f+g)(x)=f(x)+g(x) \text{ and } (fg)(x)=f(x)g(x)\]


The additive identity is the constant function 0(x)=0 and the multiplicative identity is the constant function 1(x)=1. The additive inverse of f(x) is (-f)(x):=-f(x).

Example 4: The set of real-valued continuous functions with domain D is a ring under the same operations in Example 3.

Example 5: The set of real-valued (infinitely) differentiable functions with domain D is a ring under the same operations in Example 3.

Examples of Ring Homomorphisms

Example 6: Let R be the set of real-valued function with domain D as in Example 3. Let c \in D, then the evaluation homomorphism \text{ev}_c: R \rightarrow \mathbb{R} at c is defined by \text{ev}_c(f)=f(c). This is in fact a ring homomorphism.

Proof: Let f,g \in R,

    \[\text{ev}_c(f+g)=(f+g)(c)=f(c)+g(c)=\text{ev}_c(f)+\text{ev}_c(g)\]


    \[\text{ev}_c(f\cdot g)=(f\cdot g)(c)=f(c)\cdot g(c) = \text{ev}_c(f)\cdot\text{ev}_c(g)\]


and \text{ev}_c(1)=1(c)=1.

Example 7: Let \iota : \mathbb{C} \rightarrow \mathbb{C} be defined by conjugation, i.e. for any z=a+bi \in \mathbb{C}, i(z)=\overline{z}=\overline{a+bi}=a-bi. Then, i is a ring homomorphism.

Proof: Let z_1=a_1+b_1i and z_2=a_2+b_2i, then,

    \begin{align*} \iota(z_1+z_2) &=\iota(a_1+b_1i+a_2+b_2i) \\ &=\iota((a_1+a_2)+(b_1+b_2)i) \\ &=a_1+a_2-(b_1+b_2)i \\ &=a_1-b_1i+a_2-b_2i \\ &=\iota(z_1)+\iota(z_2) \end{align*}


    \begin{align*} \iota(z_1z_2)&=\iota((a_1+b_1i)(a_2+b_2i)) \\ &=\iota(a_1a_2-b_1b_2+(a_1b_2+a_2b_1)i) \\ &=a_1a_2-b_1b_2-(a_1b_2+a_2b_1)i \\ &= a_1(a_2-b_2i)-b_1i(a_2-b_2i) \\ &= (a_1-b_1i)(a_2-b_2i) \\ &=\iota(z_1)\iota(z_2) \end{align*}


and \iota(1)=1-0i=1.

Example 8: Let \varphi: \mathbb{C} \rightarrow M_2(\mathbb{R} be defined by:

    \[\varphi(a+bi) = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}\]



Proof: Let z_1=a_1+b_1i and z_2=a_2+b_2i, then

    \begin{align*} \varphi(z_1+z_2)&=\varphi((a_1+a_2)+(b_1+b_2)i) \\ &= \begin{pmatrix} a_1+a_2 & b_1+b_2 \\ - (b_1+b_2) & a_1+a_2 \end{pmatrix} \\ &= \begin{pmatrix} a_1+a_2 & b_1+b_2 \\ - b_1-b_2 & a_1+a_2 \end{pmatrix} \\ &= \begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{pmatrix} + \begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{pmatrix} \\ &= \varphi(z_1)+\varphi(z_2) \end{align*}


    \begin{align*} \varphi(z_1z_2)&=\varphi((a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i) \\ &= \begin{pmatrix} a_1a_2-b_1b_2 & a_2b_1+a_1b_2 \\ - (a_2b_1+a_1b_2) & a_1a_2-b_1b_2 \end{pmatrix} \\ &= \begin{pmatrix} a_1a_2-b_1b_2 & a_2b_1+a_1b_2 \\ - a_2b_1-a_1b_2) & a_1a_2-b_1b_2 \end{pmatrix} \\ &= \begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{pmatrix} \cdot \begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{pmatrix} \\ &= \varphi(z_1)\cdot\varphi(z_2) \end{align*}


    \[\varphi(1)=\begin{pmatrix} 1 & 0 \\ -0 & 1 \end{pmatrix} = I_2\]



Note that for the proof of the multiplication condition for the ring homomorphism, one may consider computing \varphi(z_1z_2) and \varphi(z_1)\varphi(z_2) and pretend as if they knew exactly how the multiplication decomposes by connecting the ending results by equal signs.