Question

In the proof of Case (2) of Theorem 4.1, show that R[x] is a subring of S[x] that contains R.

Short answer

It is proved that$\mathrm{R}\left[\mathrm{x}\right]$ is a subring of $\mathrm{S}\left[\mathrm{x}\right]$.

Step-by-step solution

Definition of integral domain 

The commutative ring with identity with no zero divisor is known as integral domain.

Show that R[x] is a subring of S[x] that contains R 

Consider theorem 4.1 as if $\mathrm{R}$is a ring then there exist ring $T$containing an element $x$that is not in $R$and has these properties;

$R$is a subring of$T$ , and$\mathrm{xa}=\mathrm{ax}$ for every$a\in R$

Consider$R$ does not have identity and consider that$\mathrm{R}\left[\mathrm{x}\right]$ is not a subring of$S$ .

Consider that$S\left[x\right]=T$ the polynomial in$S\left[x\right]$ . But$R$ is a subring of $T$, which is a contradiction of the assumption that$R\left[x\right]$ is not a subring of $S$. This implies that$R\left[x\right]$ is a subring of $S\left[x\right]$.

Therefore, it is proved that$R\left[x\right]$ is a subring of $S\left[x\right]$.