Question

Let S and Tbe subrings of a ring R. In (a) and (b), if the answer is "yes," prove it. If the answer is "no," give a counterexample.

(a) Is $\mathit{S}\mathbf{\cap }\mathit{T}$ a subring of R?

(b) Is $\mathit{S}\mathbf{\cup }\mathit{T}$a subring of R?

Short answer

(a). It is proved that $S\cap T$ is a subring of R.

(b). The set $S\cup T$ is not a subring of R, for example, role="math" localid="1648203115158" $2+3=5\notin 2\mathrm{\mathbb{Z}}\cup 3\mathrm{\mathbb{Z}}$, where $n\mathrm{\mathbb{Z}}=\left\{nk|k\in \mathrm{\mathbb{Z}}\right\}$.

Step-by-step solution

Solution to Part (a)

The given set $S\cap T$, let$a,b\in S\cap T$ then, by definition of intersection$a,b\in S$ and $a,b\in T$.

Now, use the definition of subring, we have,

$a-b\in S,T$and$a,b\in S,T$

Which implies that,$a-b\in S\cap T$ and $a,b\in S\cap T$.

Hence, is a subringof R.

Solution to Part (b)

The given set $S\cup T$ is not a subring of R.

Assume that, role="math" localid="1648208075509" $\mathrm{\mathbb{Z}},2\mathrm{\mathbb{Z}},\mathrm{and}3\mathrm{\mathbb{Z}}$, where $n\mathrm{\mathbb{Z}}=\left\{nk|k\in \mathrm{\mathbb{Z}}\right\}$.

Now, apply the definition of subring, that$\mathrm{\mathbb{Z}}$ with ordinary addition and multiplication is a ring and role="math" localid="1648208616064" $2\mathrm{\mathbb{Z}},3\mathrm{\mathbb{Z}}$, are its subring. Such that, $2,3\in 2\mathrm{\mathbb{Z}}\cup 3\mathrm{\mathbb{Z}}$, then,

$2+3=5\notin 2\mathrm{\mathbb{Z}}\cup 3\mathrm{\mathbb{Z}}$.

Which shows that,$2\mathrm{\mathbb{Z}}\cup 3\mathrm{\mathbb{Z}}$ is not a subring. Hence,$S\cup T$ is not a subring of R.