Question

Let $\mathit{T}$be the ring in Example 8. Let $\mathit{S}\mathbf{=}\left\{f\in T|f\left(2\right)=0\right\}$. Prove that $\mathit{S}$is a subring of$\mathit{T}$ .

Short answer

It is proved that$S$ is a subring of$T$

Step-by-step solution

Substring

Consider R is a ring and S is a subset of R, then S is a subring of R if it satisfies the following axioms:

1. S is closed under addition.
2. S is closed under multiplication.
3. $0_R\in S$
   
4. If $a\in S$, then the solution of the equation is $a+x=0_R$in S .

Prove that S is a subring of T

It is given that $S=\left\{f\in T|f\left(2\right)=0\right\}$. Prove that$S$is a subring of $T$.

1. Since $f\left(2\right)=0$, this implies that the identity element $f\left(x\right)=0$for all $x\in T$is also an element of $S$.
2. Consider $\left(g,h\right)\in S$, therefore,$$\begin{gathered} g\left(2\right)+h\left(2\right)=0+0 \\ =0 \end{gathered}$$

This implies that. $g+h\in S$

3. Consider $f\in S$, $-f\left(2\right)=0$as $f\left(2\right)=0$, and $f\left(x\right)-f\left(x\right)=0$Hence, $f+x=0$is a solution of $S$.

4. Consider$g,h\in S$, therefore, $$\begin{gathered} g\left(2\right)h\left(2\right)=0\left(0\right) \\ =0 \end{gathered}$$

This implies that$gh\in S$ .

Hence, it is proved that$S$ is a subring of $T$.