To show a non-empty subset S of R is a subring if \(\displaystyle{a},{b}\in{S}\Rightarrow{a}-{b}\in{S}{\quad\text{and}\quad}{a}{b}\in{S}\)

Clearly S is non-empty as 0 belongs to Ring R so, ax = 0.

let \(\displaystyle{x},{y}\in{S}\Rightarrow\) ax=0, ay=0

is is needed to show that \(\displaystyle{x}-{y}\in{S}\)

consider, \(\displaystyle{a}{x}-{a}{y}={a}{\left({x}-{y}\right)}\in{S}\)

as \(\displaystyle{x}-{y}\in{R}\)

Now show \(\displaystyle{x}{y}\in{S}\)

cnsider, \(\displaystyle{\left({a}{x}\right)}{\left({a}{y}\right)}={0}\cdot{0}={0}\)

\(\displaystyle{a}^{{2}}{\left({x}{y}\right)}={b}{\left({x}{y}\right)}={0}{\left[\begin{array}{c} {a}^{=}{b}\\{x}{y}\in{R}\end{array}\right]}\)

thus, \(\displaystyle{x}{y}\in{S}\)

Therefore, S is a subring of ring R.

Clearly S is non-empty as 0 belongs to Ring R so, ax = 0.

let \(\displaystyle{x},{y}\in{S}\Rightarrow\) ax=0, ay=0

is is needed to show that \(\displaystyle{x}-{y}\in{S}\)

consider, \(\displaystyle{a}{x}-{a}{y}={a}{\left({x}-{y}\right)}\in{S}\)

as \(\displaystyle{x}-{y}\in{R}\)

Now show \(\displaystyle{x}{y}\in{S}\)

cnsider, \(\displaystyle{\left({a}{x}\right)}{\left({a}{y}\right)}={0}\cdot{0}={0}\)

\(\displaystyle{a}^{{2}}{\left({x}{y}\right)}={b}{\left({x}{y}\right)}={0}{\left[\begin{array}{c} {a}^{=}{b}\\{x}{y}\in{R}\end{array}\right]}\)

thus, \(\displaystyle{x}{y}\in{S}\)

Therefore, S is a subring of ring R.